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2 years ago

For the following exercises, determine whether each of the following relations is a function.

Mathematics

1 answer:

irina1246 [14]2 years ago

3 0

Answer:

The relation y=2x+8 is a function because the two different output values of the function have the one unique input value.

Step-by-step explanation:

The given relation is y=2x+8.

It is required to determine the relation is a function.

This solution can be obtained by using the concept of function. First write the given relation y=2x+8 and determine whether the given relation is a function.

Step 1 of 1

The given relation is y=2x+8.

This relation shows that the two different output values of the function have one unique input value and this relation is a linear function.

Hence, the relation is a function.

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Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

Drupady [299]

Answer:

$V = \frac{\pi^2}{8}$

$V = 1.23245$

Step-by-step explanation:

Given

$y = \cos 2x$

$y = 0; x = 0; x = \frac{\pi}{4}$

Required

Determine the volume of the solid generated

Using the disk method approach, we have:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

Where

$y = R(x) = \cos 2x$

$a = \frac{\pi}{4}; b =0$

So:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

Where

$y = R(x) = \cos 2x$

$a = \frac{\pi}{4}; b =0$

So:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {(\cos 2x)^2} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx$

Apply the following half angle trigonometry identity;

$\cos^2(x) = \frac{1}{2}[1 + \cos(2x)]$

So, we have:

$\cos^2(2x) = \frac{1}{2}[1 + \cos(2*2x)]$

$\cos^2(2x) = \frac{1}{2}[1 + \cos(4x)]$

Open bracket

$\cos^2(2x) = \frac{1}{2} + \frac{1}{2}\cos(4x)$

So, we have:

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {[\frac{1}{2} + \frac{1}{2}\cos(4x)]} \, dx$

Integrate

$V = \pi [\frac{x}{2} + \frac{1}{8}\sin(4x)]\limits^{\frac{\pi}{4}}_0$

Expand

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [\frac{0}{2} + \frac{1}{8}\sin(4*0)])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [0 + 0])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})])$

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}\sin(\pi)])$

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So:

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}*0])$

$V = \pi *[{\frac{\pi}{8}]$

$V = \frac{\pi^2}{8}$

$V = \frac{3.14^2}{8}$

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