| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.20 |
| Score | 0% | 64% |
If side x = 10cm, side y = 15cm, and side z = 6cm what is the perimeter of this triangle?
| 33cm | |
| 21cm | |
| 40cm | |
| 31cm |
The perimeter of a triangle is the sum of the lengths of its sides:
p = x + y + z
p = 10cm + 15cm + 6cm = 31cm
To multiply binomials, use the FOIL method. Which of the following is not a part of the FOIL method?
Last |
|
Odd |
|
Inside |
|
First |
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.
Solve 7b - 4b = -7b - 2x - 7 for b in terms of x.
| 2\(\frac{3}{4}\)x - 1\(\frac{1}{4}\) | |
| 2\(\frac{1}{2}\)x + 1 | |
| \(\frac{1}{7}\)x - \(\frac{1}{2}\) | |
| -\(\frac{7}{13}\)x + \(\frac{8}{13}\) |
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.
7b - 4x = -7b - 2x - 7
7b = -7b - 2x - 7 + 4x
7b + 7b = -2x - 7 + 4x
14b = 2x - 7
b = \( \frac{2x - 7}{14} \)
b = \( \frac{2x}{14} \) + \( \frac{-7}{14} \)
b = \(\frac{1}{7}\)x - \(\frac{1}{2}\)
If the area of this square is 4, what is the length of one of the diagonals?
| 7\( \sqrt{2} \) | |
| 4\( \sqrt{2} \) | |
| 6\( \sqrt{2} \) | |
| 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 = \( \sqrt{a} \) = \( \sqrt{4} \) = 2
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 = 22 + 22
c2 = 8
c = \( \sqrt{8} \) = \( \sqrt{4 x 2} \) = \( \sqrt{4} \) \( \sqrt{2} \)
c = 2\( \sqrt{2} \)
Solve for a:
a2 + 10a - 6 = 4a + 1
| 6 or -1 | |
| -1 or -7 | |
| 3 or -4 | |
| 1 or -7 |
The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:
a2 + 10a - 6 = 4a + 1
a2 + 10a - 6 - 1 = 4a
a2 + 10a - 4a - 7 = 0
a2 + 6a - 7 = 0
Next, factor the quadratic equation:
a2 + 6a - 7 = 0
(a - 1)(a + 7) = 0
For this expression to be true, the left side of the expression must equal zero. Therefore, either (a - 1) or (a + 7) must equal zero:
If (a - 1) = 0, a must equal 1
If (a + 7) = 0, a must equal -7
So the solution is that a = 1 or -7