| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.22 |
| Score | 0% | 64% |
Simplify (2a)(6ab) + (8a2)(2b).
| 28ab2 | |
| 80a2b | |
| 4ab2 | |
| 28a2b |
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.
(2a)(6ab) + (8a2)(2b)
(2 x 6)(a x a x b) + (8 x 2)(a2 x b)
(12)(a1+1 x b) + (16)(a2b)
12a2b + 16a2b
28a2b
If the area of this square is 9, what is the length of one of the diagonals?
| 4\( \sqrt{2} \) | |
| 7\( \sqrt{2} \) | |
| 3\( \sqrt{2} \) | |
| \( \sqrt{2} \) |
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 = \( \sqrt{a} \) = \( \sqrt{9} \) = 3
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 = 32 + 32
c2 = 18
c = \( \sqrt{18} \) = \( \sqrt{9 x 2} \) = \( \sqrt{9} \) \( \sqrt{2} \)
c = 3\( \sqrt{2} \)
If angle a = 28° and angle b = 36° what is the length of angle c?
| 108° | |
| 116° | |
| 122° | |
| 101° |
The sum of the interior angles of a triangle is 180°:
180° = a° + b° + c°
c° = 180° - a° - b°
c° = 180° - 28° - 36° = 116°
Which of the following is not true about both rectangles and squares?
the lengths of all sides are equal |
|
the perimeter is the sum of the lengths of all four sides |
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all interior angles are right angles |
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the area is length x width |
A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2).
Solve for y:
-7y + 7 < -4 - 2y
| y < 2\(\frac{1}{5}\) | |
| y < -4 | |
| y < 1\(\frac{4}{5}\) | |
| y < -\(\frac{7}{8}\) |
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.
-7y + 7 < -4 - 2y
-7y < -4 - 2y - 7
-7y + 2y < -4 - 7
-5y < -11
y < \( \frac{-11}{-5} \)
y < 2\(\frac{1}{5}\)