ASVAB Math Knowledge Practice Test 208667 Results

Your Results Global Average
Questions 5 5
Correct 0 3.35
Score 0% 67%

Review

1

If side a = 8, side b = 8, what is the length of the hypotenuse of this right triangle?

64% Answer Correctly
\( \sqrt{29} \)
\( \sqrt{128} \)
\( \sqrt{98} \)
\( \sqrt{117} \)

Solution

According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:

c2 = a2 + b2
c2 = 82 + 82
c2 = 64 + 64
c2 = 128
c = \( \sqrt{128} \)


2

On this circle, line segment AB is the:

72% Answer Correctly

circumference

radius

chord

diameter


Solution

A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter. A chord is a line segment that connects any two points along its perimeter. The diameter of a circle is the length of a chord that passes through the center of the circle and equals twice the circle's radius (2r).


3

The endpoints of this line segment are at (-2, -5) and (2, 3). What is the slope of this line?

46% Answer Correctly
-1
2
-\(\frac{1}{2}\)
-2\(\frac{1}{2}\)

Solution

The slope of this line is the change in y divided by the change in x. The endpoints of this line segment are at (-2, -5) and (2, 3) so the slope becomes:

m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(3.0) - (-5.0)}{(2) - (-2)} \) = \( \frac{8}{4} \)
m = 2


4

If a = 3, b = 1, c = 9, and d = 4, what is the perimeter of this quadrilateral?

88% Answer Correctly
24
17
31
20

Solution

Perimeter is equal to the sum of the four sides:

p = a + b + c + d
p = 3 + 1 + 9 + 4
p = 17


5

Simplify (5a)(6ab) + (9a2)(7b).

66% Answer Correctly
93a2b
-33ab2
33a2b
176a2b

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.

(5a)(6ab) + (9a2)(7b)
(5 x 6)(a x a x b) + (9 x 7)(a2 x b)
(30)(a1+1 x b) + (63)(a2b)
30a2b + 63a2b
93a2b