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3 years ago

The cylinder shown has a volume of π in3. Find the volume of a cone with the same base and height as the cylinder.

Mathematics

2 answers:

malfutka [58]3 years ago

5 0

Answer:

Step-by-step explanation:

Alright, lets get started.

Suppose the height of cylinder is h and radius of base is r.

The volume of cylinder will be : $\pi r^2h$

The cone is of same height means h and same base means radius will be r.

The formula of volume of cone is : $\frac{1}{3} \pi r^2h$

It means the volume of cone is one third of the volume of cylinder.

The volume of cylinder is given as π.

So, the volume of cone will be : $\frac{\pi }{3}$ : Answer

Hope it will help :)

Over [174]3 years ago

4 0

The volume of a cylinder is 3 times larger than a cylinder. So find your answer for the cylinder and multiply it by 1/3 to get your answer. Hope this helps! Study on!

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the perimeter of equilateral triangular Garden is 18 ft. The side lengths are 3x feet. what is the value of x?

Yuliya22 [10]

X = 2

18 / 3 = 2

2 * 6 = 6

6 + 6 + 6 = 18

Hope this helps.

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3 years ago

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yan [13]

280 eels

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eel = 56/200

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(y+1)2_ (x + 2)2 > 1?

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Answer:

(–2, 6)

Step-by-step explanation:

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3 years ago

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Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

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