| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.61 |
| Score | 0% | 52% |
The endpoints of this line segment are at (-2, 4) and (2, -2). What is the slope of this line?
| \(\frac{1}{2}\) | |
| 2 | |
| 3 | |
| -1\(\frac{1}{2}\) |
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, 4) and (2, -2) so the slope becomes:
m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(-2.0) - (4.0)}{(2) - (-2)} \) = \( \frac{-6}{4} \)On this circle, line segment CD is the:
diameter |
|
chord |
|
radius |
|
circumference |
A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter. A chord is a line segment that connects any two points along its perimeter. The diameter of a circle is the length of a chord that passes through the center of the circle and equals twice the circle's radius (2r).
Solve for z:
z2 + 7z - 2 = 2z + 4
| 1 or -6 | |
| 5 or -4 | |
| -5 or -5 | |
| 5 or 5 |
The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:
z2 + 7z - 2 = 2z + 4
z2 + 7z - 2 - 4 = 2z
z2 + 7z - 2z - 6 = 0
z2 + 5z - 6 = 0
Next, factor the quadratic equation:
z2 + 5z - 6 = 0
(z - 1)(z + 6) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (z - 1) or (z + 6) must equal zero:
If (z - 1) = 0, z must equal 1
If (z + 6) = 0, z must equal -6
So the solution is that z = 1 or -6
Simplify (5a)(9ab) + (6a2)(3b).
| 63a2b | |
| 126a2b | |
| 27ab2 | |
| 126ab2 |
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)(9ab) + (6a2)(3b)
(5 x 9)(a x a x b) + (6 x 3)(a2 x b)
(45)(a1+1 x b) + (18)(a2b)
45a2b + 18a2b
63a2b
Solve for c:
8c - 9 < 5 + 9c
| c < -14 | |
| c < -3\(\frac{1}{2}\) | |
| c < \(\frac{1}{9}\) | |
| c < 1 |
To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the < sign and the answer on the other.
8c - 9 < 5 + 9c
8c < 5 + 9c + 9
8c - 9c < 5 + 9
-c < 14
c < \( \frac{14}{-1} \)
c < -14