# Math 1 Unit 2 Practice Test

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| By Klrhinehart3843
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This is a practice test for the Unit 2 Assessment in Mrs. Rhinehart, Mrs. Rucker and Mrs. Gee's Yearlong Math 1 classes.

• 1.

### MM1A2a.  Simplify sqrt(27)

• A.

3

• B.

3sqrt(3)

• C.

9sqrt(3)

• D.

9

B. 3sqrt(3)
Explanation
The square root of 27 can be simplified as 3 multiplied by the square root of 3. This is because 27 can be factored into 3 multiplied by 3 multiplied by 3, which can be written as 3^3. Taking the square root of 3^3 gives us 3 multiplied by the square root of 3.

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• 2.

### MM1A2a.  Simplify -2sqrt(12m2n)

• A.

-8msqrt(3n)

• B.

4sqrt(3m^2n)

• C.

-4msqrt(3n)

• D.

8msqrt(3n)

C. -4msqrt(3n)
Explanation
To simplify -2sqrt(12m^2n), we can break down the square root of 12 into its factors. The square root of 12 can be simplified to 2sqrt(3). We can then bring the -2 outside of the square root. So, the simplified expression becomes -2 * 2sqrt(3m^2n), which simplifies to -4msqrt(3n).

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• 3.

### MM1A2b.  Add 5sqrt(8) + 12sqrt(2)

• A.

22sqrt(2)

• B.

17sqrt(10)

• C.

19sqrt(2)

• D.

-7sqrt(6)

A. 22sqrt(2)
Explanation
To solve the given expression, we can simplify each term separately and then add them together. The first term, 5sqrt(8), can be simplified by factoring out the square root of 4 from 8, resulting in 5 * 2sqrt(2), which simplifies to 10sqrt(2). The second term, 12sqrt(2), remains the same. Now, we can add the simplified terms together: 10sqrt(2) + 12sqrt(2) = 22sqrt(2). Therefore, the correct answer is 22sqrt(2).

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• 4.

### MM1A2b.  Divide [-14sqrt(12)]/[7sqrt(6)]

• A.

-2sqrt(2)

• B.

-21sqrt(6)

• C.

2sqrt(2)

• D.

21sqrt(6)

A. -2sqrt(2)
Explanation
To divide the given expression, we can simplify the numerator and denominator separately. The square root of 12 can be simplified as 2sqrt(3), and the square root of 6 can be simplified as sqrt(6). Dividing -14sqrt(12) by 7sqrt(6) gives us (-14/7)(2sqrt(3)/sqrt(6)), which simplifies to -2sqrt(2). Therefore, the correct answer is -2sqrt(2).

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• 5.

### MM1A2c.  Add (7x3 + 6x2 - 1) + (-3x2 - 9x + 5)

• A.

4x^5 - 3x^3 + 4

• B.

4x^5 - 3x^2 + 4

• C.

7x^3 + 3x^2 - 9x + 4

• D.

7x^3 + 3x^2 + 9x - 4

C. 7x^3 + 3x^2 - 9x + 4
Explanation
The given question asks to add the two polynomials (7x^3 + 6x^2 - 1) and (-3x^2 - 9x + 5). To add polynomials, we combine like terms. In this case, the like terms are the terms with the same exponent. Adding the coefficients of the like terms, we get 7x^3 + 3x^2 - 9x + 4. Therefore, the correct answer is 7x^3 + 3x^2 - 9x + 4.

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• 6.

### MM1A2c.  Multiply (5x + 2)(2x - 3)

• A.

21x - 6

• B.

10x^2 - 11x - 6

• C.

21x + 6

• D.

10x^2 + 19x - 6

B. 10x^2 - 11x - 6
Explanation
The given expression is a multiplication of two binomials. To find the product, we can use the distributive property. Multiply each term of the first binomial (5x + 2) by each term of the second binomial (2x - 3). Simplifying the expression, we get 10x^2 - 11x - 6. Therefore, the correct answer is 10x^2 - 11x - 6.

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• 7.

### MM1A2d.  Which of the following tools are used to expand binomial expressions using the Binomial Theorem?

• A.

Algebra Tiles

• B.

Parent Functions

• C.

Pascal's Triangle

• D.

None of the Above

C. Pascal's Triangle
Explanation
Pascal's Triangle is used to expand binomial expressions using the Binomial Theorem. It is a triangular array of numbers where each number is the sum of the two numbers directly above it. The coefficients of the expanded binomial expression can be found by looking at the corresponding row in Pascal's Triangle. Algebra Tiles and Parent Functions are not used for expanding binomial expressions using the Binomial Theorem.

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• 8.

### MM1A2d.  Expand (x + 7)3

• A.

X^3 + 2x^2 + x

• B.

X^2 + 7x + 49

• C.

7x^2 + 49x + 343

• D.

X^3 + 21x^2 + 147x + 343

D. X^3 + 21x^2 + 147x + 343
Explanation
The given expression is (x + 7)3. To expand this expression, we use the binomial expansion formula. The formula states that (a + b)n = an + (nC1)(a^(n-1))(b^1) + (nC2)(a^(n-2))(b^2) + ... + (nCn-1)(a^1)(b^(n-1)) + bn. In this case, a = x, b = 7, and n = 3. Plugging in these values into the formula, we get x^3 + 3(x^2)(7) + 3(x)(7^2) + 7^3, which simplifies to x^3 + 21x^2 + 147x + 343. Therefore, the correct answer is x^3 + 21x^2 + 147x + 343.

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• 9.

### MM1A2e.  Subtract [(9x + 3)/2]-[(3x-5)/2]

• A.

6x - 1

• B.

1.5x + 2

• C.

3x + 4

• D.

6x + 8

C. 3x + 4
Explanation
The given expression involves subtracting two fractions with the same denominator. To subtract fractions, we need to have a common denominator. In this case, the common denominator is 2.

Next, we can subtract the numerators of the fractions. The numerator of the first fraction is (9x + 3) and the numerator of the second fraction is (3x - 5).

Subtracting the numerators, we get (9x + 3) - (3x - 5) = 9x + 3 - 3x + 5 = 6x + 8.

Finally, we have the result of the subtraction as (6x + 8). Therefore, the correct answer is 6x + 8.

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• 10.

### MM1A2e.  Divide [(x2 + 3x +2)/3x+3] and [(x+2)/6x]

• A.

2x

• B.

2(x + 3)/(x + 2)

• C.

6x

A. 2x
Explanation
To divide two fractions, we need to multiply the first fraction by the reciprocal of the second fraction. In this case, the reciprocal of (x+2)/(6x) is 6x/(x+2). So, we multiply (x^2 + 3x + 2)/(3x + 3) by 6x/(x+2). Simplifying the expression, we get 2x. Therefore, the answer is 2x.

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• 11.

### MM1A2f.  Factor x2 - 10x - 24

• A.

(x - 8)(x + 3)

• B.

(x - 8)(x - 3)

• C.

(x - 12)(x - 2)

• D.

(x - 12)(x + 2)

D. (x - 12)(x + 2)
Explanation
The correct answer is (x - 12)(x + 2). This is the correct factorization of the given expression x^2 - 10x - 24. By using the FOIL method, we can expand (x - 12)(x + 2) to get x^2 - 10x - 24, which matches the given expression. Therefore, (x - 12)(x + 2) is the correct answer.

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• 12.

### MM1A2f.  Factor 4x2 - 81

• A.

(2x - 81)(2x + 81)

• B.

(2x - 9)(2x + 9)

• C.

(4x - 9)(4x + 9)

• D.

(2x - 9)(2x - 9)

B. (2x - 9)(2x + 9)
Explanation
The given expression is a difference of squares, which can be factored using the formula a^2 - b^2 = (a + b)(a - b). In this case, a = 2x and b = 9. Plugging in these values, we get (2x)^2 - 9^2 = (2x + 9)(2x - 9). Therefore, the correct answer is (2x - 9)(2x + 9).

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• 13.

### MM1A2g.  Which expression represents the length and width of the image below?

• A.

(x^2 + 3x) & (x^2 + 2x)

• B.

(x + 6) & (x + 1)

• C.

(x^2 + 3) & (x^2 + 2)

• D.

(x + 3) & (x + 2)

D. (x + 3) & (x + 2)
• 14.

### MM1A2g.  A picture frame has an area of 48in2.  A 4 x 6 picture fits in the frame.  The frame is equal width on all sides.  What is the width of the frame?

• A.

1 inch

• B.

2 inches

• C.

3 inches

• D.

None of the above

A. 1 inch
Explanation
The area of the picture frame is 48in2, and a 4 x 6 picture fits in the frame. This means that the dimensions of the frame are larger than the dimensions of the picture by the same amount on all sides. Since the picture is 4 x 6, the dimensions of the frame must be 4 + 2x and 6 + 2x, where x represents the width of the frame. The area of the frame can be calculated as (4 + 2x)(6 + 2x). Since the area of the frame is 48in2, we can set up the equation (4 + 2x)(6 + 2x) = 48 and solve for x. The only value of x that satisfies this equation is 1, which means the width of the frame is 1 inch.

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• 15.

### MM1A3a.  Solve the quadratic equation:  x2 + 8x = -7

• A.

X = 1 & x = 7

• B.

X = -1 & x = -7

• C.

X = 2 & x = 6

• D.

X = -2 & x = -6

B. X = -1 & x = -7
Explanation
The correct answer is x = -1 & x = -7. To solve the quadratic equation x^2 + 8x = -7, we can rearrange the equation to x^2 + 8x + 7 = 0. To factorize this equation, we need to find two numbers that multiply to 7 and add up to 8. The numbers are 1 and 7. Therefore, the factored form of the equation is (x + 1)(x + 7) = 0. Setting each factor equal to zero, we get x + 1 = 0 and x + 7 = 0. Solving these equations, we find x = -1 and x = -7.

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• 16.

### MM1A3a.  What is/are the solution(s) to the function below?

• A.

(0,3)

• B.

(-1, 0)

• C.

(-1, 0) & (-3, 0)

• D.

(1, 0) & (3, 0)

C. (-1, 0) & (-3, 0)
Explanation
The given answer, (-1, 0) & (-3, 0), represents the solutions to the function. This means that when the function is evaluated at x = -1 and x = -3, the output is 0. Therefore, these are the x-values that make the function equal to 0.

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• Current Version
• Mar 21, 2023
Quiz Edited by
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• Sep 20, 2009
Quiz Created by
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