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yaroslaw [1]
3 years ago
11

What is the volume of this cylinder? *

Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
5 0

Answer   V≈351.86

Step-by-step explanation: V=πr2h=π·42·7≈351.85838

The formula for the volume of a cylinder is V=Bh  or  V=πr2h .

The radius of the cylinder is 4 cm and the height is 7 cm.

Substitute 4 for r and 7 for h in the formula V=πr2h .

V= pie (4)² (7)

v= pie ( 14) (49)

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From the sum of 2 and -11, subtract the sum of 2 and -6<br>​
jolli1 [7]

Answer: -5

Step-by-step explanation: 2 + -11 = -9 and subtract -4 from it since it’s the sum of 2 + -6. -9 - -4 is -9 + 4 so -5

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4 years ago
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Through (5,1) , slope 4/5
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Answer:

Step-by-step explanation:

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3 years ago
Write the equation of a circle that has a center of (-8, 0)<br> and a diameter of 8.
Sveta_85 [38]

Answer:

(x+8)^2+y^2=16

Step-by-step explanation:

The equation for a circle in center-radius form is

(x-h)^2+(y-k)^2=r^2

where (h,k) is the center and r is the radius.

We are given the diameter is 8 so the radius is 8/2=4.

We are also given (h,k) is (-8,0).

The equation for the circle is

(x--8)^2+(y-0)^2=4^2

(x+8)^2+y^2=16

8 0
3 years ago
To the nearest tenth which choice is length of ac​
xz_007 [3.2K]

Answer:

there is no photo

Step-by-step explanation:

but to answer the question for example: 0.04 would round down to the nearest tenth 0.0 and 0.05 would round up to 0.1      . So if the number is 5 or above round up or if it is 4 or below round down, hope this helps

3 0
3 years ago
Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a
Crank

I'll assume the ODE is actually

y''+(x-2)y'+y=0

Look for a series solution centered at x=2, with

y=\displaystyle\sum_{n\ge0}c_n(x-2)^n

\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n

\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n

with c_0=y(2)=2 and c_1=y'(2)=0.

Substituting the series into the ODE gives

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0

\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}

  • If n=2k for integers k\ge0, then

k=0\implies n=0\implies c_0=c_0

k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}

k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}

k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}

and so on, with

c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}

  • If n=2k+1, we have c_{2k+1}=0 for all k\ge0 because c_1=0 causes every odd-indexed coefficient to vanish.

So we have

y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}

Recall that

e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}

The solution we found can then be written as

y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k

\implies\boxed{y(x)=2e^{-(x-2)^2/2}}

6 0
4 years ago
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