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sladkih [1.3K]
4 years ago
10

How do you right 0.4 (repeating decimal) as a fraction in simplest form? Also could you do 0.21 (repeating decimal)

Mathematics
1 answer:
Dovator [93]4 years ago
4 0
0.4= 2/5 not sure what u mean by repeating decimal
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Mr. Gomez took his family to the zoo and spent $30.50 on drinks and food. He spent $14 on drinks that cost $1.75 each. He spent
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How do you graph equations?
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4 years ago
James's Sister is arranging rows of beads. She has 56 yellow beads and 120 blue beads. She wants to arrange them so that all the
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22

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6 0
3 years ago
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)])

\sin \pi = 0

So:

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

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

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

or

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

V = 1.23245

4 0
3 years ago
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