ASVAB Math Knowledge Practice Test 898122 Results

Your Results Global Average
Questions 5 5
Correct 0 3.31
Score 0% 66%

Review

1

Breaking apart a quadratic expression into a pair of binomials is called:

74% Answer Correctly

normalizing

squaring

factoring

deconstructing


Solution

To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.


2

The dimensions of this cube are height (h) = 7, length (l) = 8, and width (w) = 5. What is the volume?

83% Answer Correctly
40
150
504
280

Solution

The volume of a cube is height x length x width:

v = h x l x w
v = 7 x 8 x 5
v = 280


3

Simplify (9a)(4ab) + (7a2)(9b).

65% Answer Correctly
99a2b
99ab2
208ab2
208a2b

Solution

To multiply monomials, multiply the coefficients (the numbers that come before the variables) of each term, add the exponents of like variables, and multiply the different variables together.

(9a)(4ab) + (7a2)(9b)
(9 x 4)(a x a x b) + (7 x 9)(a2 x b)
(36)(a1+1 x b) + (63)(a2b)
36a2b + 63a2b
99a2b


4

Which of the following is not required to define the slope-intercept equation for a line?

42% Answer Correctly

y-intercept

slope

\({\Delta y \over \Delta x}\)

x-intercept


Solution

A line on the coordinate grid can be defined by a slope-intercept equation: y = mx + b. For a given value of x, the value of y can be determined given the slope (m) and y-intercept (b) of the line. The slope of a line is change in y over change in x, \({\Delta y \over \Delta x}\), and the y-intercept is the y-coordinate where the line crosses the vertical y-axis.


5

If the area of this square is 4, what is the length of one of the diagonals?

68% Answer Correctly
5\( \sqrt{2} \)
2\( \sqrt{2} \)
6\( \sqrt{2} \)
7\( \sqrt{2} \)

Solution

To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:

a = s2

so the length of one side of the square is:

s = \( \sqrt{a} \) = \( \sqrt{4} \) = 2

The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:

c2 = a2 + b2
c2 = 22 + 22
c2 = 8
c = \( \sqrt{8} \) = \( \sqrt{4 x 2} \) = \( \sqrt{4} \) \( \sqrt{2} \)
c = 2\( \sqrt{2} \)