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Tanzania [10]
2 years ago
15

Round off 431.3608 to the nearest hundreth​

Mathematics
2 answers:
FromTheMoon [43]2 years ago
3 0
This would be 431.36
pychu [463]2 years ago
3 0

Answer:

431.36

Step-by-step explanation:

The number in the thousandths is smaller than 5 so the hundredth will stay the same.

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Cameron and Sydney are running for class president. Cameron won and received 25% more votes than Sydney. There were a total of 7
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Cameron would have gotten 182 votes. Sydney would have gotten 547.
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Johan sold 9 of his video games online. The next day, he sold 27 Video games. He collected a total of $900 if Johann charged the
Setler79 [48]

Answer:

$25

Step-by-step explanation:

First you would add 27 and 9 to find the total number of games he sold, next you would divide 900 by 36 to get 25, then you would do 25 times 36 to check your answer.

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5m+100=2m+10 . what is the value of m ?
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The value of m is -30 because once you get m on the one side the equation is 3m=-90 and once you divide -90 by 3 you get m=-30.
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3 years ago
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2) On Monday, the stock rose 6.25 points. Wednesday, the stock dropped 9 points. At the end of the week, the stock dropped anoth
Darina [25.2K]

Answer:

-6.25

Step-by-step explanation:

6.25 - 9= -2.75.

-2.75-3.5= -6.25

5 0
3 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
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