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3 years ago

Help with this question!Please

Mathematics

2 answers:

Sedbober [7]3 years ago

5 0

Answer:

x = 14 for the first one, x = 60 for the second one

Step-by-step explanation:

tamaranim1 [39]3 years ago

3 0

Step-by-step explanation:

We know the measure of a straight line is 180° . So that ,

> 5x - 10 + 120 = 180

> 5x = 60 + 10

> 5x = 70

> x = 70/5

> x = 14

Solution 2 :-

> x + 50 + 70 = 180

> x = 180 - 120

> x = 60

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Step-by-step explanation:

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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.

We can find the area like this:

$A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy$

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

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)$

And we can find the volume with this formula:

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

$V= \pi \int_{0}^2 64-32x^2 +4x^4 dx$

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

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \frac{\pi}{2} \int_{0}^8 y dy$

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

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3 years ago

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Answer:

Step-by-step explanation:

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2 years ago

What must be true of PQ?

yulyashka [42]

Interesting question
Usually when you look at something like that construction, you think that AB has been bisected by PQ and that the two segments are perpendicular. They are perpendicular but nowhere is that stated. So the answer is C because all the other answers are wrong.

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8 0

3 years ago