| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.26 |
| Score | 0% | 65% |
Simplify (y + 6)(y + 2)
| y2 + 4y - 12 | |
| y2 - 4y - 12 | |
| y2 + 8y + 12 | |
| y2 - 8y + 12 |
To multiply binomials, use the FOIL method. FOIL stands for First, Outside, Inside, Last and refers to the position of each term in the parentheses:
(y + 6)(y + 2)
(y x y) + (y x 2) + (6 x y) + (6 x 2)
y2 + 2y + 6y + 12
y2 + 8y + 12
A right angle measures:
90° |
|
360° |
|
45° |
|
180° |
A right angle measures 90 degrees and is the intersection of two perpendicular lines. In diagrams, a right angle is indicated by a small box completing a square with the perpendicular lines.
If the area of this square is 1, what is the length of one of the diagonals?
| 2\( \sqrt{2} \) | |
| 8\( \sqrt{2} \) | |
| \( \sqrt{2} \) | |
| 3\( \sqrt{2} \) |
To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:
a = s2
so the length of one side of the square is:
s = \( \sqrt{a} \) = \( \sqrt{1} \) = 1
The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:
c2 = a2 + b2
c2 = 12 + 12
c2 = 2
c = \( \sqrt{2} \)
If side a = 5, side b = 1, what is the length of the hypotenuse of this right triangle?
| \( \sqrt{89} \) | |
| \( \sqrt{26} \) | |
| \( \sqrt{18} \) | |
| \( \sqrt{106} \) |
According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:
c2 = a2 + b2
c2 = 52 + 12
c2 = 25 + 1
c2 = 26
c = \( \sqrt{26} \)
Solve for b:
-4b - 5 < \( \frac{b}{-5} \)
| b < -1\(\frac{7}{9}\) | |
| b < -\(\frac{6}{11}\) | |
| b < \(\frac{7}{29}\) | |
| b < -1\(\frac{6}{19}\) |
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.
-4b - 5 < \( \frac{b}{-5} \)
-5 x (-4b - 5) < b
(-5 x -4b) + (-5 x -5) < b
20b + 25 < b
20b + 25 - b < 0
20b - b < -25
19b < -25
b < \( \frac{-25}{19} \)
b < -1\(\frac{6}{19}\)