ASVAB Math Knowledge Practice Test 759801 Results

Your Results Global Average
Questions 5 5
Correct 0 2.91
Score 0% 58%

Review

1

Factor y2 - 2y - 15

54% Answer Correctly
(y - 5)(y - 3)
(y + 5)(y - 3)
(y + 5)(y + 3)
(y - 5)(y + 3)

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 -15 as well and sum (Inside, Outside) to equal -2. For this problem, those two numbers are -5 and 3. Then, plug these into a set of binomials using the square root of the First variable (y2):

y2 - 2y - 15
y2 + (-5 + 3)y + (-5 x 3)
(y - 5)(y + 3)


2

The endpoints of this line segment are at (-2, 7) and (2, -3). What is the slope of this line?

46% Answer Correctly
1
-2\(\frac{1}{2}\)
2\(\frac{1}{2}\)
-3

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, -3) so the slope becomes:

m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(-3.0) - (7.0)}{(2) - (-2)} \) = \( \frac{-10}{4} \)
m = -2\(\frac{1}{2}\)


3

If a = c = 1, b = d = 2, what is the area of this rectangle?

80% Answer Correctly
40
2
8
72

Solution

The area of a rectangle is equal to its length x width:

a = l x w
a = a x b
a = 1 x 2
a = 2


4

For this diagram, the Pythagorean theorem states that b2 = ?

47% Answer Correctly

c2 + a2

c - a

c2 - a2

a2 - c2


Solution

The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, \(c = \sqrt{a + b}\)


5

Simplify (7a)(2ab) + (2a2)(7b).

65% Answer Correctly
28a2b
b2
2b
81a2b

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.

(7a)(2ab) + (2a2)(7b)
(7 x 2)(a x a x b) + (2 x 7)(a2 x b)
(14)(a1+1 x b) + (14)(a2b)
14a2b + 14a2b
28a2b