| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.26 |
| Score | 0% | 65% |
Which of the following is not a part of PEMDAS, the acronym for math order of operations?
exponents |
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pairs |
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division |
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addition |
When solving an equation with two variables, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)
Order the following types of angle from least number of degrees to most number of degrees.
acute, obtuse, right |
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right, obtuse, acute |
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right, acute, obtuse |
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acute, right, obtuse |
An acute angle measures less than 90°, a right angle measures 90°, and an obtuse angle measures more than 90°.
The endpoints of this line segment are at (-2, -3) and (2, -1). What is the slope of this line?
| -3 | |
| \(\frac{1}{2}\) | |
| 3 | |
| -\(\frac{1}{2}\) |
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, -3) and (2, -1) so the slope becomes:
m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(-1.0) - (-3.0)}{(2) - (-2)} \) = \( \frac{2}{4} \)Which of the following is not true about both rectangles and squares?
the lengths of all sides are equal |
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all interior angles are right angles |
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the area is length x width |
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the perimeter is the sum of the lengths of all four sides |
A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2).
Factor y2 - 11y + 28
| (y - 7)(y + 4) | |
| (y - 7)(y - 4) | |
| (y + 7)(y + 4) | |
| (y + 7)(y - 4) |
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 28 as well and sum (Inside, Outside) to equal -11. For this problem, those two numbers are -7 and -4. Then, plug these into a set of binomials using the square root of the First variable (y2):
y2 - 11y + 28
y2 + (-7 - 4)y + (-7 x -4)
(y - 7)(y - 4)