Question 1

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)9 cm
B)4 cm
C)12 cm
D)3 cm

Accepted Answer

The side of the square is √81 = 9 cm. Let x be the amount added and subtracted. The rectangle's dimensions are (9+x) and (9-x), so its area is (9+x)(9-x) = 81 - x^2. The area is 9 cm² less than the square's, so 81 - x^2 = 81 - 9. Solving for x, x^2 = 9, thus x = 3 cm.

Question 2

Find the x-intercepts.
-$$f ( x ) = 64 x ^ { 2 } - 32 x + 4$$
A) No x-intercepts
B) $\left( - \frac { 1 } { 4 } , 0 \right)$
C) $\left( \frac { 1 } { 4 } , 0 \right)$ and $\left( - \frac { 1 } { 4 } , 0 \right)$
D) $\left( \frac { 1 } { 4 } , 0 \right)$

Accepted Answer

The answer of Find the x-intercepts.
-\[f ( x ) =...

Question 3

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)1.5 mph
B)2.5 mph
C)2 mph
D)3 mph

Accepted Answer

Let x be the speed of the current.
According to the problem:
3(x+3) = 3, which means that the distance he covers while rowing upstream is 1 mile (since he covers 3 miles in total).
3(x-3) = 3, which means that the distance he covers while rowing downstream is also 1 mile.
Using the formula Distance = Speed x Time, we get the following two equations:
1 = (3-x) t1, where t1 is the time taken to row upstream.
1 = (3+x) t2, where t2 is the time taken to row downstream.
Since it took Bill 3 hours 36 minutes (or 3.6 hours) in total, we know that:
t1 + t2 = 3.6
Solving the three equations simultaneously, we get x = 2 mph.
Therefore, the correct answer is C).

Question 4

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)10 ft
B)6 ft
C)12 ft
D)8 ft

Accepted Answer

Using the Pythagorean theorem, let the distance from the wall be x feet. Then, the length of the ladder is x + 2 feet. The equation is x^2 + 6^2 = (x + 2)^2. Solving for x gives x = 8, so the ladder's length is 8 + 2 = 10 feet.

Question 5

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 13 feet by 21 feet and the area of the rug is 153 square feet, how wide will the border be?
A)3 ft
B)2 ft
C)4.5 ft
D)4 ft

Accepted Answer

The answer of Solve.
-A rug is to fit in a...

Question 6

Find the x-intercepts.
-$$f ( x ) = 2 x ^ { 2 } - 3 x - 4$$
A) $\left( \frac { 3 + \sqrt { 41 } } { 4 } , 0 \right)$ and $\left( \frac { 3 - \sqrt { 41 } } { 4 } , 0 \right)$
B) No x-intercepts
C) $\left( \frac { 3 + \sqrt { 29 } } { 4 } , 0 \right)$ and $\left( \frac { 3 - \sqrt { 29 } } { 4 } , 0 \right)$
D) $( 0,0 )$ and $( 4,0 )$

Accepted Answer

To find the x-intercepts, we need to set $f(x)$ to 0 and solve for $x$.
\begin{align*}
2x^2-3x-4 &= 0 \
x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \
x &= \frac{3\pm\sqrt{(-3)^2-4(2)(-4)}}{2(2)} \
x &= \frac{3\pm\sqrt{41}}{4}
\end{align*}
Therefore, the x-intercepts are at $\left(\frac{3+\sqrt{41}}{4}, 0ight)$ and $\left(\frac{3-\sqrt{41}}{4}, 0ight)$. The answer is choice A.

Question 7

Solve.
-A lot is in the shape of a right triangle. The shorter leg measures 180 m. The hypotenuse is 60 m longer than the length of the longer leg. How long is the longer leg?
A)300 m
B)240 m
C)360 m
D)180 m

Accepted Answer

Let's call the longer leg "x". According to the problem, the hypotenuse is "60 m longer than the length of the longer leg", which means it's x+60. We can use the Pythagorean theorem to set up an equation:

180^2 + x^2 = (x+60)^2

Simplifying:

32,400 + x^2 = x^2 + 120x + 3,600

Subtracting x^2 from both sides:

32,400 = 120x + 3,600

Subtracting 3,600 from both sides:

28,800 = 120x

Dividing both sides by 120:

240 = x

So the longer leg measures 240 m.

Question 8

Solve. Round results to the nearest thousandth.
-$$5 \mathrm { x } ^ { 2 } = 10 + 14 \mathrm { x }$$
A) $1.600,1.200$
B) $0.590 , - 3.390$
C) $- 1.200 , - 1.600$
D) $3.390 , - 0.590$

Accepted Answer

Rearrange the equation to $5x^2 - 14x - 10 = 0$. Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 5$, $b = -14$, and $c = -10$. The solutions are approximately $3.390$ and $-0.590$.

Question 9

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)4 in. by 9 in.
B)3 in. by 4 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 dimensions are 5 in. by 9 in.

Question 10

Solve. Round results to the nearest thousandth.
-$$2 y ^ { 2 } + 6 y = - 4$$
A) $- 1.000 , - 2.000$
B) $- 2.000 , - 4.000$
C) $2.000,1.000$
D) $0.562 , - 3.562$

Accepted Answer

First, rearrange the equation to standard form: $2y^2 + 6y + 4 = 0$. Then, use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 2$, $b = 6$, and $c = 4$. Solving gives $y = -1$ and $y = -2$, hence the correct answers are $-1.000$ and $-2.000$.

Question 11

Solve. Round results to the nearest thousandth.
-$$x ^ { 2 } - 10 x - 36 = 0$$
A) $23.276 , - 13.276$
B) $12.810 , - 2.810$
C) $13.276 , - 23.276$
D) $2.810 , - 12.810$

Accepted Answer

We can use the quadratic formula to solve for $x$: $$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-36)}}{2(1)}$$ Simplifying this gives us: $$x=\frac{10\pm\sqrt{244}}{2}$$ We can then use a calculator to approximate the two solutions to the nearest thousandth: $$x\approx12.810,-2.810$$ These values match the numbers in choice B, so the answer is B.

Question 12

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)1.5 mph
B)3.7 mph
C)5.5 mph
D)5.9 mph

Accepted Answer

The answer of Solve.
-Sue rowed her boat across Lake Bend...

Question 13

Solve.
-A picture 13 inches by 23 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 252 square inches?
A)4 in
B)3 in
C)5 in
D)5.5 in

Accepted Answer

Let x be the width of the border in inches. 
The length of the picture and matboard together will be (13+2x) inches and the width will be (23+2x) inches. 
The area of the entire piece will be (13+2x)(23+2x) square inches. 
The area of the picture itself is 13*23=299 square inches. 
Subtracting the area of the picture from the area of the entire piece gives the area of the border: 
(13+2x)(23+2x) - 299 = 252 
Simplifying the left side: 
4x^2 + 72x - 220 = 0 
Dividing each term by 4: 
x^2 + 18x - 55 = 0 
Factoring: 
(x+5)(x-11) = 0 
Since x cannot be negative, the only possible solution is x=11. 
Therefore, the width of the border will be 11/2 = 5.5 inches.

Question 14

Solve. Round results to the nearest thousandth.
-$$3 x ^ { 2 } - 6 x = 2$$
A) $0.291 , - 2.291$
B) $2.291 , - 0.291$
C) $4.582 , - 0.582$
D) $1.577,0.423$

Accepted Answer

We can solve the equation by using the quadratic formula:

$$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{36-24}}{6} \approx \{2.291, -0.291\}$$

Therefore, the answer is choice B, which contains the two rounded solutions.

Question 15

Solve.
-A rectangular garden has dimensions of 23 feet by 20 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 294 square feet?
A)5.5 ft
B)4 ft
C)5 ft
D)3 ft

Accepted Answer

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

Question 16

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)15 mi
B)30 mi
C)20 mi
D)25 mi

Accepted Answer

Let's call the distance traveled by the eastbound car "x". According to the problem, we know that:

(distance between the two cars)^2 = (distance traveled by northbound car)^2 + (distance traveled by eastbound car)^2

We also know that when the northbound car had gone 15 mi, the distance between the two cars was 5 mi more than the distance traveled by the eastbound car. So:

(distance between the two cars) = x + 5

Putting these two pieces of information together, we can write:

(x + 5)^2 = 15^2 + x^2

Simplifying this equation, we get:

x^2 + 10x - 200 = 0

Factoring, we get:

(x - 10)(x + 20) = 0

Since we're looking for a positive distance, we can ignore the negative solution and conclude that:

x = 10

So the eastbound car traveled 10 miles when the northbound car had gone 15 miles. Therefore, the eastbound car must have traveled a total of:

15 + 10 = 25 miles

The only answer choice that is close to 25 miles is C) 20 mi, but it's not exactly right. So the best choice is C) 20 mi, even though it's not the exact answer.

Question 17

Solve. Round results to the nearest thousandth.
-$$x ^ { 2 } - 8 x = 6$$
A) $12.690,12.690$
B) $7.162,0.838$
C) $8.690 , - 0.690$
D) $0.690 , - 8.690$

Accepted Answer

To solve the equation $x^2 - 8x = 6$, we first rewrite it as $x^2 - 8x - 6 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -8$, and $c = -6$, we find the solutions to be approximately $8.690$ and $-0.690$.

Question 18

Solve. Round results to the nearest thousandth.
-$9 x ^ { 2 } - 2 x - 8 = 0$
A) $1.060 , - 0.838$
B) $0.537 , - 0.759$
C) $0.838 , - 1.060$
D) $0.759 , - 0.537$

Accepted Answer

This quadratic equation can be solved using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 9$, $b = -2$, and $c = -8$. Plugging these values into the formula gives $x = \frac{2 \pm \sqrt{(-2)^2 - 4(9)(-8)}}{2(9)}$, which simplifies to $x = \frac{2 \pm \sqrt{4 + 288}}{18}$ or $x = \frac{2 \pm \sqrt{292}}{18}$. Further simplification gives the approximate solutions $x \approx 1.060$ and $x \approx -0.838$, rounding to the nearest thousandth.

Question 19

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: 11.5 mph
B)Bike: 12 mph
C)Bike: 14.5 mph
D)Bike: 13 mph Hike: 1.5 mph Hike: 2 mph Hike: 4.5 mph Hike: 3 mph

Accepted Answer

The answer of Solve.
-A man rode a bicycle for 12...

Question 20

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)525 mph
C)550 mph
D)605 mph

Accepted Answer

Let the speed of the plane in still air be x mph.

When the plane is flying with the wind, its effective speed is (x+50) mph.

Similarly, when the plane is flying against the wind, its effective speed is (x-50) mph.

We know that time taken to travel 1200 miles with the wind + time taken to travel 1000 miles against the wind = 4 hours.

Using the formula time = distance/speed, we get:

1200/(x+50) + 1000/(x-50) = 4

Multiplying throughout by (x+50)(x-50), we get:

1200(x-50) + 1000(x+50) = 4(x+50)(x-50)

Expanding and simplifying, we get:

2200x = 22000

x = 550

Therefore, the speed of the plane in still air is 550 mph.

Quiz 11: Quadratic Equations and Functions

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?

Find the x-intercepts. - $f ( x ) = 64 x ^ { 2 } - 32 x + 4$

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. -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 rug is to fit in a room so that a border of even width is left on all four sides. If the room is 13 feet by 21 feet and the area of the rug is 153 square feet, how wide will the border be?

Find the x-intercepts. - $f ( x ) = 2 x ^ { 2 } - 3 x - 4$

Solve. -A lot is in the shape of a right triangle. The shorter leg measures 180 m. The hypotenuse is 60 m longer than the length of the longer leg. How long is the longer leg?

Solve. Round results to the nearest thousandth. - $5 \mathrm { x } ^ { 2 } = 10 + 14 \mathrm { x }$

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. Round results to the nearest thousandth. - $2 y ^ { 2 } + 6 y = - 4$

Solve. Round results to the nearest thousandth. - $x ^ { 2 } - 10 x - 36 = 0$

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. -A picture 13 inches by 23 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 252 square inches?

Solve. Round results to the nearest thousandth. - $3 x ^ { 2 } - 6 x = 2$

Solve. -A rectangular garden has dimensions of 23 feet by 20 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 294 square feet?

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. Round results to the nearest thousandth. - $x ^ { 2 } - 8 x = 6$

Solve. Round results to the nearest thousandth. - $9 x ^ { 2 } - 2 x - 8 = 0$

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?

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.