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)
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
The answer is 27 I believe
Answer:
Required center of mass 
Step-by-step explanation:
Given semcircles are,
whose radious are 1 and 4 respectively.
To find center of mass,
, let density at any point is
and distance from the origin is r be such that,
where k is a constant.
Mass of the lamina=m=
where A is the total region and D is curves.
then,

- Now, x-coordinate of center of mass is
. in polar coordinate 




![=3k\big[-\cos\theta\big]_{0}^{\pi}](https://tex.z-dn.net/?f=%3D3k%5Cbig%5B-%5Ccos%5Ctheta%5Cbig%5D_%7B0%7D%5E%7B%5Cpi%7D)
![=3k\big[-\cos\pi+\cos 0\big]](https://tex.z-dn.net/?f=%3D3k%5Cbig%5B-%5Ccos%5Cpi%2B%5Ccos%200%5Cbig%5D)

Then, 
- y-coordinate of center of mass is
. in polar coordinate 




![=3k\big[\sin\theta\big]_{0}^{\pi}](https://tex.z-dn.net/?f=%3D3k%5Cbig%5B%5Csin%5Ctheta%5Cbig%5D_%7B0%7D%5E%7B%5Cpi%7D)
![=3k\big[\sin\pi-\sin 0\big]](https://tex.z-dn.net/?f=%3D3k%5Cbig%5B%5Csin%5Cpi-%5Csin%200%5Cbig%5D)

Then, 
Hence center of mass 
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