ASVAB Math Knowledge Practice Test 162086 Results

Your Results Global Average
Questions 5 5
Correct 0 2.86
Score 0% 57%

Review

1

Which of the following expressions contains exactly two terms?

82% Answer Correctly

binomial

quadratic

monomial

polynomial


Solution

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


2

A cylinder with a radius (r) and a height (h) has a surface area of:

53% Answer Correctly

π r2h

4π r2

2(π r2) + 2π rh

π r2h2


Solution

A cylinder is a solid figure with straight parallel sides and a circular or oval cross section with a radius (r) and a height (h). The volume of a cylinder is π r2h and the surface area is 2(π r2) + 2π rh.


3

The endpoints of this line segment are at (-2, 7) and (2, -1). What is the slope of this line?

46% Answer Correctly
-2
2
-3
-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, 7) and (2, -1) so the slope becomes:

m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(-1.0) - (7.0)}{(2) - (-2)} \) = \( \frac{-8}{4} \)
m = -2


4

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

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

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\( \sqrt{2} \)


5

Solve -7c - 6c = -5c + 5y + 6 for c in terms of y.

34% Answer Correctly
2y - 2\(\frac{1}{3}\)
\(\frac{4}{13}\)y - \(\frac{9}{13}\)
-5\(\frac{1}{2}\)y - 3
y - 2\(\frac{1}{2}\)

Solution

To solve this equation, isolate the variable for which you are solving (c) on one side of the equation and put everything else on the other side.

-7c - 6y = -5c + 5y + 6
-7c = -5c + 5y + 6 + 6y
-7c + 5c = 5y + 6 + 6y
-2c = 11y + 6
c = \( \frac{11y + 6}{-2} \)
c = \( \frac{11y}{-2} \) + \( \frac{6}{-2} \)
c = -5\(\frac{1}{2}\)y - 3