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Complete the following questions

• 1.

### Use factoring to solve 3x2 – 13x – 10 = 0

• A.

X = 3/2 and x = 5

• B.

X = 2/3 and x = –5

• C.

X = –2/3 and x = –5

• D.

X = –2/3 and x = 5

D. X = –2/3 and x = 5
Explanation
To solve the quadratic equation 3x^2 - 13x - 10 = 0 using factoring, we need to find two numbers that multiply to give -30 (the product of the coefficient of x^2 and the constant term) and add up to -13 (the coefficient of x). The numbers that satisfy these conditions are -15 and 2. Therefore, we can rewrite the equation as (3x + 2)(x - 5) = 0. Setting each factor equal to zero, we get 3x + 2 = 0 and x - 5 = 0. Solving these equations gives x = -2/3 and x = 5 as the solutions.

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• 2.

### A skateboard company models its profit with the function P(x) = –2x2 + 13x – 15, where x of the number, in thousands that the company sells, and P(x) is the profit in tens of thousands of dollars.  How many skateboards must the company sell to break even? Use factoring to solve

• A.

At 1500 and 5000 skateboards

• B.

At 150 and 500 skateboards

• C.

At 1.5 and 5 skateboards

• D.

At 1500 and 50 000 skateboards

D. At 1500 and 50 000 skateboards
Explanation
The question asks for the number of skateboards the company must sell to break even. To find this, we need to set the profit function P(x) equal to zero and solve for x. By factoring the quadratic equation -2x^2 + 13x - 15 = 0, we can determine that the solutions are x = 1.5 and x = 5. However, since the company sells in thousands, we need to multiply these values by 1000. Therefore, the company must sell 1500 and 5000 skateboards to break even. The answer provided, "At 1500 and 50,000 skateboards," is incorrect as it includes an extra zero in the second value.

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• 3.

### Which table of values could be used to solve –2x2 – 5x + 3 = 0?

• A.

Table A

• B.

Table B

• C.

Table C

• D.

Table D

A. Table A
Explanation
Table A could be used to solve the equation –2x2 – 5x + 3 = 0 because it likely contains a set of x-values and their corresponding y-values that satisfy the equation when plugged into it. By substituting the x-values from Table A into the equation and calculating the corresponding y-values, one can determine if any of the y-values are equal to zero, indicating a solution to the equation.

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• 4.

### Kevin threw a ball into the air.  The height of the ball is modelled by the function h(t) = –5t2+35t, where h(t) is the height of the ball in metres and t is the time in seconds.  When does the ball hit the ground?

• A.

1s

• B.

3.5s

• C.

7s

• D.

61.25s

C. 7s
Explanation
The ball hits the ground when its height is equal to zero. To find the time when this occurs, we can set the height function h(t) equal to zero and solve for t. In this case, the equation -5t^2 + 35t = 0 represents the height of the ball. By factoring out a common factor of t, we get t(-5t + 35) = 0. This equation is satisfied when t = 0 or -5t + 35 = 0. Solving the second equation, we find t = 7. Therefore, the ball hits the ground at 7 seconds.

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• 5.

### Use the graph of the function below to solve the corresponding quadratic equation.

• A.

X = –16 and x = 0

• B.

X = 0.8 and x = –4

• C.

X = 0.8 and x = 0

• D.

X = 4 and x = 1.25

B. X = 0.8 and x = –4
Explanation
The correct answer is x = 0.8 and x = –4. This means that the quadratic equation represented by the graph has two solutions, one at x = 0.8 and the other at x = –4. These values of x correspond to the points where the graph intersects the x-axis.

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• 6.

• A.

2100

• B.

2021

• C.

2121

• D.

2210

B. 2021
• 7.

### A seagull drops a sea urchin onto the rocks below from a height of 80m.  The function h(t) = –5t2+80 models the height of the urchin, in metres, and t is the time in seconds.  When will the urchin hit the rocks?

• A.

–4s

• B.

4s

• C.

5s

• D.

80s

B. 4s
Explanation
The function h(t) = -5t^2 + 80 models the height of the urchin, where t is the time in seconds. To find when the urchin hits the rocks, we need to find the value of t when h(t) = 0. Setting -5t^2 + 80 = 0, we can solve for t. Rearranging the equation, we get 5t^2 = 80. Dividing both sides by 5, we have t^2 = 16. Taking the square root of both sides, we get t = ±4. Since time cannot be negative in this context, the urchin will hit the rocks after 4 seconds. Therefore, the correct answer is 4s.

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• 8.

### Which graph could you use to solve –2x2­­­ – 10x – 5 = 7

• A.

Graph A

• B.

Graph B

• C.

Graph C

• D.

Graph D

C. Graph C
Explanation
Graph C could be used to solve the equation –2x^2 – 10x – 5 = 7 because it appears to be a quadratic graph. The equation is in the form of a quadratic equation, and graph C shows a parabolic shape, which is characteristic of quadratic functions. By analyzing the graph, one can determine the x-intercepts or solutions of the equation.

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• 9.

### The area, in m2 of a rectangular deck is modelled by the function A(x) = 84x – 2x2.  What dimensions will give the maximum area for the deck?

• A.

21 x 21 m

• B.

21 x 42 m

• C.

21 x 63 m

• D.

21 x 882 m

B. 21 x 42 m
Explanation
The function A(x) represents the area of the rectangular deck, where x is the length of one side of the deck. The function A(x) = 84x - 2x^2 is a quadratic function with a negative coefficient for the x^2 term. This means that the graph of the function is an upside-down parabola. The maximum area of the deck will occur at the vertex of this parabola. To find the x-coordinate of the vertex, we can use the formula x = -b/2a, where a = -2 and b = 84. Plugging in these values, we get x = -84/(2*(-2)) = 21. Therefore, the dimensions that will give the maximum area for the deck are 21 x 42 m.

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• Current Version
• Aug 28, 2023
Quiz Edited by
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• Oct 13, 2009
Quiz Created by
Seixeiroda

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