| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.02 |
| Score | 0% | 60% |
A bread recipe calls for 2\(\frac{1}{2}\) cups of flour. If you only have \(\frac{1}{8}\) cup, how much more flour is needed?
| 1\(\frac{1}{2}\) cups | |
| 2\(\frac{7}{8}\) cups | |
| 2\(\frac{3}{8}\) cups | |
| 2\(\frac{3}{4}\) cups |
The amount of flour you need is (2\(\frac{1}{2}\) - \(\frac{1}{8}\)) cups. Rewrite the quantities so they share a common denominator and subtract:
(\( \frac{20}{8} \) - \( \frac{1}{8} \)) cups
\( \frac{19}{8} \) cups
2\(\frac{3}{8}\) cups
What is \( \sqrt{\frac{25}{4}} \)?
| 2\(\frac{1}{2}\) | |
| \(\frac{6}{7}\) | |
| 4 | |
| 2\(\frac{2}{3}\) |
To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:
\( \sqrt{\frac{25}{4}} \)
\( \frac{\sqrt{25}}{\sqrt{4}} \)
\( \frac{\sqrt{5^2}}{\sqrt{2^2}} \)
\( \frac{5}{2} \)
2\(\frac{1}{2}\)
A circular logo is enlarged to fit the lid of a jar. The new diameter is 70% larger than the original. By what percentage has the area of the logo increased?
| 35% | |
| 22\(\frac{1}{2}\)% | |
| 30% | |
| 32\(\frac{1}{2}\)% |
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 70% the radius (and, consequently, the total area) increases by \( \frac{70\text{%}}{2} \) = 35%
A factor is a positive __________ that divides evenly into a given number.
mixed number |
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improper fraction |
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integer |
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fraction |
A factor is a positive integer that divides evenly into a given number. For example, the factors of 8 are 1, 2, 4, and 8.
What is \( 4 \)\( \sqrt{175} \) - \( 5 \)\( \sqrt{7} \)
| 20\( \sqrt{175} \) | |
| -1\( \sqrt{1225} \) | |
| 15\( \sqrt{7} \) | |
| 20\( \sqrt{7} \) |
To subtract these radicals together their radicands must be the same:
4\( \sqrt{175} \) - 5\( \sqrt{7} \)
4\( \sqrt{25 \times 7} \) - 5\( \sqrt{7} \)
4\( \sqrt{5^2 \times 7} \) - 5\( \sqrt{7} \)
(4)(5)\( \sqrt{7} \) - 5\( \sqrt{7} \)
20\( \sqrt{7} \) - 5\( \sqrt{7} \)
Now that the radicands are identical, you can subtract them:
20\( \sqrt{7} \) - 5\( \sqrt{7} \)