| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.29 |
| Score | 0% | 66% |
A(n) __________ is two expressions separated by an equal sign.
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equation |
An equation is two expressions separated by an equal sign. The key to solving equations is to 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.
Simplify (3a)(6ab) + (7a2)(3b).
| 39ab2 | |
| 39a2b | |
| -3a2b | |
| 90ab2 |
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.
(3a)(6ab) + (7a2)(3b)
(3 x 6)(a x a x b) + (7 x 3)(a2 x b)
(18)(a1+1 x b) + (21)(a2b)
18a2b + 21a2b
39a2b
This diagram represents two parallel lines with a transversal. If b° = 141, what is the value of y°?
| 163 | |
| 21 | |
| 141 | |
| 147 |
For parallel lines with a transversal, the following relationships apply:
Applying these relationships starting with b° = 141, the value of y° is 141.
If the area of this square is 1, what is the length of one of the diagonals?
| 7\( \sqrt{2} \) | |
| 9\( \sqrt{2} \) | |
| 4\( \sqrt{2} \) | |
| \( \sqrt{2} \) |
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 = \( \sqrt{a} \) = \( \sqrt{1} \) = 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 = \( \sqrt{2} \)
Solve for y:
-2y + 1 = \( \frac{y}{2} \)
| -7\(\frac{1}{5}\) | |
| \(\frac{2}{5}\) | |
| 10\(\frac{4}{5}\) | |
| -3\(\frac{1}{2}\) |
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.
-2y + 1 = \( \frac{y}{2} \)
2 x (-2y + 1) = y
(2 x -2y) + (2 x 1) = y
-4y + 2 = y
-4y + 2 - y = 0
-4y - y = -2
-5y = -2
y = \( \frac{-2}{-5} \)
y = \(\frac{2}{5}\)