Mathematics, 11.03.2022 03:20 djmccarter

100 POINTS PLEASE HELP GIVING BRAINLIEST There are two right circular cylinders. The radius of the first cylinder is 3 centimeters, and its height is 5 centimeters. The radius of the second cylinder is 15 centimeters, but its height is also 5 centimeters. What is the ratio of the volume of the second cylinder to the volume of the first cylinder?

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Answer from: daniii000

9:1

Step-by-step explanation:

The formula for the volume of a cylinder is πr² · h (pi · radius² · height) (use 3.14 for pi).

The two right circular cylinders have the same height. The larger cylinder has a radius that is three times the size of the smaller radius.

4 · 3 = 12

However, in the equation for volume, the radius is squared. Because of this, the larger cylinder is 9 times the size of the smaller cylinder because 3² = 9.

You can check your work by finding the volume of the two cylinders using the volume formula πr² · h:

3.14 · 4² · 5 = 251.2

3.14 · 12² · 5 = 2,260.8

The cylinder with a radius of 12 cm is 9 times the size of the cylinder with a radius of 4 cm.

2.260.8 ÷ 9 = 251.2

Answer from: THEOUTCOMER

9 : 1

Step-by-step explanation:

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant

Considering the first cylinder,

Radius = 4 cm

Height = 5 cm

Volume = π × 4² × 5 = 80π cm³

Considering the second cylinder,

Radius = 12 cm

Height = 5 cm

Volume = π × 12² × 5 = 720π cm³

The ratio of the volume of the larger cylinder to the volume of the smaller cylinder is

720π/80π = 9/1

Answer from: Quest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

Answer from: Quest

answer: 43/157

this is the answer and if you try to prove me wrong your a meanie

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100 POINTS PLEASE HELP GIVING BRAINLIEST There are two right circular cylinders. The radius of the...