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3 years ago

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Two similar cylinders have heights of 75 cm & 25 cm. What is the ratio of the VOLUME of the larger cylinder to the VOLUME of

the smaller cylinder?

1 answer:

castortr0y [4]3 years ago

6 0

I'm not going to say I'm 100% correct.

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Bookmarks Return Submit 1 3.3 Financing with Simple Interest Brittany's car purchasing decisions 1 erd Brittany has decided to g

guajiro [1.7K]

5.12% means

5.12/100 = 0.0512 (each year)

PreOwned Vehicle Cost = 31,100

For 5 years simple interest, the value would be:

31100 * 0.0512 * 5 = $7961.6

Total have to pay: 31,100 + 7961.6 = $39,061.6

4 0

1 year ago

What is Integration ?

jek_recluse [69]

Answer:

Integration is the act of bringing together smaller components into a single system that functions as one. These links usually are established between the components of the process and control layer of each system to promote the free flow of data across systems.

7 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B32%7D%20" id="TexFormula1" title=" \sqrt{32} " alt=" \sqrt{32} " align="absmiddle

hammer [34]

The answer to sqrt 32, simplified is 4sqrt2.

4 0

3 years ago

C Campus Student

Alla [95]

52 is your answer

Hope this helps

4 0

3 years ago

Find the area of the shaded region

o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

$\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)$

$A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}$

5 0

2 years ago