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Rufina [12.5K]
3 years ago
13

Please awnser this question in the image

Mathematics
2 answers:
tatiyna3 years ago
8 0

Answer:

The answer is B! :)

Step-by-step explanation:

Paraphin [41]3 years ago
7 0
The answer is b because -radical 7 is -2,64575
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Create an equivalent system of these equations and test your solution <br> x + y = 1<br> x 3y =9
ad-work [718]
An equvilent equation
remember you can do anything to an equation as long asyou do it to both sides


assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent



ok, solve initial

x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation


-x-y=-1
<u>x-3y=9 +</u>
0x-4y=8

-4y=8
divide both sides by -4
y=-2

sub back
x+y=1
x-2=1
add 2
x=3


x=3
y=-2
(3,-2)

if we test it in other one

2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep

2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep


solution is (3,-2)
4 0
3 years ago
Read 2 more answers
The volume of a rectangular prism is 2,058 cubic cm. The length of the prism is 3 times the width. The height is twice the width
liraira [26]
Let the width be w, length = 3w and height = 2w

Volume = length x width x height = w x 3w x 2w = 6w^3
6w^3 = 2,058
w^3 = 2,058/6 = 343
w = ∛343 = 7

width = 7 cm
8 0
3 years ago
What is the perimeter of △GHI?
ZanzabumX [31]
The answer is 27 I believe
5 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
Mr. Kelly has 50 coins, all of which are either nickels or dimes. They have a
Ilia_Sergeevich [38]

So i think its $4.35.

Step-by-step explanation:

I did addition to find out this answer my calculations seem to be correct.

$3.85 + 50 cents = 4 dollars and 35 cents

3 0
3 years ago
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