Question

extension problems 9 - 11 solve. 9. the volume of a sphere is a function of its radius. find the volume of air in a standard volleyball with a diameter of 8.5 inches, given v = $\frac{4}{3}pi r^{3}$. 10. an educational study suggest that “a student’s test grade is related to the amount of time spent on practice prior to testing.” can you determine whether or not this statement represents a function or a relation? use a table, mapping or set of ordered pairs to defend your conclusion. 11. draw a circle whose radius is 3 and centered at (1, 2). a) shade the circle’s interior. b) state the domain. c) state the range.

Explanation:

9.

Step1: Find the radius

Given diameter $d = 8.5$ inches, radius $r=\frac{d}{2}=\frac{8.5}{2}=4.25$ inches.

Step2: Calculate the volume

Using the formula $V=\frac{4}{3}\pi r^{3}$, substitute $r = 4.25$ into it.
$V=\frac{4}{3}\pi(4.25)^{3}=\frac{4}{3}\pi\times76.765625\approx\frac{4\times76.765625\pi}{3}\approx\frac{307.0625\pi}{3}\approx 321.56$ cubic - inches.

Let the amount of practice time be the input (independent variable) and the test - grade be the output (dependent variable). For a function, each input value (amount of practice time) must be associated with exactly one output value (test - grade). In real - life, it is possible that two students can spend the same amount of time practicing but get different test - grades. So, this is a relation but not a function. We can represent this as a set of ordered pairs $(x,y)$ where $x$ is the practice time and $y$ is the test - grade. For example, if student A practices for 5 hours and gets a grade of 80, and student B practices for 5 hours and gets a grade of 75, we have the ordered pairs $(5,80)$ and $(5,75)$. Since the input value of 5 is associated with two different output values, it is not a function.

Step1: Draw the circle

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius. Here, $a = 1$, $b = 2$, and $r = 3$. The equation of the circle is $(x - 1)^{2}+(y - 2)^{2}=9$. To draw the circle, plot the center point $(1,2)$ on the coordinate plane. Then, using a compass with a radius of 3 units, draw the circle around the center point and shade the interior.

Step2: Find the domain

The left - most $x$ value of the circle occurs when $x=1 - 3=-2$ and the right - most $x$ value occurs when $x=1 + 3 = 4$. So the domain is $[-2,4]$.

Step3: Find the range

The bottom - most $y$ value of the circle occurs when $y=2-3=-1$ and the top - most $y$ value occurs when $y=2 + 3=5$. So the range is $[-1,5]$.

Answer:

Approximately $321.56$ cubic inches

10.