| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.96 |
| Score | 0% | 59% |
Simplify (6a)(9ab) - (5a2)(3b).
| 39a2b | |
| 120ab2 | |
| 120a2b | |
| -39ab2 |
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.
(6a)(9ab) - (5a2)(3b)
(6 x 9)(a x a x b) - (5 x 3)(a2 x b)
(54)(a1+1 x b) - (15)(a2b)
54a2b - 15a2b
39a2b
If angle a = 23° and angle b = 31° what is the length of angle d?
| 157° | |
| 126° | |
| 119° | |
| 121° |
An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite:
d° = b° + c°
To find angle c, remember that the sum of the interior angles of a triangle is 180°:
180° = a° + b° + c°
c° = 180° - a° - b°
c° = 180° - 23° - 31° = 126°
So, d° = 31° + 126° = 157°
A shortcut to get this answer is to remember that angles around a line add up to 180°:
a° + d° = 180°
d° = 180° - a°
d° = 180° - 23° = 157°
If the area of this square is 9, what is the length of one of the diagonals?
| 6\( \sqrt{2} \) | |
| 3\( \sqrt{2} \) | |
| 7\( \sqrt{2} \) | |
| \( \sqrt{2} \) |
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 side a = 9, side b = 6, what is the length of the hypotenuse of this right triangle?
| \( \sqrt{68} \) | |
| \( \sqrt{17} \) | |
| \( \sqrt{117} \) | |
| \( \sqrt{73} \) |
According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:
c2 = a2 + b2
c2 = 92 + 62
c2 = 81 + 36
c2 = 117
c = \( \sqrt{117} \)
Solve for c:
-4c + 1 = \( \frac{c}{-7} \)
| 1\(\frac{8}{19}\) | |
| 1\(\frac{1}{15}\) | |
| \(\frac{7}{27}\) | |
| \(\frac{15}{16}\) |
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.
-4c + 1 = \( \frac{c}{-7} \)
-7 x (-4c + 1) = c
(-7 x -4c) + (-7 x 1) = c
28c - 7 = c
28c - 7 - c = 0
28c - c = 7
27c = 7
c = \( \frac{7}{27} \)
c = \(\frac{7}{27}\)