ASVAB Math Knowledge Practice Test 805417 Results

Your Results Global Average
Questions 5 5
Correct 0 3.47
Score 0% 69%

Review

1

On this circle, a line segment connecting point A to point D is called:

46% Answer Correctly

diameter

chord

circumference

radius


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).


2

A coordinate grid is composed of which of the following?

91% Answer Correctly

y-axis

all of these

x-axis

origin


Solution

The coordinate grid is composed of a horizontal x-axis and a vertical y-axis. The center of the grid, where the x-axis and y-axis meet, is called the origin.


3

Which of the following expressions contains exactly two terms?

83% Answer Correctly

binomial

polynomial

quadratic

monomial


Solution

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


4

If the area of this square is 9, what is the length of one of the diagonals?

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

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 = \( \sqrt{a} \) = \( \sqrt{9} \) = 3

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 = 32 + 32
c2 = 18
c = \( \sqrt{18} \) = \( \sqrt{9 x 2} \) = \( \sqrt{9} \) \( \sqrt{2} \)
c = 3\( \sqrt{2} \)


5

Solve for y:
y - 7 > -6 + 2y

55% Answer Correctly
y > 1\(\frac{1}{5}\)
y > 4
y > \(\frac{1}{7}\)
y > -1

Solution

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the > sign and the answer on the other.

y - 7 > -6 + 2y
y > -6 + 2y + 7
y - 2y > -6 + 7
-y > 1
y > \( \frac{1}{-1} \)
y > -1