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kiruha [24]
3 years ago
8

Jesse made a model of a mountain range for class. to do this, he bought a cylinder of cardboard and cut it into three cones. eac

h cone is 5 inches tall and has a diameter of 4 inches. what was the volume of the original cylinder jesse bought in terms of \piπ?
Mathematics
1 answer:
aniked [119]3 years ago
4 0
By definition, the volume of the cone is given by:
 V = (1/3) (π) (r ^ 2) (h)
 Substituting values we have:
 V = (1/3) (π) ((4/2) ^ 2) (5)
 V = (20/3) (π)
 Then, the volume of the cylinder is:
 Vc = 3 * V
 Vc = 3 * (20/3) (π)
 Vc = 20π
 Answer:
 
the volume of the original cylinder jesse bought is:
 
Vc = 20π
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olganol [36]

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2 years ago
Im toooo bad at math plz help
balandron [24]

Answer:

-3/22

Step-by-step explanation:

Mulitiply straight across.  Num * num/ denom * denom.  

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If 3t-7=5t, then 6t =
Natalija [7]
The answer is:  " -21 " .
__________________________________________
We are asked to solve for: "6t" ; which is: "6 * t" .
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Given:  3t − 7 = 5t ;  Solve for "t" ; then solve for "6t" ;

     3t − 7 = 5t ;   Subtract "3t" from each side of the equation;
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    3t − 7 − 3t  =  5t − 3t ;

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Divide EACH side of the equation by "2" ; to isolate "t" on one side of the equation; and to solve for "t" ;

-7/2 = 2t / 2 ;

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6 0
3 years ago
Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
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\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
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4 0
2 years ago
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8090 [49]

Answer:

False

Step-by-step explanation:

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7 0
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