# Thread: Polynomial equations

1. ## Polynomial equations

1. The volume of a cylinder is 2400cm^3 and the height is 6cm. Find the radius of the cylinder to the nearest tenth by first isolating r, and then by substituting known values.

2. Expand and simplify the expression π(R+r)(R-r) to explain why it represents the area of a ring. Draw a diagram of the ring and identify R and r.

and can somebody give me a website to learn how to solve these types of problems it would be much appreciated.

2. Originally Posted by Skoz
1. The volume of a cylinder is 2400cm^3 and the height is 6cm. Find the radius of the cylinder to the nearest tenth by first isolating r, and then by substituting known values.
the volume of a cylinder is

$V = \pi r^2 h$

$r$ is the radius, and $h$ is the height.

solve that for $r$, and then plug in $V = 2400$ and $h = 6$

2. Expand and simplify the expression π(R+r)(R-r) to explain why it represents the area of a ring. Draw a diagram of the ring and identify R and r.

and can somebody give me a website to learn how to solve these types of problems it would be much appreciated.
we have the difference of two squares here:

$\pi (R + r)(R - r) = \pi (R^2 - r^2) = \pi R^2 - \pi r^2$

now what can you say? did you draw the diagram? the inner radius is r and the outer radius is R

3. Originally Posted by Skoz
1. The volume of a cylinder is 2400cm^3 and the height is 6cm. Find the radius of the cylinder to the nearest tenth by first isolating r, and then by substituting known values.

2. Expand and simplify the expression π(R+r)(R-r) to explain why it represents the area of a ring. Draw a diagram of the ring and identify R and r.

and can somebody give me a website to learn how to solve these types of problems it would be much appreciated.
1) $V=\pi r^2h$

$2400=\pi r^2(6)$

$r^2=\frac{2400}{6\pi}$

$r=\sqrt{\frac{2400}{6\pi}}\approx 11.3$ cm.

2) $\pi(R+r)(R-r)=\pi(R^2-r^2)$

A ring or annulus with outer radius R and inner radius r has area $\pi(R^2-r^2)$