| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.85 |
| Score | 0% | 57% |
Which of the following is not required to define the slope-intercept equation for a line?
y-intercept |
|
slope |
|
\({\Delta y \over \Delta x}\) |
|
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.
Solve 3b + 6b = 5b - 5z - 2 for b in terms of z.
| 5\(\frac{1}{2}\)z + 1 | |
| -3z - 4 | |
| 1\(\frac{2}{5}\)z + \(\frac{1}{5}\) | |
| \(\frac{1}{2}\)z + \(\frac{3}{4}\) |
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.
3b + 6z = 5b - 5z - 2
3b = 5b - 5z - 2 - 6z
3b - 5b = -5z - 2 - 6z
-2b = -11z - 2
b = \( \frac{-11z - 2}{-2} \)
b = \( \frac{-11z}{-2} \) + \( \frac{-2}{-2} \)
b = 5\(\frac{1}{2}\)z + 1
If a = c = 3, b = d = 4, what is the area of this rectangle?
| 12 | |
| 15 | |
| 72 | |
| 36 |
The area of a rectangle is equal to its length x width:
a = l x w
a = a x b
a = 3 x 4
a = 12
If BD = 11 and AD = 17, AB = ?
| 2 | |
| 3 | |
| 14 | |
| 6 |
The entire length of this line is represented by AD which is AB + BD:
AD = AB + BD
Solving for AB:AB = AD - BDFactor y2 - 3y + 2
| (y + 2)(y + 1) | |
| (y - 2)(y + 1) | |
| (y + 2)(y - 1) | |
| (y - 2)(y - 1) |
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 2 as well and sum (Inside, Outside) to equal -3. For this problem, those two numbers are -2 and -1. Then, plug these into a set of binomials using the square root of the First variable (y2):
y2 - 3y + 2
y2 + (-2 - 1)y + (-2 x -1)
(y - 2)(y - 1)