PARCC Algebra Practice Test 2 — Questions and Answers

Question 1: 1. Simplify the expression (4x + 22x) / (2x)

6</li>
2+2x</li>
2x2x</li>
2x+1</li> (Correct answer)

Correct answer: 2x+1</li>

First, recognize that 2^(2x) can be rewritten as (2^2)^x or 4^x. So the expression becomes (4^x + 4^x) / (2^x), which simplifies to 2 * 4^x / 2^x. Since 4^x is (2^2)^x or 2^(2x), the expression is 2 * 2^(2x) / 2^x. Using exponent rules, this simplifies to 2^(1 + 2x - x), resulting in 2^(x+1).

Question 2: 2. Simplify the expression (2x2-5x-12)/(2x2-4x-16).

(x-6)/2(x-2) </li>
(x-6)/2(x+2)</li>
(2x+3)/2(x-2) </li>
(2x+3)/2(x+2) </li> (Correct answer)

Correct answer: (2x+3)/2(x+2) </li>

To simplify the expression, factor both the numerator and the denominator. The numerator, 2x^2 - 5x - 12, factors into (2x+3)(x-4). The denominator, 2x^2 - 4x - 16, factors into 2(x^2 - 2x - 8), which further factors into 2(x-4)(x+2). Canceling the common factor (x-4) from both the numerator and denominator leaves the simplified expression (2x+3) / [2(x+2)].

Question 3: 3. Suppose that the function f(x) is a quadratic function with roots at x=2-3i and x=2+3i. Find f(x).

f(x)=x2-4x-5</li>
f(x)=x2-4x+13</li> (Correct answer)
f(x)=x2-6ix-5</li>
f(x)=x2-6ix+13</li>

Correct answer: f(x)=x2-4x+13</li>

If the roots are x=2-3i and x=2+3i, the factors of the quadratic function are (x - (2-3i)) and (x - (2+3i)). Multiplying these factors gives (x - 2 + 3i)(x - 2 - 3i). This is a difference of squares, ( (x-2) + 3i ) ( (x-2) - 3i ), which simplifies to (x-2)^2 - (3i)^2. Expanding yields x^2 - 4x + 4 - 9i^2, and since i^2 = -1, the function is x^2 - 4x + 4 + 9, or f(x) = x^2 - 4x + 13.

Question 4: 4. Solve the inequality for x. Select all that apply. <ol list type="a"> 4x3+10x2-24x<0

x<-4</li> (Correct answer)
-4<x<0</li>
0<x<3/2</li> (Correct answer)
x>3/2</li>

Correct answer: x<-4</li>

First, factor the polynomial 4x^3 + 10x^2 - 24x by taking out the common factor 2x, resulting in 2x(2x^2 + 5x - 12). Then, factor the quadratic expression to get 2x(2x-3)(x+4). The roots are x=0, x=3/2, and x=-4. By testing intervals around these roots, the expression 4x^3 + 10x^2 - 24x is less than 0 when x < -4 and when 0 < x < 3/2.

Question 5: 5. A baseball is thrown up in the air from an initial height of 6 feet. Its height above the ground (in feet) t seconds after being thrown is given by the function h(t)=-16t^2+46t+6. How long will it take (in seconds) for the baseball to hit the ground?

2 seconds</li>
5/2 seconds</li>
3 seconds</li>
4 seconds</li> (Correct answer)

Correct answer: 4 seconds</li>

To find when the baseball hits the ground, set the height function h(t) to 0: -16t^2 + 46t + 6 = 0. Dividing by -2 simplifies the equation to 8t^2 - 23t - 3 = 0. Factoring this quadratic equation yields (8t+1)(t-3) = 0, which gives solutions t = -1/8 and t = 3. Since time cannot be negative, the baseball hits the ground after 3 seconds. (Note: There appears to be a discrepancy between the calculated answer of 3 seconds and the provided correct answer of 4 seconds for this specific function.)

Question 6: 6. Solve the equation for x. Select all that apply. <ol list type="a"> log2(8x-x2 )=4

x=-8</li>
x=0</li>
x=4</li> (Correct answer)
x=8</li>

Correct answer: x=4</li>

To solve the logarithmic equation log₂(8x - x²) = 4, convert it to its exponential form: 8x - x² = 2^4. This simplifies to 8x - x² = 16. Rearranging the terms into a standard quadratic equation gives x² - 8x + 16 = 0. This equation is a perfect square trinomial, (x-4)² = 0, which yields the solution x = 4. This solution is valid as 8(4) - 4² = 16, which is greater than 0.

Question 7: 7. Calculate the average rate of change of f between x=1 and x=4. <ol list type="a"> f(x)=x3+3x+1

6</li>
20/3</li>
24</li> (Correct answer)
72</li>

Correct answer: 24</li>

The average rate of change of a function f(x) between x=a and x=b is calculated using the formula [f(b) - f(a)] / (b - a). First, evaluate f(1) = 1^3 + 3(1) + 1 = 5. Next, evaluate f(4) = 4^3 + 3(4) + 1 = 64 + 12 + 1 = 77. Substituting these values into the formula gives (77 - 5) / (4 - 1) = 72 / 3, which equals 24.

Question 8: 8. Simplify the expression (x3-3x2+2x-6)/(x2-9).

1</li>
(x-3)/(x+3)</li>
(x2+2)/(x-3)</li>
(x2+2)/(x+3)</li> (Correct answer)

Correct answer: (x2+2)/(x+3)</li>

To simplify the expression, factor both the numerator and the denominator. The numerator, x^3 - 3x^2 + 2x - 6, can be factored by grouping as x^2(x-3) + 2(x-3), which becomes (x^2+2)(x-3). The denominator, x^2 - 9, is a difference of squares and factors into (x-3)(x+3). Canceling the common factor (x-3) from both the numerator and denominator leaves the simplified expression (x^2+2) / (x+3).

Question 9: 9. Suppose that angle θ is in Quadrant I and cos θ = 12/13. Find tan θ.

tan θ = 1/13</li>
tan θ = 13</li>
tan θ = 5/12</li> (Correct answer)
tan θ = 12/5</li>

Correct answer: tan θ = 5/12</li>

Given cos θ = 12/13 and that angle θ is in Quadrant I, we can use the Pythagorean identity sin²θ + cos²θ = 1. Substituting the value of cos θ, we get sin²θ + (12/13)² = 1, which means sin²θ = 1 - 144/169 = 25/169. Since θ is in Quadrant I, sin θ is positive, so sin θ = 5/13. Finally, tan θ = sin θ / cos θ = (5/13) / (12/13) = 5/12.

Question 10: 10. Which expression is equivalent to 6√x+10x?

2(3x-1+5x)</li>
2(3x1/2+5x)</li> (Correct answer)
2x(3x-1+5)</li>
2x(3x1/2+5)</li>

Correct answer: 2(3x1/2+5x)</li>

The expression 6√x + 10x can be rewritten using fractional exponents as 6x^(1/2) + 10x. To find an equivalent expression, look for common factors. Both terms share a factor of 2. Factoring out 2 from both terms yields 2(3x^(1/2) + 5x). This matches option B, as x^(1/2) is equivalent to √x.

PARCC Algebra Practice Test 2

1.
Simplify the expression (4x + 22x) / (2x)