| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.10 |
| Score | 0% | 62% |
If AD = 18 and BD = 16, AB = ?
| 2 | |
| 14 | |
| 8 | |
| 6 |
The entire length of this line is represented by AD which is AB + BD:
AD = AB + BD
Solving for AB:AB = AD - BDSimplify 5a x 4b.
| 20\( \frac{a}{b} \) | |
| 9ab | |
| 20ab | |
| 20a2b2 |
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.
5a x 4b = (5 x 4) (a x b) = 20ab
Simplify (8a)(2ab) - (9a2)(2b).
| 34a2b | |
| 110a2b | |
| 110ab2 | |
| -2a2b |
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.
(8a)(2ab) - (9a2)(2b)
(8 x 2)(a x a x b) - (9 x 2)(a2 x b)
(16)(a1+1 x b) - (18)(a2b)
16a2b - 18a2b
-2a2b
The endpoints of this line segment are at (-2, 4) and (2, -2). What is the slope-intercept equation for this line?
| y = -2x + 0 | |
| y = x - 1 | |
| y = 1\(\frac{1}{2}\)x - 4 | |
| y = -1\(\frac{1}{2}\)x + 1 |
The slope-intercept equation for a line is y = mx + b where m is the slope and b is the y-intercept of the line. From the graph, you can see that the y-intercept (the y-value from the point where the line crosses the y-axis) is 1. 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, 4) and (2, -2) so the slope becomes:
m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(-2.0) - (4.0)}{(2) - (-2)} \) = \( \frac{-6}{4} \)Plugging these values into the slope-intercept equation:
y = -1\(\frac{1}{2}\)x + 1
If the length of AB equals the length of BD, point B __________ this line segment.
bisects |
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intersects |
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trisects |
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midpoints |
A line segment is a portion of a line with a measurable length. The midpoint of a line segment is the point exactly halfway between the endpoints. The midpoint bisects (cuts in half) the line segment.