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zloy xaker [14]
3 years ago
12

Find exact volume of a sphere with a radius of 10in

Mathematics
1 answer:
Elodia [21]3 years ago
3 0

To find the volume of a sphere, we can apply the following formula:

(4

/3)πr^3

We know the radius is 10, so we can plug that value in.

(4/3) * pi * 10^3

<h3>V = 4188.79020479in^3</h3>

The question says "exact" volume, so I've included all of the digits in the tenths and beyond place, without rounding.

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Step-by-step explanation:

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3 years ago
Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )

Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

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16a - 9 I think that’s the answer hope this helped
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