| Your Results | Global Average | |
|---|---|---|
| Questions | 5 | 5 |
| Correct | 0 | 3.03 |
| Score | 0% | 61% |
What is 7z2 x 7z6?
| 49z2 | |
| 49z6 | |
| 49z8 | |
| 49z4 |
To multiply terms with exponents, the base of both exponents must be the same. In this case they are so multiply the coefficients and add the exponents:
7z2 x 7z6
(7 x 7)z(2 + 6)
49z8
A circular logo is enlarged to fit the lid of a jar. The new diameter is 45% larger than the original. By what percentage has the area of the logo increased?
| 32\(\frac{1}{2}\)% | |
| 15% | |
| 35% | |
| 22\(\frac{1}{2}\)% |
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 45% the radius (and, consequently, the total area) increases by \( \frac{45\text{%}}{2} \) = 22\(\frac{1}{2}\)%
A machine in a factory has an error rate of 9 parts per 100. The machine normally runs 24 hours a day and produces 7 parts per hour. Yesterday the machine was shut down for 3 hours for maintenance.
How many error-free parts did the machine produce yesterday?
| 133.8 | |
| 98.7 | |
| 195.3 | |
| 201.6 |
The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:
\( \frac{9}{100} \) x 7 = \( \frac{9 \times 7}{100} \) = \( \frac{63}{100} \) = 0.63 errors per hour
So, in an average hour, the machine will produce 7 - 0.63 = 6.37 error free parts.
The machine ran for 24 - 3 = 21 hours yesterday so you would expect that 21 x 6.37 = 133.8 error free parts were produced yesterday.
What is -3y6 + y6?
| -4y6 | |
| 4y6 | |
| -2y6 | |
| -2y36 |
To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:
-3y6 + 1y6
(-3 + 1)y6
-2y6
If \( \left|y - 3\right| \) + 5 = -6, which of these is a possible value for y?
| 8 | |
| 0 | |
| -8 | |
| -4 |
First, solve for \( \left|y - 3\right| \):
\( \left|y - 3\right| \) + 5 = -6
\( \left|y - 3\right| \) = -6 - 5
\( \left|y - 3\right| \) = -11
The value inside the absolute value brackets can be either positive or negative so (y - 3) must equal - 11 or --11 for \( \left|y - 3\right| \) to equal -11:
| y - 3 = -11 y = -11 + 3 y = -8 | y - 3 = 11 y = 11 + 3 y = 14 |
So, y = 14 or y = -8.