Question 1

Solve. Round results to the nearest thousandth.
-$10 x ^ { 2 } - 3 x - 15 = 0$
A) $1.384 , - 1.084$
B) $1.084 , - 1.384$
C) $0.779 , - 1.079$
D) $1.079 , - 0.779$

Accepted Answer

This quadratic equation can be solved using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 10$, $b = -3$, and $c = -15$. Plugging these values into the formula gives $x = \frac{3 \pm \sqrt{(-3)^2 - 4(10)(-15)}}{2(10)}$, which simplifies to $x = \frac{3 \pm \sqrt{9 + 600}}{20}$ or $x = \frac{3 \pm \sqrt{609}}{20}$. This further simplifies to $x \approx 1.384$ and $x \approx -1.084$, when rounded to the nearest thousandth.

Question 2

Solve.
-A lot is in the shape of a right triangle. The shorter leg measures 120 m. The hypotenuse is 40 m longer than the length of the longer leg. How long is the longer leg?
A)160 m
B)120 m
C)240 m
D)200 m

Accepted Answer

Using the Pythagorean theorem, let x be the length of the longer leg. The hypotenuse is x + 40. The equation is $120^2 + x^2 = (x + 40)^2$. Solving for x gives x = 160 m.

Question 3

Solve.
-A rectangular garden has dimensions of 23 feet by 16 feet. A gravel path of equal width is to be built around the garden. How wide can the path be if there is enough gravel for 612 square feet?
A)8.5 ft
B)6 ft
C)8 ft
D)7 ft

Accepted Answer

The answer of Solve.
-A rectangular garden has dimensions of 23...

Question 4

Solve.
-A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10 miles per hour faster than his rate walking, what was each rate?
A)Bike: 12 mph
B)Bike: 14.5 mph
C)Bike: 11.5 mph
D)Bike: 13 mph Hike: 2 mph Hike: 4.5 mph Hike: 1.5 mph Hike: 3 mph

Accepted Answer

The man's biking speed is 12 mph and hiking speed is 2 mph. Using the formula time = distance/speed, the biking time is 12 miles / 12 mph = 1 hour, and the hiking time is 8 miles / 2 mph = 4 hours, totaling 5 hours.

Question 5

Find the x-intercepts.
-$$f ( x ) = 11 x ^ { 2 } - 3 x - 6$$
A) No x-intercepts
B) $( 9,0 )$ and $( 6,0 )$
C) $\left( \frac { 3 + \sqrt { 273 } } { 22 } , 0 \right)$ and $\left( \frac { 3 - \sqrt { 273 } } { 22 } , 0 \right)$
D) $\left( \frac { 3 + \sqrt { 261 } } { 22 } , 0 \right)$ and $\left( \frac { 3 - \sqrt { 261 } } { 22 } , 0 \right)$

Accepted Answer

To find the x-intercepts of the quadratic equation $f(x) = 11x^2 - 3x - 6$, we set $f(x) = 0$ and solve for $x$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 11$, $b = -3$, and $c = -6$. Plugging these values into the formula gives $x = \frac{3 \pm \sqrt{(-3)^2 - 4(11)(-6)}}{2(11)}$, which simplifies to $x = \frac{3 \pm \sqrt{9 + 264}}{22}$ or $x = \frac{3 \pm \sqrt{273}}{22}$.

Question 6

Solve.
-Bill can row 3 mph in still water. It takes him 3 hours 36 minutes to go 3 miles upstream and return. Find the speed of the current.
A)3 mph
B)2 mph
C)2.5 mph
D)1.5 mph

Accepted Answer

The answer of Solve.
-Bill can row 3 mph in still...

Question 7

Solve. Round results to the nearest thousandth.
-$3 y ^ { 2 } - 4 y = 1$
A) $1.000,0.333$
B) $0.215 , - 1.549$
C) $1.549 , - 0.215$
D) $3.098 , - 0.430$

Accepted Answer

First, rearrange the equation to standard form: $3y^2 - 4y - 1 = 0$. Then, use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 3$, $b = -4$, and $c = -1$. This gives $y = \frac{4 \pm \sqrt{(-4)^2 - 4(3)(-1)}}{2(3)} = \frac{4 \pm \sqrt{16 + 12}}{6} = \frac{4 \pm \sqrt{28}}{6}$, which simplifies to $y = 1.549$ and $y = -0.215$, rounded to the nearest thousandth.

Question 8

Solve.
-A square has an area of 49 square inches. If the same amount is added to the length and removed from the width, the resulting rectangle has an area of 45 square inches. Find the dimensions of the rectangle.
A)3 in. by 4 in.
B)4 in. by 9 in.
C)5 in. by 10 in.
D)5 in. by 9 in.

Accepted Answer

The side of the square is 7 inches (since 7^2 = 49). If x is added to the length and removed from the width, the equation becomes (7 + x)(7 - x) = 45. Solving gives x = 2, so the rectangle's dimensions are 5 in. by 9 in.

Question 9

Solve.
-The area of a square is 81 square centimeters. If the same amount is added to one dimension and removed from the other, the resulting rectangle has an area 9 square centimeters less than the area of the square. How much is Added and subtracted?
A)12 cm
B)4 cm
C)3 cm
D)9 cm

Accepted Answer

The side of the square is √81 = 9 cm. Let x be the amount added and subtracted. The new dimensions are (9 + x) and (9 - x), and their product is 81 - 9 = 72. So, (9 + x)(9 - x) = 72. Simplifying gives x^2 = 9, so x = 3.

Question 10

Solve. Round results to the nearest thousandth.
-$$9 x ^ { 2 } = 2 + 17 x$$
A) $2.000 , - 0.111$
B) $1.763,0.126$
C) $- 0.126 , - 1.763$
D) $0.111 , - 2.000$

Accepted Answer

First, rearrange the equation to standard quadratic form: $9x^2 - 17x - 2 = 0$. Then, use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 9$, $b = -17$, and $c = -2$. Solving gives $x \approx 2.000$ and $x \approx -0.111$.

Question 11

Solve.
-A rug is to fit in a room so that a border of even width is left on all four sides. If the room is 18 feet by 20 feet and the area of the rug is 120 square feet, how wide will the border be?
A)6 ft
B)5 ft
C)6.5 ft
D)4 ft

Accepted Answer

The area of the room is 18 ft * 20 ft = 360 square feet. The area of the rug is 120 square feet, so the area of the border is 360 - 120 = 240 square feet. Let x be the width of the border. The dimensions of the rug will then be (18 - 2x) by (20 - 2x). The area of the rug is also given by the equation (18 - 2x)(20 - 2x) = 120. Simplifying this equation gives a quadratic equation in terms of x. Solving for x, we find that x = 4 ft, which means the border will be 4 feet wide.

Question 12

Solve.
-Two cars leave an intersection. One car travels north; the other east. When the car traveling north had gone 15 mi, the distance between the cars was 5 mi more than the distance traveled by the car heading east. How far had The eastbound car traveled?
A)25 mi
B)20 mi
C)30 mi
D)15 mi

Accepted Answer

The problem can be solved using the Pythagorean theorem, where the distance between the two cars forms the hypotenuse of a right triangle. Let x be the distance traveled by the eastbound car. According to the problem, the hypotenuse (distance between the cars) is x + 5 miles, and the distance traveled north is 15 miles. The equation is $15^2 + x^2 = (x + 5)^2$. Solving for x gives us 20 miles.

Question 13

Solve.
-A picture 9 inches by 18 inches is to be mounted on a piece of matboard so that there is an even amount of mat all around the picture. How wide will the border be if the area of the border is 124 square inches?
A)4.5 in
B)4 in
C)3 in
D)2 in

Accepted Answer

The answer of Solve.
-A picture 9 inches by 18 inches...

Question 14

Solve.
-A ladder is resting against a wall. The top of the ladder touches the wall at a height of 6 ft. Find the length of the ladder if the length is 2 ft more than its distance from the wall.
A)8 ft
B)6 ft
C)10 ft
D)12 ft

Accepted Answer

Using the Pythagorean theorem, let the distance from the wall be x. Then, the length of the ladder is x + 2. We have x^2 + 6^2 = (x + 2)^2. Solving this equation gives x = 8, so the ladder is 10 ft long.

Question 15

Solve.
-A jet plane traveling at a constant speed goes 1200 miles with the wind, then turns around and travels for 1000 miles against the wind. If the speed of the wind is 50 mph and the total flight took 4 hours, find the speed of the Plane in still air.
A)435 mph
B)550 mph
C)605 mph
D)525 mph

Accepted Answer

The answer of Solve.
-A jet plane traveling at a constant...

Question 16

Solve. Round results to the nearest thousandth.
-$$x ^ { 2 } + 10 x - 19 = 0$$
A) $1.633 , - 11.633$
B) $11.633 , - 1.633$
C) $13.958 , - 3.958$
D) $3.958 , - 13.958$

Accepted Answer

This quadratic equation can be solved using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 10$, and $c = -19$. Plugging these values in gives $x = \frac{-10 \pm \sqrt{10^2 - 4(1)(-19)}}{2(1)} = \frac{-10 \pm \sqrt{100 + 76}}{2} = \frac{-10 \pm \sqrt{176}}{2} = \frac{-10 \pm 13.266}{2}$, which results in $x = 1.633$ and $x = -11.633$, rounded to the nearest thousandth.

Question 17

Solve. Round results to the nearest thousandth.
-$$x ^ { 2 } + 6 x = 2$$
A) $- 0.354 , - 5.646$
B) $6.317 , - 0.317$
C) $0.317 , - 6.317$
D) $- 2.683 , - 2.683$

Accepted Answer

The answer of Solve. Round results to the nearest thousandth.
-\[x...

Question 18

Solve.
-Sue rowed her boat across Lake Bend and back in 3 hours. If her rate returning was 2 mph less than the rate going, and if the distance each way was 7 miles, find her rate going.
A)5.5 mph
B)5.9 mph
C)1.5 mph
D)3.7 mph

Accepted Answer

Let's call Sue's rate going "x" mph. Then her rate returning is "x-2" mph (since it's 2 mph less).

We know that the distance each way is 7 miles, so Sue travels a total of 14 miles (7 miles going and 7 miles returning).

We also know that she completes this entire round trip in 3 hours.

Using the formula:

distance = rate x time

we can set up two equations:

7 = x*time going

7 = (x-2)*time returning

We can solve for "time going" in the first equation:

time going = 7/x

And we can solve for "time returning" in the second equation:

time returning = 7/(x-2)

We know that the total time for the round trip is 3 hours, so we can set up another equation:

time going + time returning = 3

Substituting our expressions for "time going" and "time returning":

7/x + 7/(x-2) = 3

Multiplying both sides by x(x-2) to clear the denominators:

7(x-2) + 7x = 3x(x-2)

Expanding and simplifying:

14x - 14 = 3x^2 - 6x

3x^2 - 20x + 14 = 0

We can use the quadratic formula to solve for x:

x = (20 ± sqrt(20^2 - 4(3)(14))) / (2(3))

x ≈ 5.9 or x ≈ 1.5

We know that Sue's rate going can't be 1.5 mph (since the rate returning would be x-2 = -0.5 mph, which doesn't make sense). So the only reasonable choice is B) 5.9 mph.

Question 19

Solve. Round results to the nearest thousandth.
-$4 z ^ { 2 } - 6 z = 1$
A) $1.309,0.191$
B) $0.151 , - 1.651$
C) $1.651 , - 0.151$
D) $3.302 , - 0.302$

Accepted Answer

The quadratic equation $4z^2 - 6z - 1 = 0$ can be solved using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = 4$, $b = -6$, and $c = -1$. Solving gives the roots $z = 1.651$ and $z = -0.151$.

Question 20

Find the x-intercepts.
-$f ( x ) = 25 x ^ { 2 } - 10 x + 1$
A) $\left( \frac { 1 } { 5 } , 0 \right)$ and $\left( \frac { 1 } { 5 } , 0 \right)$
B) $\left( - \frac { 1 } { 5 } , 0 \right)$
C) $\left( \frac { 1 } { 5 } , 0 \right)$
D) No x-intercepts

Accepted Answer

The x-intercepts are the points where the graph of the function crosses the x-axis. To find the x-intercepts, we set $f(x) = 0$ and solve for $x$.$$25x^2 - 10x + 1 = 0$$$$(5x - 1)^2 = 0$$$$5x - 1 = 0$$$$x = \frac{1}{5}$$Therefore, the x-intercept is $(1/5, 0)$.

Quiz 11: Quadratic Equations and Functions

Solve. Round results to the nearest thousandth. - $10 x ^ { 2 } - 3 x - 15 = 0$

Solve. -A lot is in the shape of a right triangle. The shorter leg measures 120 m. The hypotenuse is 40 m longer than the length of the longer leg. How long is the longer leg?

Solve. -A rectangular garden has dimensions of 23 feet by 16 feet. A gravel path of equal width is to be built around the garden. How wide can the path be if there is enough gravel for 612 square feet?

Solve. -A man rode a bicycle for 12 miles and then hiked an additional 8 miles. The total time for the trip was 5 hours. If his rate when he was riding a bicycle was 10 miles per hour faster than his rate walking, what was each rate?

Find the x-intercepts. - $f ( x ) = 11 x ^ { 2 } - 3 x - 6$

Solve. -Bill can row 3 mph in still water. It takes him 3 hours 36 minutes to go 3 miles upstream and return. Find the speed of the current.

Solve. Round results to the nearest thousandth. - $3 y ^ { 2 } - 4 y = 1$

Solve. -A square has an area of 49 square inches. If the same amount is added to the length and removed from the width, the resulting rectangle has an area of 45 square inches. Find the dimensions of the rectangle.

Solve. -The area of a square is 81 square centimeters. If the same amount is added to one dimension and removed from the other, the resulting rectangle has an area 9 square centimeters less than the area of the square. How much is Added and subtracted?

Solve. Round results to the nearest thousandth. - $9 x ^ { 2 } = 2 + 17 x$

Solve. -A rug is to fit in a room so that a border of even width is left on all four sides. If the room is 18 feet by 20 feet and the area of the rug is 120 square feet, how wide will the border be?

Solve. -Two cars leave an intersection. One car travels north; the other east. When the car traveling north had gone 15 mi, the distance between the cars was 5 mi more than the distance traveled by the car heading east. How far had The eastbound car traveled?

Solve. -A picture 9 inches by 18 inches is to be mounted on a piece of matboard so that there is an even amount of mat all around the picture. How wide will the border be if the area of the border is 124 square inches?

Solve. -A ladder is resting against a wall. The top of the ladder touches the wall at a height of 6 ft. Find the length of the ladder if the length is 2 ft more than its distance from the wall.

Solve. -A jet plane traveling at a constant speed goes 1200 miles with the wind, then turns around and travels for 1000 miles against the wind. If the speed of the wind is 50 mph and the total flight took 4 hours, find the speed of the Plane in still air.

Solve. Round results to the nearest thousandth. - $x ^ { 2 } + 10 x - 19 = 0$

Solve. Round results to the nearest thousandth. - $x ^ { 2 } + 6 x = 2$

Solve. -Sue rowed her boat across Lake Bend and back in 3 hours. If her rate returning was 2 mph less than the rate going, and if the distance each way was 7 miles, find her rate going.

Solve. Round results to the nearest thousandth. - $4 z ^ { 2 } - 6 z = 1$

Find the x-intercepts. - $f ( x ) = 25 x ^ { 2 } - 10 x + 1$