| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 2.94 |
| Score | 0% | 59% |
Solve 2 + (5 + 3) ÷ 4 x 2 - 42
| 2 | |
| 2\(\frac{1}{4}\) | |
| -10 | |
| 1 |
Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):
2 + (5 + 3) ÷ 4 x 2 - 42
P: 2 + (8) ÷ 4 x 2 - 42
E: 2 + 8 ÷ 4 x 2 - 16
MD: 2 + \( \frac{8}{4} \) x 2 - 16
MD: 2 + \( \frac{16}{4} \) - 16
AS: \( \frac{8}{4} \) + \( \frac{16}{4} \) - 16
AS: \( \frac{24}{4} \) - 16
AS: \( \frac{24 - 64}{4} \)
\( \frac{-40}{4} \)
-10
Solve for \( \frac{2!}{4!} \)
| 4 | |
| \( \frac{1}{120} \) | |
| \( \frac{1}{12} \) | |
| 120 |
A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:
\( \frac{2!}{4!} \)
\( \frac{2 \times 1}{4 \times 3 \times 2 \times 1} \)
\( \frac{1}{4 \times 3} \)
\( \frac{1}{12} \)
If \( \left|a - 1\right| \) - 6 = -1, which of these is a possible value for a?
| 7 | |
| -11 | |
| -4 | |
| -5 |
First, solve for \( \left|a - 1\right| \):
\( \left|a - 1\right| \) - 6 = -1
\( \left|a - 1\right| \) = -1 + 6
\( \left|a - 1\right| \) = 5
The value inside the absolute value brackets can be either positive or negative so (a - 1) must equal + 5 or -5 for \( \left|a - 1\right| \) to equal 5:
| a - 1 = 5 a = 5 + 1 a = 6 | a - 1 = -5 a = -5 + 1 a = -4 |
So, a = -4 or a = 6.
Simplify \( \sqrt{8} \)
| 8\( \sqrt{4} \) | |
| 8\( \sqrt{2} \) | |
| 2\( \sqrt{2} \) | |
| 5\( \sqrt{4} \) |
To simplify a radical, factor out the perfect squares:
\( \sqrt{8} \)
\( \sqrt{4 \times 2} \)
\( \sqrt{2^2 \times 2} \)
2\( \sqrt{2} \)
A circular logo is enlarged to fit the lid of a jar. The new diameter is 30% larger than the original. By what percentage has the area of the logo increased?
| 30% | |
| 15% | |
| 25% | |
| 37\(\frac{1}{2}\)% |
The area of a circle is given by the formula A = πr2 where r is the radius of the circle. The radius of a circle is its diameter divided by two so A = π(\( \frac{d}{2} \))2. If the diameter of the logo increases by 30% the radius (and, consequently, the total area) increases by \( \frac{30\text{%}}{2} \) = 15%