Question 1

Solve.
-$$t ^ { 3 } + 64$$
A) $( t - 4 ) \left( t ^ { 2 } + 4 t + 16 \right)$
B) $( t - 64 ) \left( t ^ { 2 } - 1 \right)$
C) $( t + 4 ) \left( t ^ { 2 } + 16 \right)$
D) $( t + 4 ) \left( t ^ { 2 } - 4 t + 16 \right)$

Accepted Answer

The expression $t^3 + 64$ can be factored as a sum of cubes: $t^3 + 4^3$. The sum of cubes formula is $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. Applying this formula with $a = t$ and $b = 4$, we get $(t + 4)(t^2 - 4t + 16)$.

Question 2

Solve.
-$25 y ^ { 4 } - 81$
A) $\left( 5 y ^ { 2 } - 9 \right) ^ { 2 }$
B) $\left( 25 y ^ { 2 } + 1 \right) \left( y ^ { 2 } - 81 \right)$
C) $\left( 5 y ^ { 2 } + 9 \right) \left( 5 y ^ { 2 } - 9 \right)$
D) $\left( 5 y ^ { 2 } + 9 \right) ^ { 2 }$

Accepted Answer

The given expression can be factorized using the difference of squares formula as follows:
$$25y^4 - 81 = (5y^2)^2 - 9^2 = (5y^2 +9)(5y^2-9)$$
Therefore, the best choice is C.

Question 3

Solve.
-Linda invests $25,000 for one year. Part is invested at 5%, another part at 6%, and the rest at 8%. The total income from all 3 investments is $1600. The income from the 5% and 6% investments is
The same as the income from the 8% investment. Find the amount invested at each rate.
A) At 5%: $10,000; at 6%: $5000; at 8%: $10,000
B) At 5%: $8000; at 6%: $10,000; at 8%: $7000
C) At 5%: $10,000; at 6%: $10,000; at 8%: $5000
D) At 5%: $5000; at 6%: $10,000; at 8%: $10,000

Accepted Answer

Let the amounts invested at 5%, 6%, and 8% be x, y, and z respectively. We have three equations based on the given information:1) x + y + z = 25,000 (total investment)2) 0.05x + 0.06y + 0.08z = 1,600 (total income)3) 0.05x + 0.06y = 0.08z (income from 5% and 6% investments equals income from 8% investment)Solving these equations simultaneously, we find that x = $10,000, y = $5,000, and z = $10,000, which corresponds to choice A.

Question 4

Solve.
-$$a ^ { 2 } - 100$$
A) $( a - 10 ) ^ { 2 }$
B) $\left( a ^ { 2 } + 10 \right) \left( a ^ { 2 } - 10 \right)$
C) $( a + 10 ) ( a - 10 )$
D) $( a + 100 ) ( a - 100 )$

Accepted Answer

This is a difference of squares, so it can be factored as $(a+10)(a-10)$. Option A is the square of a binomial, but with a different constant value than the one given in the expression. Option B is not a valid factorization since it does not simplify to the given expression. Option D is also a difference of squares, but with the wrong constant value. Therefore, the best choice is C.

Question 5

Solve.
-$$15 x ^ { 2 } - 65 x - 50$$
A) $5 ( 3 x - 2 ) ( x + 5 )$
B) $5 ( 3 x + 2 ) ( x - 5 )$
C) $( 3 x + 2 ) ( 5 x - 25 )$
D) $( 15 x + 10 ) ( x - 5 )$

Accepted Answer

The answer of Solve.
-\[15 x ^ { 2 } -...

Question 6

Solve.
-A basketball fieldhouse seats 15,000. Courtside seats sell for $10 $\text { endzone for }$
A) 4000 courtside; 3000 endzone; 8000 balcony
B) 3000 courtside; 3000 endzone; 8,000 balcony
C) 3000 courtside; 2000 endzone; 10,000 balcony
D) 3200 courtside; 1800 endzone; 10,000 balcony

Accepted Answer

The total number of seats must add up to 15,000. Option C is the only one where the sum of courtside, endzone, and balcony seats equals 15,000.

Question 7

Solve.
-$$64 x ^ { 2 } + 25$$
A) $( 8 x + 5 ) ^ { 2 }$
B) $( 8 x + 5 ) ( 8 x - 5 )$
C) $( 8 x - 5 ) ^ { 2 }$
D) Prime

Accepted Answer

The expression $64x^2 + 25$ is a sum of squares, which cannot be factored over the real numbers. Therefore, it is considered prime.

Question 8

Solve.
-$$3 x ^ { 2 } + 19 x - 40$$
A) $( 3 x + 8 ) ( x - 5 )$
B) $( x - 8 ) ( 3 x - 5 )$
C) $( 3 x - 45 ) ( 3 x - 5 )$
D) $( x + 8 ) ( 3 x - 5 )$

Accepted Answer

The quadratic $3x^2 + 19x - 40$ factors into $(x + 8)(3x - 5)$ because expanding this product gives the original quadratic.

Question 9

Solve using a system of two equations in two unknowns.
-The manager of a bulk foods establishment sells a trail mix for $7 per pound and premium cashews for $15 per pound. The manager wishes to make a 120-pound trail mix-cashew mixture
That will sell for $9 per pound. How many pounds of each should be used?
A) 75 lb of trail mix, 45 lb of cashews
B) 60 lb of trail mix, 60 lb of cashews
C) 30 lb of trail mix, 90 lb of cashews
D) 90 lb of trail mix, 30 lb of cashews

Accepted Answer

Let x be the number of pounds of trail mix and y be the number of pounds of premium cashews.

From the problem, we know:

- x + y = 120 (the total mixture weighs 120 pounds) 
- 7x + 15y = 9(120) (the value of the mixture is $9 per pound)

Simplifying the second equation:

- 7x + 15y = 1080

Multiplying the first equation by 7 and subtracting it from the second equation:

- 7x + 15y - 7x - 7y = 1080 - 840 
- 8y = 240 
- y = 30

Substituting y = 30 into the first equation:

- x + 30 = 120 
- x = 90

Therefore, we need 90 pounds of trail mix and 30 pounds of premium cashews to make a 120-pound mixture that sells for $9 per pound.

Question 10

Solve.
-$3 x ^ { 2 } - 21 x + 36$
A) $( 3 x - 9 ) ( x - 4 )$
B) Prime
C) $3 ( x - 3 ) ( x - 4 )$
D) $3 ( x - 12 ) ( x + 1 )$

Accepted Answer

The correct factorization of $3x^2 - 21x + 36$ is found by first factoring out the greatest common factor, which is 3, resulting in $3(x^2 - 7x + 12)$. Then, finding two numbers that multiply to 12 and add to -7, which are -3 and -4, gives the factorization $3(x - 3)(x - 4)$.

Question 11

Solve using a system of two equations in two unknowns.
-A chemist needs 100 milliliters of a 70% solution but has only 66% and 86% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
A) 90 milliliters of 66%; 30 milliliters of 86%
B) 90 milliliters of 66%; 20 milliliters of 86%
C) 80 milliliters of 66%; 20 milliliters of 86%
D) 20 milliliters of 66%; 80 milliliters of 86%

Accepted Answer

Let x be the amount of the 66% solution and y be the amount of the 86% solution. We can set up the following system of equations:
x + y = 100 (since the chemist needs 100 milliliters of the final solution)
0.66x + 0.86y = 0.7(100) (since the final solution must be 70%)
Solving this system of equations gives x = 80 and y = 20, so the chemist needs 80 milliliters of the 66% solution and 20 milliliters of the 86% solution. Therefore, the answer is choice C.

Question 12

Solve.
-$15 t ^ { 3 } + 75 t ^ { 2 } - 5 t - 25$
A) $5 t ( 3 t + 1 ) ( t - 5 )$
B) $\left( 3 t ^ { 3 } - 1 \right) \left( t ^ { 2 } + 5 \right)$
C) $5 \left( 3 t ^ { 2 } - 1 \right) ( t + 5 )$
D) $5 \left( 3 t ^ { 2 } + 1 \right) ( t - 5 )$

Accepted Answer

The answer of Solve.
-\(15 t ^ { 3 } +...

Question 13

Solve.
-$$6 t ^ { 2 } + 13 t + 6$$
A) $( 3 t + 2 ) ( 2 t + 3 )$
B) $( 6 t + 2 ) ( t + 3 )$
C) $( 3 t - 2 ) ( 2 t - 3 )$
D) Prime

Accepted Answer

We need to factorize the quadratic expression $6t^2+13t+6$.
We can notice that $6 	imes 1 = 6$ and $6 + 1 = 7$, which could potentially help us in factoring the expression. We want to find two numbers that multiply to give us $6$ and add up to $13$. These numbers are $6$ and $1$.
Therefore, we can write $6t^2+13t+6$ as:
$$6t^2+6t+7t+6$$
We can then group the first two terms together and the last two terms together:
$$(6t^2+6t)+(7t+6)$$
We can factor out a $6t$ from the first group and a $2$ from the second group:
$$6t(t+1)+2(3t+2)$$
We can now see that we have a common factor of $(3t+2)$ that we can factor out:
$$(3t+2)(2t+3)$$
Therefore, the answer is choice A.

Question 14

Solve.
-$$m ^ { 2 } n ^ { 2 } + 4 m n - 12$$
A) $( \mathrm { mn } - 6 \mathrm { n } ) ( \mathrm { mn } + 2 \mathrm {~m} )$
B) $( m n + 6 m ) ( m n - 2 n )$
C) $( m n + 6 ) ( m n - 2 )$
D) $( m n - 6 ) ( m n + 2 )$

Accepted Answer

The answer of Solve.
-\[m ^ { 2 } n ^...

Question 15

Solve.
-Michaelʹs bank contains only nickels, dimes, and quarters. There are 59 coins in all, valued at $4.75. The number of nickels is 1 short of being three times the sum of the number of dimes and quarters
Together. How many dimes are in the bank?
A) 44
B) 7
C) 8
D) 13

Accepted Answer

Let's use variables to represent the unknowns:

N = number of nickels 
D = number of dimes 
Q = number of quarters

From the problem, we have the following information:

1. N + D + Q = 59 (the total number of coins is 59) 
2. 0.05N + 0.10D + 0.25Q = 4.75 (the total value of the coins is $4.75) 
3. N = 3(D + Q) - 1 (the number of nickels is 1 less than 3 times the sum of dimes and quarters)

Now we can use these equations to solve for D:

Substitute equation 3 into equation 1:

(3(D + Q) - 1) + D + Q = 59

Simplify:

4D + 4Q = 60

Divide both sides by 4:

D + Q = 15

We can use this result to solve for Q:

Q = 15 - D

Substitute equations 3 and 4 into equation 2:

0.05(3(D + Q) - 1) + 0.10D + 0.25Q = 4.75

Simplify:

0.15D + 0.25Q = 5.00

Substitute Q = 15 - D:

0.15D + 0.25(15 - D) = 5.00

Simplify:

0.10D + 3.75 = 5.00

0.10D = 1.25

D = 12.5

Since D must be a whole number, the closest answer choice is C) 8.

Question 16

Solve.
-Michaelʹs bank contains only nickels, dimes, and quarters. There are 54 coins in all, valued at $4.30. The number of nickels is 2 short of being three times the sum of the number of dimes and quarters
Together. How many dimes are in the bank?
A) 6
B) 40
C) 8
D) 13

Accepted Answer

The total number of coins is 54, and their total value is $4.30. Let's denote the number of nickels as N, dimes as D, and quarters as Q. We have three equations based on the information given:1. $N + D + Q = 54$ (total number of coins)2. $0.05N + 0.10D + 0.25Q = 4.30$ (total value of coins)3. $N = 3(D + Q) - 2$ (relationship between the number of nickels, dimes, and quarters)From equation 3, we can express N in terms of D and Q. Substituting the value of N from equation 3 into equations 1 and 2, we can solve the system of equations for D and Q. Solving these equations, we find that the number of dimes, D, is 8.

Question 17

Solve using a system of two equations in two unknowns.
-Dmitri needs 5 liters of a 7% solution of sulfuric acid for a research project in molecular biology. He has two unlimited supplies of sulfuric acid solution: 5% solution and 10% solution. How many
Liters of each solution should Dmitri use?
A) 5% solution: 3.5 L, 10% solution: 1.5 L
B)5% solution: 3 L, 10% solution: 2 L
C) 5% solution: 2.5 L, 10% solution: 2.5 L
D)5% solution: 2 L, 10% solution: 3 L

Accepted Answer

Let x be the number of liters of the 5% solution and y be the number of liters of the 10% solution. Then we have the system of equations:

x + y = 5 (since Dmitri needs a total of 5 liters)
0.05x + 0.1y = 0.07(5) (since the final solution needs to be 7% sulfuric acid)

Simplifying the second equation, we get:

0.05x + 0.1y = 0.35

Multiplying both sides by 10 to eliminate decimals, we get:

5x + 10y = 35

Now we can use the first equation to solve for one of the variables in terms of the other. For example, solve for y:

y = 5 - x

Substitute this expression for y into the second equation to get:

5x + 10(5 - x) = 35

Simplify and solve for x:

5x + 50 - 10x = 35
-5x = -15
x = 3

So Dmitri needs 3 liters of the 5% solution. Substituting this back into the equation y = 5 - x, we get:

y = 5 - 3
y = 2

So Dmitri needs 2 liters of the 10% solution. Therefore, the best choice is B: 5% solution: 3 L, 10% solution: 2 L.

Quiz 8: A Transition

Solve. - $t ^ { 3 } + 64$

Solve. - $25 y ^ { 4 } - 81$

Solve. -Linda invests $25,000 for one year. Part is invested at 5%, another part at 6%, and the rest at 8%. The total income from all 3 investments is $1600. The income from the 5% and 6% investments is The same as the income from the 8% investment. Find the amount invested at each rate.

Solve. - $a ^ { 2 } - 100$

Solve. - $15 x ^ { 2 } - 65 x - 50$

Solve. -A basketball fieldhouse seats 15,000. Courtside seats sell for $10 $\text { endzone for }$

Solve. - $64 x ^ { 2 } + 25$

Solve. - $3 x ^ { 2 } + 19 x - 40$

Solve. - $3 x ^ { 2 } - 21 x + 36$

Solve using a system of two equations in two unknowns. -A chemist needs 100 milliliters of a 70% solution but has only 66% and 86% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.

Solve. - $15 t ^ { 3 } + 75 t ^ { 2 } - 5 t - 25$

Solve. - $6 t ^ { 2 } + 13 t + 6$

Solve. - $m ^ { 2 } n ^ { 2 } + 4 m n - 12$

Solve. -Michaelʹs bank contains only nickels, dimes, and quarters. There are 59 coins in all, valued at $4.75. The number of nickels is 1 short of being three times the sum of the number of dimes and quarters Together. How many dimes are in the bank?

Solve. -Michaelʹs bank contains only nickels, dimes, and quarters. There are 54 coins in all, valued at $4.30. The number of nickels is 2 short of being three times the sum of the number of dimes and quarters Together. How many dimes are in the bank?

Solve using a system of two equations in two unknowns. -Dmitri needs 5 liters of a 7% solution of sulfuric acid for a research project in molecular biology. He has two unlimited supplies of sulfuric acid solution: 5% solution and 10% solution. How many Liters of each solution should Dmitri use?

Quiz 8: A Transition

Solve. - t3+64t ^ { 3 } + 64t3+64

Solve. - 25y4−8125 y ^ { 4 } - 8125y4−81

Solve. -Linda invests $25,000 for one year. Part is invested at 5%, another part at 6%, and the rest at 8%. The total income from all 3 investments is $1600. The income from the 5% and 6% investments is The same as the income from the 8% investment. Find the amount invested at each rate.

Solve. - a2−100a ^ { 2 } - 100a2−100

Solve. - 15x2−65x−5015 x ^ { 2 } - 65 x - 5015x2−65x−50

Solve. -A basketball fieldhouse seats 15,000. Courtside seats sell for $10 endzone for \text { endzone for } endzone for

Solve. - 64x2+2564 x ^ { 2 } + 2564x2+25

Solve. - 3x2+19x−403 x ^ { 2 } + 19 x - 403x2+19x−40

Solve. - 3x2−21x+363 x ^ { 2 } - 21 x + 363x2−21x+36

Solve using a system of two equations in two unknowns. -A chemist needs 100 milliliters of a 70% solution but has only 66% and 86% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.

Solve. - 15t3+75t2−5t−2515 t ^ { 3 } + 75 t ^ { 2 } - 5 t - 2515t3+75t2−5t−25

Solve. - 6t2+13t+66 t ^ { 2 } + 13 t + 66t2+13t+6

Solve. - m2n2+4mn−12m ^ { 2 } n ^ { 2 } + 4 m n - 12m2n2+4mn−12

Solve. -Michaelʹs bank contains only nickels, dimes, and quarters. There are 59 coins in all, valued at $4.75. The number of nickels is 1 short of being three times the sum of the number of dimes and quarters Together. How many dimes are in the bank?

Solve. -Michaelʹs bank contains only nickels, dimes, and quarters. There are 54 coins in all, valued at $4.30. The number of nickels is 2 short of being three times the sum of the number of dimes and quarters Together. How many dimes are in the bank?

Solve using a system of two equations in two unknowns. -Dmitri needs 5 liters of a 7% solution of sulfuric acid for a research project in molecular biology. He has two unlimited supplies of sulfuric acid solution: 5% solution and 10% solution. How many Liters of each solution should Dmitri use?

Solve. - $t ^ { 3 } + 64$

Solve. - $25 y ^ { 4 } - 81$

Solve. - $a ^ { 2 } - 100$

Solve. - $15 x ^ { 2 } - 65 x - 50$

Solve. -A basketball fieldhouse seats 15,000. Courtside seats sell for $10 $\text { endzone for }$

Solve. - $64 x ^ { 2 } + 25$

Solve. - $3 x ^ { 2 } + 19 x - 40$

Solve. - $3 x ^ { 2 } - 21 x + 36$

Solve. - $15 t ^ { 3 } + 75 t ^ { 2 } - 5 t - 25$

Solve. - $6 t ^ { 2 } + 13 t + 6$

Solve. - $m ^ { 2 } n ^ { 2 } + 4 m n - 12$