Ask question

Ask question

All categories

8_murik_8 [283]

3 years ago

15

What is the value of X in this figure? Answers: 6 6√3 12√2 12√3

1 answer:

Lady bird [3.3K]3 years ago

4 0

Answer:

$12\sqrt{2}$

Step-by-step explanation:

$\sqrt{12^2+12^2}$

$12^2+12^2=12^2\cdot \:2$

$\sqrt{12^2\cdot \:2}$

$\sqrt{2}\sqrt{12^2}$

$12\sqrt{2}$

You might be interested in

Tina can grow 2 plants with every seed packet. How many seed packets does Tina need to

AlladinOne [14]

Answer:

7 is the answer

Step-by-step explanation:

2x7=14

8 0

3 years ago

Read 2 more answers

Write an equation of the line that passes through the points.<br><br> (0,1),(−2,−5)

Arada [10]

Answer:

y=4/-2

Step-by-step explanation:

I think that is the answer

6 0

2 years ago

Read 2 more answers

What happens to the graph of f(x) when it is multiplied<br> by a number between 0 and 1?

Dovator [93]

Answer:

Therefore multiplying x by a number between 0 and 1 creates a horizontal stretch, which looks like a vertical compression. That is, the parabola will be shorter/wideStep-by-step explanation:

5 0

3 years ago

Read 2 more answers

What does B.O.D.M.A.S. or B.E.D.M.A.S. mean?

kakasveta [241]

B E D M A S MEANS BRACKETS EXPONENTS DIVISION MULTIPLICATION ADDITION AND SUBTRACTION

8 0

3 years ago

Read 2 more answers

The boundary of a lamina consists of the semicircles y = 1 − x2 and y = 16 − x2 together with the portions of the x-axis that jo

oksano4ka [1.4K]

Answer:

Required center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

Step-by-step explanation:

Given semcircles are,

$y=\sqrt{1-x^2}, y=\sqrt{16-x^2}$ whose radious are 1 and 4 respectively.

To find center of mass, $(\bar{x},\bar{y})$ , let density at any point is $\rho$ and distance from the origin is r be such that,

$\rho=\frac{k}{r}$ where k is a constant.

Mass of the lamina=m= $\int\int_{D}\rho dA$ where A is the total region and D is curves.

then,

$m=\int\int_{D}\rho dA=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}rdrd\theta=k\int_{}^{}(4-1)d\theta=3\pi k$

Now, x-coordinate of center of mass is $\bar{y}=\frac{M_x}{m}$ . in polar coordinate $y=r\sin\theta$

$\therefore M_x=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\sin\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\sin\thetadrd\theta$

$=3k\int_{0}^{\pi}\sin\theta d\theta$

$=3k\big[-\cos\theta\big]_{0}^{\pi}$

$=3k\big[-\cos\pi+\cos 0\big]$

$=6k$

Then, $\bar{y}=\frac{M_x}{m}=\frac{2}{\pi}$

y-coordinate of center of mass is $\bar{x}=\frac{M_y}{m}$ . in polar coordinate $x=r\cos\theta$

$\therefore M_y=\int_{0}^{\pi}\int_{1}^{4}x\rho(x,y)dA$

$=\int_{0}^{\pi}\int_{1}^{4}\frac{k}{r}(r)\cos\theta)rdrd\theta$

$=k\int_{0}^{\pi}\int_{1}^{4}r\cos\theta drd\theta$

$=3k\int_{0}^{\pi}\cos\theta d\theta$

$=3k\big[\sin\theta\big]_{0}^{\pi}$

$=3k\big[\sin\pi-\sin 0\big]$

$=0$

Then, $\bar{x}=\frac{M_y}{m}=0$

Hence center of mass $(\bar{x},\bar{y})=(\frac{2}{\pi},0)$

3 0

3 years ago