| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.63 |
| Score | 0% | 53% |
If a = c = 3, b = d = 8, and the blue angle = 53°, what is the area of this parallelogram?
| 24 | |
| 25 | |
| 42 | |
| 3 |
The area of a parallelogram is equal to its length x width:
a = l x w
a = a x b
a = 3 x 8
a = 24
Which of the following is not true about both rectangles and squares?
all interior angles are right angles |
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the lengths of all sides are equal |
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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).
Which of the following is not required to define the slope-intercept equation for a line?
\({\Delta y \over \Delta x}\) |
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slope |
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y-intercept |
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x-intercept |
A line on the coordinate grid can be defined by a slope-intercept equation: y = mx + b. For a given value of x, the value of y can be determined given the slope (m) and y-intercept (b) of the line. The slope of a line is change in y over change in x, \({\Delta y \over \Delta x}\), and the y-intercept is the y-coordinate where the line crosses the vertical y-axis.
When two lines intersect, adjacent angles are __________ (they add up to 180°) and angles across from either other are __________ (they're equal).
supplementary, vertical |
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obtuse, acute |
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vertical, supplementary |
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acute, obtuse |
Angles around a line add up to 180°. Angles around a point add up to 360°. When two lines intersect, adjacent angles are supplementary (they add up to 180°) and angles across from either other are vertical (they're equal).
Solve 6b + 4b = 9b + 6z - 6 for b in terms of z.
| -\(\frac{1}{10}\)z + \(\frac{1}{5}\) | |
| -\(\frac{1}{4}\)z - \(\frac{1}{12}\) | |
| -\(\frac{2}{3}\)z + 2 | |
| -\(\frac{2}{5}\)z - 1\(\frac{4}{5}\) |
To solve this equation, isolate the variable for which you are solving (b) on one side of the equation and put everything else on the other side.
6b + 4z = 9b + 6z - 6
6b = 9b + 6z - 6 - 4z
6b - 9b = 6z - 6 - 4z
-3b = 2z - 6
b = \( \frac{2z - 6}{-3} \)
b = \( \frac{2z}{-3} \) + \( \frac{-6}{-3} \)
b = -\(\frac{2}{3}\)z + 2