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3 years ago

11

Who can solve this question?

1 answer:

Crazy boy [7]3 years ago

8 0

X=20
y=60
z=40
I guess this is true

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First teo answer get these points

djverab [1.8K]

BoopAnswer:

Step-by-step explanation:

Nope

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2 years ago

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7 cards are drawn from a standard deck of 52 playing cards. How many different 7-card hands are possible if the drawing is done

Vladimir [108]

Answer:

The number of different 7 card hands possibility is 133784560

Step-by-step explanation:

The computation of the number of different 7 card hands possibility is shown below:

Here we use the combination as the orders of choosing it is not significant

So,

The number of the different hands possible would be

= $^{52}C_7$

= 52! ÷ (7! × (52 - 7)!)

= 133784560

Hence, the number of different 7 card hands possibility is 133784560

4 0

2 years ago

Divide 3.7÷0.006 round the answer the nearest hundredth?

Julli [10]

First divide 3.7 by 0.006
Which is 616.666
The last 6 is the thousandths place. If it is 5 or bigger then the number to the left is rounded up. If its lower than 5 you round down

Since 6 is over 5 we tound up
616.67

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3 years ago

15 is subtracted from a number and the difference is divided by 17. If the quotient is 4 and there is

GREYUIT [131]

Answer:

83

Step-by-step explanation:

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2 years ago

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