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meriva
3 years ago
6

Does anyone know how to solve this?

Mathematics
2 answers:
love history [14]3 years ago
8 0
The answer would be the second one
Marina CMI [18]3 years ago
7 0
The second one is the answer
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2 years ago
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Solve the system of equations with substitution or elimination method. please help 2x+y=4 and 6x+7y=12
GuDViN [60]

Answer:

x=2-y/2

y=12/7-6x/7

Step-by-step explanation:

8 0
2 years ago
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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
A plumber uses 16 inches of tubing to connect each washing machine in a laundry to the water source. He wants to install 18 wash
adoni [48]
You would multiply 18 machines x 16" for a total of 288 inches. You would then divide by 36 because there are 36" (or 3 feet) in a yard. The answer is 8 yards of tubing.  
6 0
3 years ago
Can someone help me out please?? I would really appreciate it.
Aleks04 [339]
Tan(30deg) = n/rt(3)
1/rt(3) = n/rt(3)—> n=1
sin(30deg) = 1/m
1/2=1/m—> m=2
Thus m=2 ,n=1
4 0
2 years ago
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