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shutvik [7]
3 years ago
11

what is the hight of a cylinder with a volume of 384tt cubic inches and a radius of 8 inches? round to the nearest tenth of an i

nch.​
Mathematics
1 answer:
skelet666 [1.2K]3 years ago
5 0

Answer:

1.9 inches

Step-by-step explanation:

We need to utilise one important formula in this question which is the volume of a cylinder formula. We need to work out the height of the cylinder given the following information that the radius is 8 inches and the volume is 384 cubic inches. We can set up an equation to find the value of the height so,

→ π × r² × h = 384

⇒ Substitute in 8 for 'r'

→ π × 8² × h = 384

⇒ Simplify

→ π × 64 × h = 384

⇒ Divide both sides by 64 to isolate π and h

→ π × h = 6

⇒ Divide both sides by π to isolate 'h' and find the value of the height

→ h = 1.9098593171

The height of a cylinder with a volume of 384 cubic inches and a radius of 8 inches is 1.9 inches

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This is one pathway to prove the identity.

Part 1

\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 2

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Part 3

\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\

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We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity \sin^2(\theta)+\cos^2(\theta) = 1 in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

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