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3 years ago

Plz help me with these questions 20 points

Mathematics

1 answer:

postnew [5]3 years ago

4 0

A.Is 59.5 C. is 35.38 E. is 134.12 B. is 69.28 D. is 0.34 F. is 37.43100

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Please solve this and only answer with image

horsena [70]

$\cot^4 A - \cot^2 A = 1\\\\\cot^4 A = 1 + \cot^2 A\\\\\frac{\cos^4 A}{\sin^4 A} = 1 + \frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A}{\sin^2 A}+\frac{\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{\sin^2 A+\cos^2 A}{\sin^2 A}\\\\\frac{\cos^4 A}{\sin^4 A} = \frac{1}{\sin^2 A}\\\\\cos^4 A = \frac{\sin^4 A}{\sin^2 A}\\\\\cos^4 A = \sin^2 A\\\\$

This then means,

$\cos^4 A + \cos^2 A = 1\\\\\sin^2 A + \cos^2 A = 1\\\\$

which is the pythagorean trig identity. This concludes the proof.

Therefore, if $\cot^4 A - \cot^2 A = 1$ , then $\cos^4 A + \cos^2 A = 1$

5 0

3 years ago

How do you find the volume of the solid generated by revolving the region bounded by the graphs

d1i1m1o1n [39]

Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.