| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.10 |
| Score | 0% | 62% |
A quadrilateral is a shape with __________ sides.
4 |
|
2 |
|
3 |
|
5 |
A quadrilateral is a shape with four sides. The perimeter of a quadrilateral is the sum of the lengths of its four sides.
Solve for x:
7x + 3 = \( \frac{x}{2} \)
| -1 | |
| 3\(\frac{3}{5}\) | |
| -\(\frac{24}{55}\) | |
| -\(\frac{6}{13}\) |
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 equal sign and the answer on the other.
7x + 3 = \( \frac{x}{2} \)
2 x (7x + 3) = x
(2 x 7x) + (2 x 3) = x
14x + 6 = x
14x + 6 - x = 0
14x - x = -6
13x = -6
x = \( \frac{-6}{13} \)
x = -\(\frac{6}{13}\)
Solve for y:
y2 + 5y - 1 = 5y + 3
| 3 or 1 | |
| 2 or -2 | |
| 2 or 2 | |
| 4 or 2 |
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:
y2 + 5y - 1 = 5y + 3
y2 + 5y - 1 - 3 = 5y
y2 + 5y - 5y - 4 = 0
y2 - 4 = 0
Next, factor the quadratic equation:
y2 - 4 = 0
(y - 2)(y + 2) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (y - 2) or (y + 2) must equal zero:
If (y - 2) = 0, y must equal 2
If (y + 2) = 0, y must equal -2
So the solution is that y = 2 or -2
On this circle, line segment CD is the:
chord |
|
circumference |
|
radius |
|
diameter |
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).
Breaking apart a quadratic expression into a pair of binomials is called:
squaring |
|
factoring |
|
normalizing |
|
deconstructing |
To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.