Deck 10: Mathematical Modeling and Variation

ملء الشاشة (f)
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سؤال
The simple interest on an investment is directly proportional to the amount of the investment.By investing $2400 in a certain bond issue,you obtained an interest payment of $111.75 after 1 year.Find a mathematical model that gives the interest I for this bond issue after 1 year in terms of the amount invested P.(Round your answer to three decimal places. ) ​

A) I=0.047PI = 0.047 P
B) I=268,200PI = 268,200 P
C) I=21.477PI = 21.477 P
D) I=2400PI = 2400 P
E) I=111.75PI = 111.75 P
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سؤال
An overhead garage door has two springs,one on each side of the door (see figure).A force of 2020 pounds is required to stretch each spring 1 foot.Because of a pulley system,the springs stretch only one-half the distance the door travels.The door moves a total of x=18x = 18 feet,and the springs are at their natural length when the door is open.Find the combined lifting force applied to the door by the springs when the door is closed.​  <strong>An overhead garage door has two springs,one on each side of the door (see figure).A force of  20  pounds is required to stretch each spring 1 foot.Because of a pulley system,the springs stretch only one-half the distance the door travels.The door moves a total of  x = 18  feet,and the springs are at their natural length when the door is open.Find the combined lifting force applied to the door by the springs when the door is closed.​   ​</strong> A)  \text { Combined lifting force } = 2 F = 356 \mathrm { lb }  B)  \text { Combined lifting force } = 2 F = 360 \mathrm { lb }  C)  \text { Combined lifting force } = 2 F = 362 \mathrm { lb }  D)  \text { Combined lifting force } = 2 F = 358 \mathrm { lb }  E)  \text { Combined lifting force } = 2 F = 364 \mathrm { lb }  <div style=padding-top: 35px>

A)  Combined lifting force =2F=356lb\text { Combined lifting force } = 2 F = 356 \mathrm { lb }
B)  Combined lifting force =2F=360lb\text { Combined lifting force } = 2 F = 360 \mathrm { lb }
C)  Combined lifting force =2F=362lb\text { Combined lifting force } = 2 F = 362 \mathrm { lb }
D)  Combined lifting force =2F=358lb\text { Combined lifting force } = 2 F = 358 \mathrm { lb }
E)  Combined lifting force =2F=364lb\text { Combined lifting force } = 2 F = 364 \mathrm { lb }
سؤال
The simple interest on an investment is directly proportional to the amount of the investment.By investing $5800 in a municipal bond,you obtained an interest payment of $221.25 after 1 year.Find a mathematical model that gives the interest I for this municipal bond after 1 year in terms of the amount invested P.(Round your answer to three decimal places. ) ​

A) I=26.215PI = 26.215 P
B) I=221.25PI = 221.25 P
C) I=0.038PI = 0.038 P
D) I=1,283,250PI = 1,283,250 P
E) I=5800PI = 5800 P
سؤال
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
Y varies inversely as x. (y=9 when x=45.)( y = 9 \text { when } x = 45 . )

A) y=9xy = \frac { 9 } { x }
B) y=405xy = \frac { 405 } { x }
C) y=x405y = \frac { x } { 405 }
D) y=45xy = \frac { 45 } { x }
E) y=405xy = 405 x
سؤال
On a yardstick with scales in inches and centimeters,you notice that 11 inches is approximately the same length as 33 centimeters.Use this information to find a mathematical model that relates centimeters y to inches x.Then use the model to find the numbers of centimeters in 60 inches and 70 inches.(Round your answer to one decimal place. ) ​

A)Model: y = 13\frac { 1 } { 3 } x;20 cm,23.3 cm
B)Model: y = 3x;180 cm,23.3 cm
C)Model: y = 3x;20 cm,210 cm
D)Model: y = 3x;180 cm,210 cm
E)Model: y = 13\frac { 1 } { 3 } x;180 cm,210 cm
سؤال
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
P varies directly as x and inversely as the square of y.(P = 32\frac { 3 } { 2 } when x = 25 and y = 10. )

A) P=xy26P = \frac { x y ^ { 2 } } { 6 }
B) P=6xyP = \frac { 6 x } { y }
C) P=6xy2P = \frac { 6 x } { y ^ { 2 } }
D) P=6y2xP = \frac { 6 y ^ { 2 } } { x }
E) P=6xy2P = 6 x y ^ { 2 }
سؤال
Property tax is based on the assessed value of a property.A house that has an assessed value of $200,000 has a property tax of $4,820.Find a mathematical model that gives the amount of property tax y in terms of the assessed value x of the property.Use the model to find the property tax on a house that has an assessed value of $230,000.(Round your answer to four decimal places. ) ​

A) y=0.0241x;$230,000y = 0.0241 x ; \$ 230,000
B) y=0.0241x;$5543y = 0.0241 x ; \$ 5543
C) y=41.4938x;$5543y = 41.4938 x ; \$ 5543
D) y=41.4938x;$9,543,568y = 41.4938 x ; \$ 9,543,568
E) y=0.0241x;$9,543,568y = 0.0241 x ; \$ 9,543,568
سؤال
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
F is jointly proportional to r and the third power of s. (F=24750 when r=18 and s=5.)( F = 24750 \text { when } r = 18 \text { and } s = 5 . )

A) F=11rs3F = \frac { 11 r } { s ^ { 3 } }
B) F=rs311F = \frac { r s ^ { 3 } } { 11 }
C) F=11rs3F = 11 r s ^ { 3 }
D) F=11rs3F = \frac { 11 } { r s ^ { 3 } }
E) F=11s3rF = \frac { 11 s ^ { 3 } } { r }
سؤال
Assume that y is directly proportional to x.Use the given x-value and y-value to find a linear model that relates y and x.​ x=5,y=24x = 5 , y = 24

A) y=245xy = - \frac { 24 } { 5 } x
B) y=24xy = 24 x
C) y=524xy = - \frac { 5 } { 24 } x
D) y=245xy = \frac { 24 } { 5 } x
E) y=524xy = \frac { 5 } { 24 } x
سؤال
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
Y is inversely proportional to x. (y=7 when x=5.)( y = 7 \text { when } x = 5 . )

A) y=5xy = \frac { 5 } { x }
B) y=x35y = \frac { x } { 35 }
C) y=35xy = 35 x
D) y=35xy = \frac { 35 } { x }
E) y=7xy = \frac { 7 } { x }
سؤال
The work W (in joules)done when lifting an object varies jointly with the mass m (in kilograms)of the object and the height h (in meters)that the object is lifted.The work done when a 120-kilogram object is lifted 1.8 meters is 2116.8 joules.How much work is done when lifting a 200-kilogram object 1.5 meters? ​

A)2960 J
B)2920 J
C)2940 J
D)2950 J
E)2930 J
سؤال
Assume that y is directly proportional to x.Use the given x-value and y-value to find a linear model that relates y and x.​ x=5,y=380x = 5 , y = 380

A)y = 76x
B)y = - 76x
C)y = 176\frac { 1 } { 76 } x
D)y = - 176\frac { 1 } { 76 } x
E)y = - 380x
سؤال
The coiled spring of a toy supports the weight of a child.The spring is compressed a distance of 1.6 inches by the weight of a 35-pound child.The toy will not work properly if its spring is compressed more than 6 inches.What is the weight of the heaviest child who should be allowed to use the toy? (Round your answer to two decimal places. ) ​

A)136.25 lb
B)126.25 lb
C)131.25 lb
D)35 lb
E)141.25 lb
سؤال
A force of 270 newtons stretches a spring 0.18 meter.What force is required to stretch the spring 0.19 meter? ​

A)295 N
B)290 N
C)285 N
D)280 N
E)270 N
سؤال
Assume that y is directly proportional to x.Use the given x-value and y-value to find a linear model that relates y and x.​ x=38,y=2400x = 38 , y = 2400 ​ ​

A)y = 120019\frac { 1200 } { 19 } x
B)y = - 2400x
C)y = 191200\frac { 19 } { 1200 } x
D)y = - 191200\frac { 19 } { 1200 } x​
E)y = - 120019\frac { 1200 } { 19 } x
سؤال
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
Z varies jointly as x and y. (z=128 when x=4 and y=8.)( z = 128 \text { when } x = 4 \text { and } y = 8 . )

A) z=4yxz = \frac { 4 y } { x }
B) z=4xyz = \frac { 4 } { x y }
C) z=xy4z = \frac { x y } { 4 }
D) z=4xyz = \frac { 4 x } { y }
E) z=4xyz = 4 x y
سؤال
When buying gasoline,you notice that 16 gallons of gasoline is approximately the same amount of gasoline as 51 liters.Use this information to find a linear model that relates liters y to gallons x.Then use the model to find the numbers of liters in 25 gallons and 45 gallons. (Round your answer to one decimal place. )

A)Model: y = 1651\frac { 16 } { 51 } x;7.8 L,14.1 L
B)Model: y = 5116\frac { 51 } { 16 } x;79.7 L,14.1 L
C)Model: y = 5116\frac { 51 } { 16 } x;7.8 L,143.4 L
D)Model: y = 5116\frac { 51 } { 16 } x;79.7 L,143.4 L
E)Model: y = 1651\frac { 16 } { 51 } x;79.7 L,143.4 L
سؤال
State sales tax is based on retail price.An item that sells for $180.99 has a sales tax of $17.4.Find a mathematical model that gives the amount of sales tax y in terms of the retail price x.Use the model to find the sales tax on a $589.99 purchase.(Round your answer to four decimal places. ) ​

A) y=10.4017x;$56.72y = 10.4017 x ; \$ 56.72
B) y=0.0961x;$56.72y = 0.0961 x ; \$ 56.72
C) y=0.0961x;$589.99y = 0.0961 x ; \$ 589.99
D) y=0.0961x;$6,137y = 0.0961 x ; \$ 6,137
E) y=10.4017x;$6,137y = 10.4017 x ; \$ 6,137
سؤال
A force of f=225f = 225 newtons stretches a spring s=0.15s = 0.15 meter (see figure).  <strong>A force of  f = 225  newtons stretches a spring  s = 0.15  meter (see figure).   How far will a force of 60 newtons stretch the spring? What force is required to stretch the spring 0.1 meter? ​</strong> A)  0.15 \mathrm {~m} ; 150 \mathrm {~N}  B)  0.15 \mathrm {~m} ; 225 \mathrm {~N}  C)  0.04 \mathrm {~m} ; 150 \mathrm {~N}  D)  0.09 \mathrm {~m} ; 285 \mathrm {~N}  E)  0.04 \mathrm {~m} ; 225 \mathrm {~N}  <div style=padding-top: 35px>  How far will a force of 60 newtons stretch the spring? What force is required to stretch the spring 0.1 meter? ​

A) 0.15 m;150 N0.15 \mathrm {~m} ; 150 \mathrm {~N}
B) 0.15 m;225 N0.15 \mathrm {~m} ; 225 \mathrm {~N}
C) 0.04 m;150 N0.04 \mathrm {~m} ; 150 \mathrm {~N}
D) 0.09 m;285 N0.09 \mathrm {~m} ; 285 \mathrm {~N}
E) 0.04 m;225 N0.04 \mathrm {~m} ; 225 \mathrm {~N}
سؤال
Assume that y is directly proportional to x.Use the given x-value and y-value to find a linear model that relates y and x.​ x=2,y=58x = 2 , y = 58

A) y=29xy = 29 x
B) y=29xy = - 29 x
C) y=58xy = 58 x
D) y=58xy = - 58 x
E) y=29y = 29
سؤال
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
V varies jointly as p and q and inversely as the square of s.(ν = 1.4 when p = 4.4,q = 7.3 and s = 1.8. )

A) v=0.141pqs2v = \frac { 0.141 p } { q s ^ { 2 } }
B) v=pq0.141s2v = - \frac { p q } { 0.141 s ^ { 2 } }
C) v=0.141pqs2v = \frac { 0.141 p q } { s ^ { 2 } }
D) v=pq0.141s2v = \frac { p q } { 0.141 s ^ { 2 } }
E) v=0.141pqs2v = - \frac { 0.141 p q } { s ^ { 2 } }
سؤال
Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } Plot the points on a rectangular coordinate system. x246810y=kx2k=2\begin{array}{l}\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\\hline\end{array}\\k = 2\end{array}

A)​ x246810y=kx21218118132150\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
B)​ x246810y=kx21212121212\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
C)​ x246810y=kx21501321181812\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\& \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\\hline\end{array}

 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​   <div style=padding-top: 35px>

D)​ x246810y=kx212181181812\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = k x ^ { 2 } & & & & & \\& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
E)​ x246810y=kx2246810\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
سؤال
Determine whether the variation model below is of the form y=kxy = k x or y=kxy = \frac { k } { x } . x
13
26
39
52
65
Y
3
6
9
12
15

A) y=kxy = k x
B) y=kxy = \frac { k } { x }
سؤال
Use the fact that the resistance of a wire carrying an electrical current is directly proportional to its length and inversely proportional to its cross-sectional area. ​
A 10-foot piece of copper wire produces a resistance of 0.2 ohm.Use the constant of proportionality k = 0.000833 to find the diameter of the wire.

(Round the answer up to three decimal places. )

A)0.23 ft
B)0.58 ft
C)0.38 ft
D)0.48 ft
E)0.73 ft
سؤال
Use the given value of k to complete the table for the direct variation model​ y=kx2y = k x ^ { 2 } . ​
Plot the points on a rectangular coordinate system.

x810121416y=kx2k=12\begin{array}{l}\begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & & & & & \\& & & & & \\\hline\end{array}\\k = \frac { 1 } { 2 }\end{array}

A)​ x810121416y=kx232507298128\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
B)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
C)​ x810121416y=kx23232323232\begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​   <div style=padding-top: 35px>

D)​ x810121416y=kx212898725032\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
E)​ x810121416y=kx23250725032\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
سؤال
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
Z varies directly as the square of x and inversely as y.(z = 36 when x = 9 and y = 3. )

A) z=3x24yz = \frac { 3 x ^ { 2 } } { 4 y }
B) z=4x3yz = \frac { 4 x } { 3 y }
C) z=4x23yz = \frac { 4 x ^ { 2 } } { 3 y }
D) z=3x24yz = - \frac { 3 x ^ { 2 } } { 4 y }
E) z=4x23yz = - \frac { 4 x ^ { 2 } } { 3 y }
سؤال
Determine whether the variation model is of the form y=kxy = k x or y=kxy = \frac { k } { x } and find k.Then write a model that relates y and x. x48121620y112131415\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 8 & 12 & 16 & 20 \\\hline y & 1 & \frac { 1 } { 2 } & \frac { 1 } { 3 } & \frac { 1 } { 4 } & \frac { 1 } { 5 } \\\hline\end{array}

A) y=4xy = 4 x
B) y=1xy = \frac { 1 } { x }
C) y=4xy = \frac { 4 } { x }
D) y=xy = x
E) y=x4y = \frac { x } { 4 }
سؤال
The frequency of vibrations of a piano string varies directly as the square root of the tension on the string and inversely as the length of the string.The middle A string has a frequency of 430 vibrations per second.Find the frequency of a string that has 1.25 times as much tension and is 1.4 times as long. ​

A)373.4 vibrations / sec
B)343.4 vibrations / sec
C)353.4 vibrations / sec
D)383.4 vibrations / sec
E)363.4 vibrations / sec
سؤال
Use the fact that the diameter of the largest particle that can be moved by a stream varies approximately directly as the square of the velocity of the stream. ​
A stream with a velocity of 15\frac { 1 } { 5 } mile per hour can move coarse sand particles about 0.07 inch in diameter.Approximate the velocity required to carry particles 0.2 inch in diameter.(Round your answer to two decimal places. )

A)About 0.84 mi/h
B)About 0.19 mi/h
C)About -0.16 mi/h
D)About 0.49 mi/h
E)About 0.34 mi/h
سؤال
Use the fact that the resistance of a wire carrying an electrical current is directly proportional to its length and inversely proportional to its cross-sectional area. ​
If #28 copper wire (which has a diameter of 0.0126 inch)has a resistance of 68.17 ohms per thousand feet,what length of #28 copper wire will produce a resistance of 30.5 ohms?

A)About 447 ft
B)About 442 ft
C)About 432 ft
D)About 452 ft
E)About 462 ft
سؤال
Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } Plot the points on a rectangular coordinate system. x810121416y=kx2k=5\begin{array}{l}\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\\hline\end{array}\\k = 5\end{array}

A)​ x810121416y=kx2564120514451965256\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
B)​ x810121416y=kx2564564564564564\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
C)​ x810121416y=kx2525651965144120564\begin{array}{|c|c|c|c|c|c|}\hline x & 8 & 10 & 12 & 14 & 16 \\\hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
D)​ x810121416y=kx25641205144120564\begin{array}{|c|c|c|c|c|c|}\hline x & 8 & 10 & 12 & 14 & 16 \\\hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
E)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\& 8 & 10 & 12 & 14 & 16 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
سؤال
Use the given value of k to complete the table for the direct variation model​ y=kx2y = k x ^ { 2 } . ​
Plot the points on a rectangular coordinate system.
x810121416y=kx2k=14\begin{array}{l}\begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & & & & & \\& & & & & \\\hline\end{array}\\k = \frac { 1 } { 4 }\end{array}

A) ​ x810121416y=kx26449362516\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>
B)​ x810121416y=kx21616161616\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>
C)​ x810121416y=kx21625362516\begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>

D)​ x810121416y=kx21625364964\begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>
E)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>
سؤال
Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } Plot the points on a rectangular coordinate system.
x810121416y=kx2\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\\hline\end{array}
k=10k = 10
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​   <div style=padding-top: 35px>

A) x4681012y=kx258518532110572\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\& \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\\hline\end{array}
B)​ x4681012y=kx25858585858\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
C)​ x4681012y=kx257211053251858\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​   <div style=padding-top: 35px>

D)​ x4681012y=kx25851853251858\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
E)​ x4681012y=kx24681012\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
سؤال
Determine whether the variation model is of the form y=kxy = k x or y=kxy = \frac { k } { x } and find k.Then write a model that relates y and x.
x510152025y2613263132265\begin{array} { | c | c | c | c | c | c | } \hline x & 5 & 10 & 15 & 20 & 25 \\\hline y & 26 & 13 & \frac { 26 } { 3 } & \frac { 13 } { 2 } & \frac { 26 } { 5 } \\\hline\end{array}

A) y=130xy = \frac { 130 } { x }
B) y=x130y = \frac { x } { 130 }
C) y=1xy = \frac { 1 } { x }
D) y=130xy = 130 x
E) y=xy = x
سؤال
An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  Depth, d Temperature, C10003.820002.130001.840001.550000.5\begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\1000 & 3.8 ^ { \circ } \\2000 & 2.1 ^ { \circ } \\3000 & 1.8 ^ { \circ } \\4000 & 1.5 ^ { \circ } \\5000 & 0.5 ^ { \circ }\end{array}
Sketch a scatter plot of the data.

A)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
B)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
C)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
D)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
E)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)   <div style=padding-top: 35px>
سؤال
Determine whether the variation model is of the form y=kxy = k x or y=kxy = \frac { k } { x } and find k.Then write a model that relates y and x. x918273645y246810\begin{array} { | c | c | c | c | c | c | } \hline x & 9 & 18 & 27 & 36 & 45 \\\hline y & 2 & 4 & 6 & 8 & 10 \\\hline\end{array}

A) y=29xy = \frac { 2 } { 9 } x
B) y=9xy = \frac { 9 } { x }
C) y=29xy = \frac { 2 } { 9 x }
D) y=92xy = \frac { 9 } { 2 x }
E) y=92xy = \frac { 9 } { 2 } x
سؤال
Use the given value of k to complete the table for the direct variation model​ y=kx2y = k x ^ { 2 } . ​
Plot the points on a rectangular coordinate system.
x810121416y=kx2k=1\begin{array}{l}\begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & & & & & \\& & & & & \\\hline\end{array}\\k = 1\end{array}

A)​ x810121416y=kx225619614410064\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>
B)​ x810121416y=kx264100144196256\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\\hline\end{array}

 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>
C)​ x810121416y=kx26464646464\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\\hline\end{array}

 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>

D)​ x810121416y=kx26410014410064\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\\hline\end{array}

 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>
E)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   <div style=padding-top: 35px>
سؤال
Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } . Plot the points on a rectangular coordinate system.
x810121416y=kx2\begin{array}{|l|l|l|l|l|l|}\hline x & 8 & 10 & 12 & 14 & 16 \\\hline y=\frac{k}{x^{2}}\\\hline\end{array}
k=20k = 20

A)​​ x810121416y=kx251615536549564\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
B)​ x810121416y=kx2516516516516516\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>

C)​ x810121416y=kx256454953615516\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\\hline\end{array}
​​  <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>

D)​ x810121416y=kx25161553615516\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
E)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
سؤال
Determine whether the variation model is of the form y=kxy = k x or y=kxy = \frac { k } { x } and find k.Then write a model that relates y and x. x510152025y1.534.567.5\begin{array} { | c | c | c | c | c | c | } \hline x & 5 & 10 & 15 & 20 & 25 \\\hline y & - 1.5 & - 3 & - 4.5 & - 6 & - 7.5 \\\hline\end{array}

A) y=310xy = - \frac { 3 } { 10 x }
B) y=103xy = \frac { 10 } { 3 } x
C) y=310xy = - \frac { 3 } { 10 } x
D) y=310xy = \frac { 3 } { 10 } x
E) y=103xy = - \frac { 10 } { 3 } x
سؤال
Use the given value of k to complete the table for the direct variation model​ y=kx2y = k x ^ { 2 } . ​
Plot the points on a rectangular coordinate system.
x810121416y=kx2k=2\begin{array}{l}\begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & & & & & \\\hline\end{array}\\k = 2\end{array}

A)​ x810121416y=kx2128200288392512\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
B)​ x810121416y=kx2810121416\begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
C)​ x810121416y=kx2128128128128128\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\\hline\end{array}

 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​   <div style=padding-top: 35px>

D)​ x810121416y=kx2512392288200128\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
E)​ x810121416y=kx2128200288200128\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​   <div style=padding-top: 35px>
سؤال
After determining whether the variation model below is of the form y=kxy = k x or y=kxy = \frac { k } { x } ,find the value of k.
x20406080100y13016019011201150\begin{array} { | c | c | c | c | c | c | } \hline x & 20 & 40 & 60 & 80 & 100 \\\hline y & \frac { 1 } { 30 } & \frac { 1 } { 60 } & \frac { 1 } { 90 } & \frac { 1 } { 120 } & \frac { 1 } { 150 } \\\hline\end{array}

A) k=120k = \frac { 1 } { 20 }
B) k=32k = \frac { 3 } { 2 }
C) k=54k = \frac { 5 } { 4 }
D) k=110k = \frac { 1 } { 10 }
E) k=23k = \frac { 2 } { 3 }
سؤال
The simple interest on an investment is directly proportional to the amount of the investment.By investing $6000 in a certain certificate of deposit,you obtained an interest payment of $276.00 after 1 year.Determine a mathematical model that gives the interest,I ,for this CD after 1 year in terms of the amount invested,P.

A) I=(0.050)PI = ( 0.050 ) P
B) I=(0.041)PI = ( 0.041 ) P
C) I=(0.049)PI = ( 0.049 ) P
D) I=(0.046)PI = ( 0.046 ) P
E) I=(0.044)PI = ( 0.044 ) P
سؤال
Find a mathematical model for the verbal statement: "m varies directly as the square of w and inversely as s."

A) m=k(ws)2m = k \left( \frac { w } { s } \right) ^ { 2 }
B) m=kw2sm = \frac { k w ^ { 2 } } { s }
C) m=kw2sm = k w ^ { 2 } s
D) m=kws2m = k w s ^ { 2 }
E) m=k(sw)2m = k \left( \frac { s } { w } \right) ^ { 2 }
سؤال
Find a mathematical model for the verbal statement: ​
"Q is jointly proportional to the cube of h and the square root of m."

A) Q=kh3mQ = k h ^ { 3 } \sqrt { m }
B) Q=kh3mQ = k \sqrt { h ^ { 3 } m }
C) Q=kh2m3Q = k h ^ { 2 } \sqrt [ 3 ] { m }
D) Q=khm23Q = k \sqrt [ 3 ] { h m ^ { 2 } }
E) Q=khm3Q = k \sqrt [ 3 ] { h m }
سؤال
Hooke's law states that the magnitude of force,F,required to stretch a spring x units beyond its natural length is directly proportional to x.If a force of 3 pounds stretches a spring from its natural length of 10 inches to a length of 10.7 inches,what force will stretch the spring to a length of 11.5 inches? Round your answer to the nearest hundredth.

A) F=5.52F = 5.52
B) F=6.43F = 6.43
C) F=5.70F = 5.70
D) F=7.29F = 7.29
E) F=6.14F = 6.14
سؤال
The sales tax on an item with a retail price of $972 is $68.04.Create a variational model that gives the retail price,y,in terms of the sales tax,x,and use it to determine the retail price of an item that has a sales tax of $82.62.

A)$1182.28
B)$1151.92
C)$1180.29
D)$1192.52
E)$1124.60
سؤال
Assume that y is directly proportional to x.If x = 28 and y = 21,determine a linear model that relates y and x.

A) y=35xy = \frac { 3 } { 5 } x
B) y=43xy = \frac { 4 } { 3 } x
C) y=34xy = \frac { 3 } { 4 } x
D) y=23xy = \frac { 2 } { 3 } x
E) y=32xy = \frac { 3 } { 2 } x
سؤال
After determining whether the variation model below is of the form y=kxy = k x or y=kxy = \frac { k } { x } ,find the value of k. x154161168175182y6669727578\begin{array} { | c | c | c | c | c | c | } \hline x & 154 & 161 & 168 & 175 & 182 \\\hline y & 66 & 69 & 72 & 75 & 78 \\\hline\end{array}

A) k=7k = 7
B) k=73k = \frac { 7 } { 3 }
C) k=17k = \frac { 1 } { 7 }
D) k=37k = \frac { 3 } { 7 }
E) k=766k = \frac { 7 } { 66 }
سؤال
Determine whether the variation model below is of the form y=kxy = k x or y=kxy = \frac { k } { x } . x
12
24
36
48
60
Y 118\frac { 1 } { 18 }136\frac { 1 } { 36 }154\frac { 1 } { 54 }172\frac { 1 } { 72 }190\frac { 1 } { 90 }

A) y=kxy = k x
B) y=kxy = \frac { k } { x }
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Deck 10: Mathematical Modeling and Variation
1
The simple interest on an investment is directly proportional to the amount of the investment.By investing $2400 in a certain bond issue,you obtained an interest payment of $111.75 after 1 year.Find a mathematical model that gives the interest I for this bond issue after 1 year in terms of the amount invested P.(Round your answer to three decimal places. ) ​

A) I=0.047PI = 0.047 P
B) I=268,200PI = 268,200 P
C) I=21.477PI = 21.477 P
D) I=2400PI = 2400 P
E) I=111.75PI = 111.75 P
I=0.047PI = 0.047 P
2
An overhead garage door has two springs,one on each side of the door (see figure).A force of 2020 pounds is required to stretch each spring 1 foot.Because of a pulley system,the springs stretch only one-half the distance the door travels.The door moves a total of x=18x = 18 feet,and the springs are at their natural length when the door is open.Find the combined lifting force applied to the door by the springs when the door is closed.​  <strong>An overhead garage door has two springs,one on each side of the door (see figure).A force of  20  pounds is required to stretch each spring 1 foot.Because of a pulley system,the springs stretch only one-half the distance the door travels.The door moves a total of  x = 18  feet,and the springs are at their natural length when the door is open.Find the combined lifting force applied to the door by the springs when the door is closed.​   ​</strong> A)  \text { Combined lifting force } = 2 F = 356 \mathrm { lb }  B)  \text { Combined lifting force } = 2 F = 360 \mathrm { lb }  C)  \text { Combined lifting force } = 2 F = 362 \mathrm { lb }  D)  \text { Combined lifting force } = 2 F = 358 \mathrm { lb }  E)  \text { Combined lifting force } = 2 F = 364 \mathrm { lb }

A)  Combined lifting force =2F=356lb\text { Combined lifting force } = 2 F = 356 \mathrm { lb }
B)  Combined lifting force =2F=360lb\text { Combined lifting force } = 2 F = 360 \mathrm { lb }
C)  Combined lifting force =2F=362lb\text { Combined lifting force } = 2 F = 362 \mathrm { lb }
D)  Combined lifting force =2F=358lb\text { Combined lifting force } = 2 F = 358 \mathrm { lb }
E)  Combined lifting force =2F=364lb\text { Combined lifting force } = 2 F = 364 \mathrm { lb }
 Combined lifting force =2F=360lb\text { Combined lifting force } = 2 F = 360 \mathrm { lb }
3
The simple interest on an investment is directly proportional to the amount of the investment.By investing $5800 in a municipal bond,you obtained an interest payment of $221.25 after 1 year.Find a mathematical model that gives the interest I for this municipal bond after 1 year in terms of the amount invested P.(Round your answer to three decimal places. ) ​

A) I=26.215PI = 26.215 P
B) I=221.25PI = 221.25 P
C) I=0.038PI = 0.038 P
D) I=1,283,250PI = 1,283,250 P
E) I=5800PI = 5800 P
I=0.038PI = 0.038 P
4
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
Y varies inversely as x. (y=9 when x=45.)( y = 9 \text { when } x = 45 . )

A) y=9xy = \frac { 9 } { x }
B) y=405xy = \frac { 405 } { x }
C) y=x405y = \frac { x } { 405 }
D) y=45xy = \frac { 45 } { x }
E) y=405xy = 405 x
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5
On a yardstick with scales in inches and centimeters,you notice that 11 inches is approximately the same length as 33 centimeters.Use this information to find a mathematical model that relates centimeters y to inches x.Then use the model to find the numbers of centimeters in 60 inches and 70 inches.(Round your answer to one decimal place. ) ​

A)Model: y = 13\frac { 1 } { 3 } x;20 cm,23.3 cm
B)Model: y = 3x;180 cm,23.3 cm
C)Model: y = 3x;20 cm,210 cm
D)Model: y = 3x;180 cm,210 cm
E)Model: y = 13\frac { 1 } { 3 } x;180 cm,210 cm
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6
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
P varies directly as x and inversely as the square of y.(P = 32\frac { 3 } { 2 } when x = 25 and y = 10. )

A) P=xy26P = \frac { x y ^ { 2 } } { 6 }
B) P=6xyP = \frac { 6 x } { y }
C) P=6xy2P = \frac { 6 x } { y ^ { 2 } }
D) P=6y2xP = \frac { 6 y ^ { 2 } } { x }
E) P=6xy2P = 6 x y ^ { 2 }
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7
Property tax is based on the assessed value of a property.A house that has an assessed value of $200,000 has a property tax of $4,820.Find a mathematical model that gives the amount of property tax y in terms of the assessed value x of the property.Use the model to find the property tax on a house that has an assessed value of $230,000.(Round your answer to four decimal places. ) ​

A) y=0.0241x;$230,000y = 0.0241 x ; \$ 230,000
B) y=0.0241x;$5543y = 0.0241 x ; \$ 5543
C) y=41.4938x;$5543y = 41.4938 x ; \$ 5543
D) y=41.4938x;$9,543,568y = 41.4938 x ; \$ 9,543,568
E) y=0.0241x;$9,543,568y = 0.0241 x ; \$ 9,543,568
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8
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
F is jointly proportional to r and the third power of s. (F=24750 when r=18 and s=5.)( F = 24750 \text { when } r = 18 \text { and } s = 5 . )

A) F=11rs3F = \frac { 11 r } { s ^ { 3 } }
B) F=rs311F = \frac { r s ^ { 3 } } { 11 }
C) F=11rs3F = 11 r s ^ { 3 }
D) F=11rs3F = \frac { 11 } { r s ^ { 3 } }
E) F=11s3rF = \frac { 11 s ^ { 3 } } { r }
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9
Assume that y is directly proportional to x.Use the given x-value and y-value to find a linear model that relates y and x.​ x=5,y=24x = 5 , y = 24

A) y=245xy = - \frac { 24 } { 5 } x
B) y=24xy = 24 x
C) y=524xy = - \frac { 5 } { 24 } x
D) y=245xy = \frac { 24 } { 5 } x
E) y=524xy = \frac { 5 } { 24 } x
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10
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
Y is inversely proportional to x. (y=7 when x=5.)( y = 7 \text { when } x = 5 . )

A) y=5xy = \frac { 5 } { x }
B) y=x35y = \frac { x } { 35 }
C) y=35xy = 35 x
D) y=35xy = \frac { 35 } { x }
E) y=7xy = \frac { 7 } { x }
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11
The work W (in joules)done when lifting an object varies jointly with the mass m (in kilograms)of the object and the height h (in meters)that the object is lifted.The work done when a 120-kilogram object is lifted 1.8 meters is 2116.8 joules.How much work is done when lifting a 200-kilogram object 1.5 meters? ​

A)2960 J
B)2920 J
C)2940 J
D)2950 J
E)2930 J
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12
Assume that y is directly proportional to x.Use the given x-value and y-value to find a linear model that relates y and x.​ x=5,y=380x = 5 , y = 380

A)y = 76x
B)y = - 76x
C)y = 176\frac { 1 } { 76 } x
D)y = - 176\frac { 1 } { 76 } x
E)y = - 380x
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13
The coiled spring of a toy supports the weight of a child.The spring is compressed a distance of 1.6 inches by the weight of a 35-pound child.The toy will not work properly if its spring is compressed more than 6 inches.What is the weight of the heaviest child who should be allowed to use the toy? (Round your answer to two decimal places. ) ​

A)136.25 lb
B)126.25 lb
C)131.25 lb
D)35 lb
E)141.25 lb
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14
A force of 270 newtons stretches a spring 0.18 meter.What force is required to stretch the spring 0.19 meter? ​

A)295 N
B)290 N
C)285 N
D)280 N
E)270 N
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15
Assume that y is directly proportional to x.Use the given x-value and y-value to find a linear model that relates y and x.​ x=38,y=2400x = 38 , y = 2400 ​ ​

A)y = 120019\frac { 1200 } { 19 } x
B)y = - 2400x
C)y = 191200\frac { 19 } { 1200 } x
D)y = - 191200\frac { 19 } { 1200 } x​
E)y = - 120019\frac { 1200 } { 19 } x
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16
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
Z varies jointly as x and y. (z=128 when x=4 and y=8.)( z = 128 \text { when } x = 4 \text { and } y = 8 . )

A) z=4yxz = \frac { 4 y } { x }
B) z=4xyz = \frac { 4 } { x y }
C) z=xy4z = \frac { x y } { 4 }
D) z=4xyz = \frac { 4 x } { y }
E) z=4xyz = 4 x y
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17
When buying gasoline,you notice that 16 gallons of gasoline is approximately the same amount of gasoline as 51 liters.Use this information to find a linear model that relates liters y to gallons x.Then use the model to find the numbers of liters in 25 gallons and 45 gallons. (Round your answer to one decimal place. )

A)Model: y = 1651\frac { 16 } { 51 } x;7.8 L,14.1 L
B)Model: y = 5116\frac { 51 } { 16 } x;79.7 L,14.1 L
C)Model: y = 5116\frac { 51 } { 16 } x;7.8 L,143.4 L
D)Model: y = 5116\frac { 51 } { 16 } x;79.7 L,143.4 L
E)Model: y = 1651\frac { 16 } { 51 } x;79.7 L,143.4 L
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State sales tax is based on retail price.An item that sells for $180.99 has a sales tax of $17.4.Find a mathematical model that gives the amount of sales tax y in terms of the retail price x.Use the model to find the sales tax on a $589.99 purchase.(Round your answer to four decimal places. ) ​

A) y=10.4017x;$56.72y = 10.4017 x ; \$ 56.72
B) y=0.0961x;$56.72y = 0.0961 x ; \$ 56.72
C) y=0.0961x;$589.99y = 0.0961 x ; \$ 589.99
D) y=0.0961x;$6,137y = 0.0961 x ; \$ 6,137
E) y=10.4017x;$6,137y = 10.4017 x ; \$ 6,137
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19
A force of f=225f = 225 newtons stretches a spring s=0.15s = 0.15 meter (see figure).  <strong>A force of  f = 225  newtons stretches a spring  s = 0.15  meter (see figure).   How far will a force of 60 newtons stretch the spring? What force is required to stretch the spring 0.1 meter? ​</strong> A)  0.15 \mathrm {~m} ; 150 \mathrm {~N}  B)  0.15 \mathrm {~m} ; 225 \mathrm {~N}  C)  0.04 \mathrm {~m} ; 150 \mathrm {~N}  D)  0.09 \mathrm {~m} ; 285 \mathrm {~N}  E)  0.04 \mathrm {~m} ; 225 \mathrm {~N}   How far will a force of 60 newtons stretch the spring? What force is required to stretch the spring 0.1 meter? ​

A) 0.15 m;150 N0.15 \mathrm {~m} ; 150 \mathrm {~N}
B) 0.15 m;225 N0.15 \mathrm {~m} ; 225 \mathrm {~N}
C) 0.04 m;150 N0.04 \mathrm {~m} ; 150 \mathrm {~N}
D) 0.09 m;285 N0.09 \mathrm {~m} ; 285 \mathrm {~N}
E) 0.04 m;225 N0.04 \mathrm {~m} ; 225 \mathrm {~N}
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Assume that y is directly proportional to x.Use the given x-value and y-value to find a linear model that relates y and x.​ x=2,y=58x = 2 , y = 58

A) y=29xy = 29 x
B) y=29xy = - 29 x
C) y=58xy = 58 x
D) y=58xy = - 58 x
E) y=29y = 29
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21
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
V varies jointly as p and q and inversely as the square of s.(ν = 1.4 when p = 4.4,q = 7.3 and s = 1.8. )

A) v=0.141pqs2v = \frac { 0.141 p } { q s ^ { 2 } }
B) v=pq0.141s2v = - \frac { p q } { 0.141 s ^ { 2 } }
C) v=0.141pqs2v = \frac { 0.141 p q } { s ^ { 2 } }
D) v=pq0.141s2v = \frac { p q } { 0.141 s ^ { 2 } }
E) v=0.141pqs2v = - \frac { 0.141 p q } { s ^ { 2 } }
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22
Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } Plot the points on a rectangular coordinate system. x246810y=kx2k=2\begin{array}{l}\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\\hline\end{array}\\k = 2\end{array}

A)​ x246810y=kx21218118132150\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​
B)​ x246810y=kx21212121212\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​
C)​ x246810y=kx21501321181812\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\& \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\\hline\end{array}

 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​

D)​ x246810y=kx212181181812\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = k x ^ { 2 } & & & & & \\& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​
E)​ x246810y=kx2246810\begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\\hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline  y = \frac { k } { x ^ { 2 } }& \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 32 } & \frac { 1 } { 50 } \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 1 } {2 }&\frac { 1 } {2 } &\frac { 1 } {2 } & \frac { 1 } {2 }&\frac { 1 } {2 } \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 1 } { 50 } & \frac { 1 } { 32 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}   ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = k x ^ { 2 } & & & & & \\ & \frac { 1 } { 2 } & \frac { 1 } { 8 } & \frac { 1 } { 18 } & \frac { 1 } { 8 } & \frac { 1 } { 2 } \\\\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 2 & 4 & 6 & 8 & 10 \\ \hline y = \frac { k } { x ^ { 2 } } &2 & 4& 6& 8&10 \\ \hline \end{array}  ​
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23
Determine whether the variation model below is of the form y=kxy = k x or y=kxy = \frac { k } { x } . x
13
26
39
52
65
Y
3
6
9
12
15

A) y=kxy = k x
B) y=kxy = \frac { k } { x }
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24
Use the fact that the resistance of a wire carrying an electrical current is directly proportional to its length and inversely proportional to its cross-sectional area. ​
A 10-foot piece of copper wire produces a resistance of 0.2 ohm.Use the constant of proportionality k = 0.000833 to find the diameter of the wire.

(Round the answer up to three decimal places. )

A)0.23 ft
B)0.58 ft
C)0.38 ft
D)0.48 ft
E)0.73 ft
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25
Use the given value of k to complete the table for the direct variation model​ y=kx2y = k x ^ { 2 } . ​
Plot the points on a rectangular coordinate system.

x810121416y=kx2k=12\begin{array}{l}\begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & & & & & \\& & & & & \\\hline\end{array}\\k = \frac { 1 } { 2 }\end{array}

A)​ x810121416y=kx232507298128\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​
B)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​
C)​ x810121416y=kx23232323232\begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​

D)​ x810121416y=kx212898725032\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​
E)​ x810121416y=kx23250725032\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.   \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 2 } \end{array}  </strong> A)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 98 & 128 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 32 & 32 & 32 & 32 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 98 & 72 & 50 & 32 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 32 & 50 & 72 & 50 & 32 \\ \hline \end{array}  ​
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26
Find a mathematical model representing the statement.(Determine the constant of proportionality. ) ​
Z varies directly as the square of x and inversely as y.(z = 36 when x = 9 and y = 3. )

A) z=3x24yz = \frac { 3 x ^ { 2 } } { 4 y }
B) z=4x3yz = \frac { 4 x } { 3 y }
C) z=4x23yz = \frac { 4 x ^ { 2 } } { 3 y }
D) z=3x24yz = - \frac { 3 x ^ { 2 } } { 4 y }
E) z=4x23yz = - \frac { 4 x ^ { 2 } } { 3 y }
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27
Determine whether the variation model is of the form y=kxy = k x or y=kxy = \frac { k } { x } and find k.Then write a model that relates y and x. x48121620y112131415\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 8 & 12 & 16 & 20 \\\hline y & 1 & \frac { 1 } { 2 } & \frac { 1 } { 3 } & \frac { 1 } { 4 } & \frac { 1 } { 5 } \\\hline\end{array}

A) y=4xy = 4 x
B) y=1xy = \frac { 1 } { x }
C) y=4xy = \frac { 4 } { x }
D) y=xy = x
E) y=x4y = \frac { x } { 4 }
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28
The frequency of vibrations of a piano string varies directly as the square root of the tension on the string and inversely as the length of the string.The middle A string has a frequency of 430 vibrations per second.Find the frequency of a string that has 1.25 times as much tension and is 1.4 times as long. ​

A)373.4 vibrations / sec
B)343.4 vibrations / sec
C)353.4 vibrations / sec
D)383.4 vibrations / sec
E)363.4 vibrations / sec
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29
Use the fact that the diameter of the largest particle that can be moved by a stream varies approximately directly as the square of the velocity of the stream. ​
A stream with a velocity of 15\frac { 1 } { 5 } mile per hour can move coarse sand particles about 0.07 inch in diameter.Approximate the velocity required to carry particles 0.2 inch in diameter.(Round your answer to two decimal places. )

A)About 0.84 mi/h
B)About 0.19 mi/h
C)About -0.16 mi/h
D)About 0.49 mi/h
E)About 0.34 mi/h
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30
Use the fact that the resistance of a wire carrying an electrical current is directly proportional to its length and inversely proportional to its cross-sectional area. ​
If #28 copper wire (which has a diameter of 0.0126 inch)has a resistance of 68.17 ohms per thousand feet,what length of #28 copper wire will produce a resistance of 30.5 ohms?

A)About 447 ft
B)About 442 ft
C)About 432 ft
D)About 452 ft
E)About 462 ft
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31
Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } Plot the points on a rectangular coordinate system. x810121416y=kx2k=5\begin{array}{l}\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\\hline\end{array}\\k = 5\end{array}

A)​ x810121416y=kx2564120514451965256\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
B)​ x810121416y=kx2564564564564564\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
C)​ x810121416y=kx2525651965144120564\begin{array}{|c|c|c|c|c|c|}\hline x & 8 & 10 & 12 & 14 & 16 \\\hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
D)​ x810121416y=kx25641205144120564\begin{array}{|c|c|c|c|c|c|}\hline x & 8 & 10 & 12 & 14 & 16 \\\hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
E)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\& 8 & 10 & 12 & 14 & 16 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}\\ k = 5 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} & \frac{5}{64} & \frac{1}{20} & \frac{5}{144} & \frac{5}{196} & \frac{5}{256}\\\\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline \\y=\frac{k}{x^{2}} &\frac{5}{64} &\frac{5}{64} & \frac{5}{64} &\frac{5}{64} &\frac{5}{64} \\ \\ \hline \end{array}  ​   C)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{256}& \frac{5}{196}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​  D)​  \begin{array}{|c|c|c|c|c|c|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}} & \frac{5}{64}& \frac{1}{20}& \frac{5}{144}& \frac{1}{20}& \frac{5}{64}\\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
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32
Use the given value of k to complete the table for the direct variation model​ y=kx2y = k x ^ { 2 } . ​
Plot the points on a rectangular coordinate system.
x810121416y=kx2k=14\begin{array}{l}\begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & & & & & \\& & & & & \\\hline\end{array}\\k = \frac { 1 } { 4 }\end{array}

A) ​ x810121416y=kx26449362516\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​
B)​ x810121416y=kx21616161616\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​
C)​ x810121416y=kx21625362516\begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​

D)​ x810121416y=kx21625364964\begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​
E)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = \frac { 1 } { 4 } \end{array} </strong> A) ​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 49 & 36 & 25 & 16 \\ & & & & & \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 16 & 16 & 16 & 16 \\ \hline \end{array}  ​   C)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 25 & 16 \\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 16 & 25 & 36 & 49 & 64 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​
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33
Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } Plot the points on a rectangular coordinate system.
x810121416y=kx2\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\\hline\end{array}
k=10k = 10
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​

A) x4681012y=kx258518532110572\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline y = \frac { k } { x ^ { 2 } } & & & & & \\& \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\\hline\end{array}
B)​ x4681012y=kx25858585858\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​
C)​ x4681012y=kx257211053251858\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​

D)​ x4681012y=kx25851853251858\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​
E)​ x4681012y=kx24681012\begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\\hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  Plot the points on a rectangular coordinate system.  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ \hline \end{array}   k = 10  ​  </strong> A)  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & & & & & \\ & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 1 } { 10 } & \frac { 5 } { 72 } \\ \hline \end{array}  B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline \\y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 72 } & \frac { 1 } { 10 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\\\ \hline \end{array}  ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 8 } & \frac { 5 } { 18 } & \frac { 5 } { 32 } & \frac { 5 } { 18 } & \frac { 5 } { 8 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 4 & 6 & 8 & 10 & 12 \\ \hline y = \frac { k } { x ^ { 2 } } & 4 & 6 & 8 & 10 & 12 \\ \hline \end{array}  ​
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Determine whether the variation model is of the form y=kxy = k x or y=kxy = \frac { k } { x } and find k.Then write a model that relates y and x.
x510152025y2613263132265\begin{array} { | c | c | c | c | c | c | } \hline x & 5 & 10 & 15 & 20 & 25 \\\hline y & 26 & 13 & \frac { 26 } { 3 } & \frac { 13 } { 2 } & \frac { 26 } { 5 } \\\hline\end{array}

A) y=130xy = \frac { 130 } { x }
B) y=x130y = \frac { x } { 130 }
C) y=1xy = \frac { 1 } { x }
D) y=130xy = 130 x
E) y=xy = x
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An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  Depth, d Temperature, C10003.820002.130001.840001.550000.5\begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\1000 & 3.8 ^ { \circ } \\2000 & 2.1 ^ { \circ } \\3000 & 1.8 ^ { \circ } \\4000 & 1.5 ^ { \circ } \\5000 & 0.5 ^ { \circ }\end{array}
Sketch a scatter plot of the data.

A)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)
B)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)
C)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)
D)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)
E)  <strong>An oceanographer took readings of the water temperatures C (in degrees Celsius)at several depths d (in meters).The data collected are shown in the table.  \begin{array} { l l } \text { Depth, } d & \text { Temperature, } C \\ 1000 & 3.8 ^ { \circ } \\ 2000 & 2.1 ^ { \circ } \\ 3000 & 1.8 ^ { \circ } \\ 4000 & 1.5 ^ { \circ } \\ 5000 & 0.5 ^ { \circ } \end{array}  Sketch a scatter plot of the data.</strong> A)   B)   C)   D)   E)
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Determine whether the variation model is of the form y=kxy = k x or y=kxy = \frac { k } { x } and find k.Then write a model that relates y and x. x918273645y246810\begin{array} { | c | c | c | c | c | c | } \hline x & 9 & 18 & 27 & 36 & 45 \\\hline y & 2 & 4 & 6 & 8 & 10 \\\hline\end{array}

A) y=29xy = \frac { 2 } { 9 } x
B) y=9xy = \frac { 9 } { x }
C) y=29xy = \frac { 2 } { 9 x }
D) y=92xy = \frac { 9 } { 2 x }
E) y=92xy = \frac { 9 } { 2 } x
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Use the given value of k to complete the table for the direct variation model​ y=kx2y = k x ^ { 2 } . ​
Plot the points on a rectangular coordinate system.
x810121416y=kx2k=1\begin{array}{l}\begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & & & & & \\& & & & & \\\hline\end{array}\\k = 1\end{array}

A)​ x810121416y=kx225619614410064\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​
B)​ x810121416y=kx264100144196256\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\\hline\end{array}

 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​
C)​ x810121416y=kx26464646464\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\\hline\end{array}

 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​

D)​ x810121416y=kx26410014410064\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\\hline\end{array}

 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​
E)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ & & & & & \\ \hline \end{array}\\ k = 1 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 256 & 196 & 144 & 100 & 64 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 196 & 256 \\ \hline \end{array}  ​ ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 64 & 64 & 64 & 64 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 64 & 100 & 144 & 100 & 64 \\ \hline \end{array}  ​ ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​
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38
Use the given value of k to complete the table for the inverse variation model​ y=kx2y = \frac { k } { x ^ { 2 } } . Plot the points on a rectangular coordinate system.
x810121416y=kx2\begin{array}{|l|l|l|l|l|l|}\hline x & 8 & 10 & 12 & 14 & 16 \\\hline y=\frac{k}{x^{2}}\\\hline\end{array}
k=20k = 20

A)​​ x810121416y=kx251615536549564\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
B)​ x810121416y=kx2516516516516516\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​

C)​ x810121416y=kx256454953615516\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\\hline\end{array}
​​  <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​

D)​ x810121416y=kx25161553615516\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
E)​ x810121416y=kx2810121416\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the inverse variation model​  y = \frac { k } { x ^ { 2 } }  . Plot the points on a rectangular coordinate system.  \begin{array}{|l|l|l|l|l|l|} \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y=\frac{k}{x^{2}}\\ \hline \end{array}   k = 20 </strong> A)​​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 5 } { 49 } & \frac { 5 } { 64 } \\ \hline \end{array}    B)​ \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   ​ C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & \frac { 5 } { 64 } & \frac { 5 } { 49 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & \frac { 5 } { 16 } & \frac { 1 } { 5 } & \frac { 5 } { 36 } & \frac { 1 } { 5 } & \frac { 5 } { 16 } \\ \hline \end{array}  ​   E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = \frac { k } { x ^ { 2 } } & 8 & 10 & 12 & 14 & 16 \\ \hline \end{array}  ​
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Determine whether the variation model is of the form y=kxy = k x or y=kxy = \frac { k } { x } and find k.Then write a model that relates y and x. x510152025y1.534.567.5\begin{array} { | c | c | c | c | c | c | } \hline x & 5 & 10 & 15 & 20 & 25 \\\hline y & - 1.5 & - 3 & - 4.5 & - 6 & - 7.5 \\\hline\end{array}

A) y=310xy = - \frac { 3 } { 10 x }
B) y=103xy = \frac { 10 } { 3 } x
C) y=310xy = - \frac { 3 } { 10 } x
D) y=310xy = \frac { 3 } { 10 } x
E) y=103xy = - \frac { 10 } { 3 } x
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40
Use the given value of k to complete the table for the direct variation model​ y=kx2y = k x ^ { 2 } . ​
Plot the points on a rectangular coordinate system.
x810121416y=kx2k=2\begin{array}{l}\begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & & & & & \\\hline\end{array}\\k = 2\end{array}

A)​ x810121416y=kx2128200288392512\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​
B)​ x810121416y=kx2810121416\begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\& & & & & \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​
C)​ x810121416y=kx2128128128128128\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\\hline\end{array}

 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​

D)​ x810121416y=kx2512392288200128\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​
E)​ x810121416y=kx2128200288200128\begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\\hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\\hline\end{array}
 <strong>Use the given value of k to complete the table for the direct variation model​  y = k x ^ { 2 }  . ​ Plot the points on a rectangular coordinate system.  \begin{array}{l} \begin{array} { | l | l | l | l | l | l | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & & & & & \\ \hline \end{array}\\ k = 2 \end{array} </strong> A)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 392 & 512 \\ \hline \end{array}  ​   B)​  \begin{array} { | c | r | r | r | r | r | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 8 & 10 & 12 & 14 & 16 \\ & & & & & \\ \hline \end{array}  ​   C)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 128 & 128 & 128 & 128 \\ \hline \end{array}  ​ ​   ​ D)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 512 & 392 & 288 & 200 & 128 \\ \hline \end{array}  ​   ​ E)​  \begin{array} { | c | c | c | c | c | c | } \hline x & 8 & 10 & 12 & 14 & 16 \\ \hline y = k x ^ { 2 } & 128 & 200 & 288 & 200 & 128 \\ \hline \end{array}  ​
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41
After determining whether the variation model below is of the form y=kxy = k x or y=kxy = \frac { k } { x } ,find the value of k.
x20406080100y13016019011201150\begin{array} { | c | c | c | c | c | c | } \hline x & 20 & 40 & 60 & 80 & 100 \\\hline y & \frac { 1 } { 30 } & \frac { 1 } { 60 } & \frac { 1 } { 90 } & \frac { 1 } { 120 } & \frac { 1 } { 150 } \\\hline\end{array}

A) k=120k = \frac { 1 } { 20 }
B) k=32k = \frac { 3 } { 2 }
C) k=54k = \frac { 5 } { 4 }
D) k=110k = \frac { 1 } { 10 }
E) k=23k = \frac { 2 } { 3 }
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42
The simple interest on an investment is directly proportional to the amount of the investment.By investing $6000 in a certain certificate of deposit,you obtained an interest payment of $276.00 after 1 year.Determine a mathematical model that gives the interest,I ,for this CD after 1 year in terms of the amount invested,P.

A) I=(0.050)PI = ( 0.050 ) P
B) I=(0.041)PI = ( 0.041 ) P
C) I=(0.049)PI = ( 0.049 ) P
D) I=(0.046)PI = ( 0.046 ) P
E) I=(0.044)PI = ( 0.044 ) P
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43
Find a mathematical model for the verbal statement: "m varies directly as the square of w and inversely as s."

A) m=k(ws)2m = k \left( \frac { w } { s } \right) ^ { 2 }
B) m=kw2sm = \frac { k w ^ { 2 } } { s }
C) m=kw2sm = k w ^ { 2 } s
D) m=kws2m = k w s ^ { 2 }
E) m=k(sw)2m = k \left( \frac { s } { w } \right) ^ { 2 }
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44
Find a mathematical model for the verbal statement: ​
"Q is jointly proportional to the cube of h and the square root of m."

A) Q=kh3mQ = k h ^ { 3 } \sqrt { m }
B) Q=kh3mQ = k \sqrt { h ^ { 3 } m }
C) Q=kh2m3Q = k h ^ { 2 } \sqrt [ 3 ] { m }
D) Q=khm23Q = k \sqrt [ 3 ] { h m ^ { 2 } }
E) Q=khm3Q = k \sqrt [ 3 ] { h m }
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45
Hooke's law states that the magnitude of force,F,required to stretch a spring x units beyond its natural length is directly proportional to x.If a force of 3 pounds stretches a spring from its natural length of 10 inches to a length of 10.7 inches,what force will stretch the spring to a length of 11.5 inches? Round your answer to the nearest hundredth.

A) F=5.52F = 5.52
B) F=6.43F = 6.43
C) F=5.70F = 5.70
D) F=7.29F = 7.29
E) F=6.14F = 6.14
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46
The sales tax on an item with a retail price of $972 is $68.04.Create a variational model that gives the retail price,y,in terms of the sales tax,x,and use it to determine the retail price of an item that has a sales tax of $82.62.

A)$1182.28
B)$1151.92
C)$1180.29
D)$1192.52
E)$1124.60
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47
Assume that y is directly proportional to x.If x = 28 and y = 21,determine a linear model that relates y and x.

A) y=35xy = \frac { 3 } { 5 } x
B) y=43xy = \frac { 4 } { 3 } x
C) y=34xy = \frac { 3 } { 4 } x
D) y=23xy = \frac { 2 } { 3 } x
E) y=32xy = \frac { 3 } { 2 } x
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48
After determining whether the variation model below is of the form y=kxy = k x or y=kxy = \frac { k } { x } ,find the value of k. x154161168175182y6669727578\begin{array} { | c | c | c | c | c | c | } \hline x & 154 & 161 & 168 & 175 & 182 \\\hline y & 66 & 69 & 72 & 75 & 78 \\\hline\end{array}

A) k=7k = 7
B) k=73k = \frac { 7 } { 3 }
C) k=17k = \frac { 1 } { 7 }
D) k=37k = \frac { 3 } { 7 }
E) k=766k = \frac { 7 } { 66 }
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49
Determine whether the variation model below is of the form y=kxy = k x or y=kxy = \frac { k } { x } . x
12
24
36
48
60
Y 118\frac { 1 } { 18 }136\frac { 1 } { 36 }154\frac { 1 } { 54 }172\frac { 1 } { 72 }190\frac { 1 } { 90 }

A) y=kxy = k x
B) y=kxy = \frac { k } { x }
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