the mass of a cylinder made of barium with a height of 2 inches depends on the radius of the cylinder as…

the mass of a cylinder made of barium with a height of 2 inches depends on the radius of the cylinder as defined by the function $m(r)=7.188pi r^{2}$. which system of equations and solution can be used to represent the radius if the mass of the cylinder is 11,000 grams? round to the nearest hundredth of an inch. $y = 7.188pi r^{2}$ and $y = 4$; $r = 0.42$ inches $y = 7.188pi$ and $y = r^{2}$; $r = 4.75$ inches $y = 7.188pi r^{2}$ and $y = 11,000$; $r = 22.07$ inches $y = 7.188r^{2}$ and $y=pi$; $r = 0.66$ inches

the mass of a cylinder made of barium with a height of 2 inches depends on the radius of the cylinder as defined by the function $m(r)=7.188pi r^{2}$. which system of equations and solution can be used to represent the radius if the mass of the cylinder is 11,000 grams? round to the nearest hundredth of an inch. $y = 7.188pi r^{2}$ and $y = 4$; $r = 0.42$ inches $y = 7.188pi$ and $y = r^{2}$; $r = 4.75$ inches $y = 7.188pi r^{2}$ and $y = 11,000$; $r = 22.07$ inches $y = 7.188r^{2}$ and $y=pi$; $r = 0.66$ inches

Answer

Answer:

C. $y = 7.188\pi r^{2}$ and $y = 11000$; $r = 22.07$ inches

Explanation:

Step1: Set up the equation

We know $M(r)=7.188\pi r^{2}$ and $M = 11000$. So we set $y = 7.188\pi r^{2}$ and $y = 11000$.

Step2: Solve for $r$

$11000=7.188\pi r^{2}$. First, $\pi\approx3.14159$. Then $7.188\times3.14159r^{2}=11000$, $22.585r^{2}=11000$, $r^{2}=\frac{11000}{22.585}\approx487.04$. Taking the square - root of both sides, $r=\sqrt{487.04}\approx22.07$ inches.