ASVAB Math Knowledge Practice Test 899082 Results

Your Results Global Average
Questions 5 5
Correct 0 3.21
Score 0% 64%

Review

1

A trapezoid is a quadrilateral with one set of __________ sides.

70% Answer Correctly

right angle

equal angle

parallel

equal length


Solution

A trapezoid is a quadrilateral with one set of parallel sides.


2

Which of the following expressions contains exactly two terms?

82% Answer Correctly

quadratic

polynomial

binomial

monomial


Solution

A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms.


3

On this circle, line segment CD is the:

46% Answer Correctly

diameter

chord

radius

circumference


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).


4

If the area of this square is 36, what is the length of one of the diagonals?

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

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 = \( \sqrt{a} \) = \( \sqrt{36} \) = 6

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 = 62 + 62
c2 = 72
c = \( \sqrt{72} \) = \( \sqrt{36 x 2} \) = \( \sqrt{36} \) \( \sqrt{2} \)
c = 6\( \sqrt{2} \)


5

Factor y2 - 4

54% Answer Correctly
(y - 2)(y - 2)
(y + 2)(y + 2)
(y - 2)(y + 2)
(y + 2)(y - 2)

Solution

To factor a quadratic expression, apply the FOIL method (First, Outside, Inside, Last) in reverse. First, find the two Last terms that will multiply to produce -4 as well and sum (Inside, Outside) to equal 0. For this problem, those two numbers are -2 and 2. Then, plug these into a set of binomials using the square root of the First variable (y2):

y2 - 4
y2 + (-2 + 2)y + (-2 x 2)
(y - 2)(y + 2)