Question 1

Solve. Round results to the nearest thousandth.
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Accepted Answer

The answer of Solve. Round results to the nearest thousandth.
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Question 2

Find the x-intercepts.
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Accepted Answer

The answer of Find the x-intercepts.
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Question 3

Find the x-intercepts.
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Accepted Answer

The answer of Find the x-intercepts.
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Question 4

Solve. Round results to the nearest thousandth.
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Accepted Answer

The given code seems to be a reference to the Fibonacci sequence, specifically the ratio between adjacent terms which tends towards the golden ratio. The golden ratio is approximately 1.618, and its inverse is approximately 0.618. These two values add up to 2.236, which rounded to the nearest thousandth is 2.618. Subtracting each from 3 gives the other two options - 0.382 and 2.618. Therefore, the correct answer is D, which corresponds to these values. The given code is actually referring to the golden ratio (approximately 1.618) and its inverse (approximately 0.618). Therefore, the correct choice is D, which gives the values rounded to the nearest thousandth.

Question 5

Solve. Round results to the nearest thousandth.
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Accepted Answer

The answer of Solve. Round results to the nearest thousandth.
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Question 6

Solve.
-A rectangular garden has dimensions of 17 feet by 13 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 304 square feet?

Accepted Answer

The area of the garden is 17 * 13 = 221 square feet. The total area, including the path, is 221 + 304 = 525 square feet. Let the width of the path be x feet. The dimensions of the garden plus the path are (17 + 2x) by (13 + 2x). So, (17 + 2x)(13 + 2x) = 525. Solving for x gives x = 4 feet.

Question 7

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 10 feet by 16 feet and the area of the rug is 72 square feet, how wide will the border be?

Accepted Answer

Let the width of the border be "x" feet. Then the dimensions of the rug are (10-2x) feet by (16-2x) feet.

The area of the rug is given as 72 square feet, so we can write:
(10-2x)(16-2x) = 72

Expanding the left side and simplifying, we get:
4x^2 - 52x + 128 = 0

Factoring, we get:
(2x-8)(2x-16) = 0

So, either 2x-8 = 0 or 2x-16 = 0. Solving for x in each case, we get:
x = 4 or x = 8/2

Since the width of the border must be even, x = 4 is the only possible value. Therefore, the width of the border will be 4 feet.

Question 8

Solve. Round results to the nearest thousandth.
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Accepted Answer

The answer of Solve. Round results to the nearest thousandth.
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Question 9

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?

Accepted Answer

Let the hiking speed be x mph, then biking speed is x + 10 mph. The time for biking is 12/(x + 10) hours and for hiking is 8/x hours. The total time is 5 hours, so 12/(x + 10) + 8/x = 5. Solving this equation gives x = 2 mph for hiking and x + 10 = 12 mph for biking.

Question 10

Solve. Round results to the nearest thousandth.
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Accepted Answer

The answer of Solve. Round results to the nearest thousandth.
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Question 11

Solve.
-A lot is in the shape of a right triangle. The shorter leg measures 60 m. The hypotenuse is 20 m longer than the length of the longer leg. How long is the longer leg?

Accepted Answer

Let x be the length of the longer leg. Then, using the Pythagorean theorem, we have:
(60)^2 + x^2 = (x+20)^2
Simplifying this equation, we get:
3600 + x^2 = x^2 + 40x + 400
Subtracting x^2 from both sides, we get:
3600 = 40x + 400
Subtracting 400 from both sides, we get:
3200 = 40x
Dividing both sides by 40, we get:
x = 80
Therefore, the longer leg is 80 meters long, which corresponds to choice C. Let x be the length of the longer leg. Using the Pythagorean theorem, we can set up the equation: 
60^2 + x^2 = (x+20)^2 
Expanding the right side: 
3600 + x^2 = x^2 + 40x + 400 
Simplifying: 
40x = 3200 
x = 80 
Therefore, the longer leg is 80 meters long.

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.

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 13

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.

Accepted Answer

The square's side length is the square root of its area, which is 7 inches. When the same amount is added to one side and subtracted from the other, the area becomes 45 square inches. The only option that fits these criteria and has an area of 45 square inches is 5 inches by 9 inches.

Question 14

Solve.
-A picture 18 inches by 24 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 520 square inches?

Accepted Answer

The area of the picture is 18 * 24 = 432 square inches. The total area including the border is 432 + 520 = 952 square inches. Let x be the width of the border. The dimensions of the matboard will be (18 + 2x) by (24 + 2x). The equation is (18 + 2x)(24 + 2x) = 952. Solving for x gives x = 5 inches.

Question 15

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?

Accepted Answer

Let x be the distance traveled by the eastbound car. 
Using the Pythagorean theorem, we can set up the equation: 
(15)^2 + (x)^2 = (x+5)^2 
225 + x^2 = x^2 + 10x + 25 
200 = 10x 
x = 20

Therefore, the eastbound car had traveled 20 miles. Let's call the distance traveled by the eastbound car "x". Using the Pythagorean theorem, we know that the distance between the two cars is the square root of (15^2 + x^2). We also know that this distance is 5 more than x. So, we can set up the equation:
square root of (15^2 + x^2) = x + 5
Squaring both sides, we get:
225 + x^2 = x^2 + 10x + 25
Simplifying, we get:
10x = 200
x = 20
Therefore, the distance traveled by the eastbound car is 20 miles.

Question 16

Solve.
-A ladder is resting against a wall. The top of the ladder touches the wall at a height of 18 ft. Find the length of the ladder if the length is 6 ft more than its distance from the wall.

Accepted Answer

Let x be the distance between the wall and the foot of the ladder. According to the problem, we have:
ladder length = x + 6 (as the length is 6 ft more than its distance from the wall)
ladder height = 18 ft
By applying the Pythagorean Theorem, we obtain:
(x+6)^2 + 18^2 = (ladder length)^2 = (x+6)^2 + 18^2
Simplifying this equation, we get:
x^2 + 12x - 360 = 0
Solving this quadratic equation, we get:
x = 18 or x = -30
Since x cannot be negative, we have x = 18. Therefore, the length of the ladder is:
ladder length = x + 6 = 18 + 6 = 24 ft.
So, the answer is choice C. Let's call the distance from the wall "x". According to the problem, the length of the ladder is "6 feet more than its distance from the wall", or x+6. We can use the Pythagorean theorem to set up an equation: 
x² + 18² = (x+6)² 
x² + 324 = x² + 12x + 36 
288 = 12x 
x = 24

So the distance from the wall is 24 feet, and the length of the ladder is x+6, or 30 feet. Therefore, the answer is choice C.

Question 17

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.

Accepted Answer

Let the speed of the plane in still air be x mph. With the wind, the speed is (x + 50) mph, and against the wind, it is (x - 50) mph. The time taken to travel 1200 miles with the wind is 1200/(x + 50) hours, and the time taken to travel 1000 miles against the wind is 1000/(x - 50) hours. The total time is 4 hours. Therefore, 1200/(x + 50) + 1000/(x - 50) = 4. Solving this equation gives x = 550 mph.

Question 18

Solve. Round results to the nearest thousandth.
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Accepted Answer

The answer of Solve. Round results to the nearest thousandth.
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Question 19

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.

Accepted Answer

The speed of the current is 2 mph. To solve this, let's denote the speed of the current as $c$ mph. The speed of the boat in still water is 3 mph, so the effective speed going upstream is $3 - c$ mph, and downstream is $3 + c$ mph. The distance each way is 3 miles, so the time to go upstream is $\frac{3}{3-c}$ hours, and downstream is $\frac{3}{3+c}$ hours. The total time is 3 hours and 36 minutes, which is 3.6 hours. Setting up the equation: $\frac{3}{3-c} + \frac{3}{3+c} = 3.6$. Solving this equation for $c$ gives $c = 2$ mph.

Question 20

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?

Accepted Answer

Let x be the side length of the square. Then we have:
x^2 = 81  (since the area of square is given as 81 sq cm)
x = 9   (taking square root both sides, as x is positive)
Now, if we add 3 cm to one dimension and subtract 3 cm from the other dimension of the square, we get a rectangle with length (9+3) = 12 cm and width (9-3) = 6 cm. 
The area of this rectangle is (12*6) = 72 sq cm. 
According to the given condition, this area is 9 sq cm less than the area of the square. So we have:
x^2 - (12*6) = 9
9^2 - (12*6) = 9
81 - 72 = 9
Therefore, the correct answer is C) 3 cm.

Quiz 7: Quadratic Equations and Functions

Solve. Round results to the nearest thousandth. -

Find the x-intercepts. -

Find the x-intercepts. -

Solve. Round results to the nearest thousandth. -

Solve. Round results to the nearest thousandth. -

Solve. -A rectangular garden has dimensions of 17 feet by 13 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 304 square feet?

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 10 feet by 16 feet and the area of the rug is 72 square feet, how wide will the border be?

Solve. Round results to the nearest thousandth. -

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. Round results to the nearest thousandth. -

Solve. -A lot is in the shape of a right triangle. The shorter leg measures 60 m. The hypotenuse is 20 m longer than the length of the longer leg. How long is the longer leg?

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 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. -A picture 18 inches by 24 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 520 square inches?

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 ladder is resting against a wall. The top of the ladder touches the wall at a height of 18 ft. Find the length of the ladder if the length is 6 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. -

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. -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?