| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.08 |
| Score | 0% | 62% |
Solve for y:
-4y + 2 = \( \frac{y}{6} \)
| \(\frac{12}{25}\) | |
| -1\(\frac{3}{7}\) | |
| -\(\frac{9}{82}\) | |
| -1\(\frac{1}{6}\) |
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.
-4y + 2 = \( \frac{y}{6} \)
6 x (-4y + 2) = y
(6 x -4y) + (6 x 2) = y
-24y + 12 = y
-24y + 12 - y = 0
-24y - y = -12
-25y = -12
y = \( \frac{-12}{-25} \)
y = \(\frac{12}{25}\)
If a = 9, b = 4, c = 4, and d = 6, what is the perimeter of this quadrilateral?
| 12 | |
| 23 | |
| 26 | |
| 31 |
Perimeter is equal to the sum of the four sides:
p = a + b + c + d
p = 9 + 4 + 4 + 6
p = 23
A quadrilateral is a shape with __________ sides.
2 |
|
4 |
|
5 |
|
3 |
A quadrilateral is a shape with four sides. The perimeter of a quadrilateral is the sum of the lengths of its four sides.
Solve -8b + 3b = 8b + 3x - 4 for b in terms of x.
| \(\frac{6}{7}\)x - \(\frac{5}{7}\) | |
| -\(\frac{5}{7}\)x - \(\frac{3}{14}\) | |
| x + \(\frac{1}{4}\) | |
| -1\(\frac{2}{9}\)x + 1 |
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.
-8b + 3x = 8b + 3x - 4
-8b = 8b + 3x - 4 - 3x
-8b - 8b = 3x - 4 - 3x
-16b = - 4
b = \( \frac{ - 4}{-16} \)
b = \( \frac{}{-16} \) + \( \frac{-4}{-16} \)
b = x + \(\frac{1}{4}\)
Which of the following statements about a parallelogram is not true?
opposite sides and adjacent angles are equal |
|
a parallelogram is a quadrilateral |
|
the perimeter of a parallelogram is the sum of the lengths of all sides |
|
the area of a parallelogram is base x height |
A parallelogram is a quadrilateral with two sets of parallel sides. Opposite sides (a = c, b = d) and angles (red = red, blue = blue) are equal. The area of a parallelogram is base x height and the perimeter is the sum of the lengths of all sides (a + b + c + d).