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3 years ago

FInd the volume of the cylinder in terms of Pi.

Mathematics

1 answer:

Luba_88 [7]3 years ago

7 0

The formula for the area of a cylinder is $\pi$ $r^{2}$ ×h

First, we do not have to worry about pi because we are leaving our answer in terms of it so we can move on to the radius squared.

The radius is halfway across a circle so our radius would be 4 and we have to square that and we will get a product of 16.

Lastly, we have to multiply 16 by 8 because 8 is the height of the cylinder

16 × 8 = 128

Therefore, the volume of the cylinder would be 128 $\pi$ in.³

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Given that lim x → 2 f ( x ) = 1 lim x → 2 g ( x ) = − 4 lim x → 2 h ( x ) = 0 limx→2f(x)=1 limx→2g(x)=-4 limx→2h(x)=0, find the

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According what I can read, I have the following statements:

$\lim_{x \to 2} f(x) = 1$

$\lim_{x \to 2} g(x) = -4$

$\lim_{x \to 2} h(x) = 0$

a) Applying properties of limits

$\lim_{x \to 2} f(x) + 5g(x) = \lim_{x \to 2} f(x) + 5 \lim_{x \to 2} g(x) = 1 + 5*-4 = -19$

b) Applying properties of limits

$\lim_{x \to 2} g(x)^{3} = {(\lim_{x \to 2} g(x))}^{3} = (-4)^{3} = -64$

c) Applying properties of limits

$\lim_{x \to 2} \sqrt{f(x)} = \sqrt{\lim_{x \to 2} f(x)} = \sqrt{1} = 1$

d) Applying properties of limits

$\lim_{x \to 2} 4*g(x)*f(x) = 4*\lim_{x \to 2} g(x)*\lim_{x \to 2} f(x) = 4*-4*1 =-16$

e) Applying properties of limits

$\lim_{x \to 2} g(x)*h(x) = \lim_{x \to 2} g(x)*\lim_{x \to 2} h(x) = -4*0 =0$

f) Applying properties of limits

$\lim_{x \to 2} g(x)*h(x)*f(x) = \lim_{x \to 2} g(x)*\lim_{x \to 2} h(x)*\lim_{x \to 2} f(x = -4*0*1 =0$

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3 years ago