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ohaa [14]
3 years ago
12

An artist creates a​ cone-shaped sculpture for an art exhibit. If the sculpture is 55 feet tall and has a base with a circumfere

nce of 30.144 ​feet, what is the volume of the​ sculpture?
Mathematics
1 answer:
yulyashka [42]3 years ago
4 0

Answer:

4164.69504π

Step-by-step explanation:

So the volume of a cone is πr^2 * h / 3

r = 15.072 / π

h = 55

15.072 / π squared * π = 227.165184π

227.165184π * 55 / 3 = 4164.69504π

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valentina_108 [34]

Answer:

a) \nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}.

b) Du_{f}(2,4,0) = -\frac{8}{\sqrt{11}}

Step-by-step explanation:

Given a function f(x,y,z), this function has the following gradient:

\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}.

(a) find the gradient of f

We have that f(x,y,z) = x\sin{yz}. So

f_{x}(x,y,z) = \sin{yz}

f_{y}(x,y,z) = xz\cos{yz}

f_{z}(x,y,z) = xy \cos{yz}.

\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}.

\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}

\nabla f(2,4,0) = \sin{0}\mathbf{i} + 0\cos{0}\mathbf{j} + 8 \cos{0}\mathbf{k}.

\nabla f(2,4,0) = 0i+0j+8k=(0,0,8)

The vector is v = i + 3j - k = (1,3,-1)

To use v as an unitary vector, we divide each component of v by the norm of v.

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So

v_{u} = (\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}})

Now, we can calculate the scalar product that is the directional derivative.

Du_{f}(2,4,0) = (0,0,8).(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}) = -\frac{8}{\sqrt{11}}

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3 years ago
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Answer:

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NeTakaya
Consecutive numbers can be written as such:
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Answer:

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Say x=100, then the number would be written as 115 in it's simplest form but you could also write it as 100+15. So I'm just using substitution.

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