If r = -3 and s = 5:
(r^-4)(s^2) = (-3^-4)(5^2) = (1/81)(25) = 25/81
If r = 5 and s = -3:
(r^-4)(s^2) = (5^-4)(3^2) = (1/625)(9) = 9/625
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 
12 ounces are in 3/4 a pound.. and 16 ounces are in a pound.. so adding 16 and 10 gives you 26, which ends up with Karla's supplies weighing 4 pounds and 10 ounces all together.
Answer:
x = 10 - 10sqrt(3)/3
Approximately 4.2265 cm
Volume = 1539.6 cm³
Step-by-step explanation:
Let the height be x
V = (20 - 2x)(40 - 2x)x
V = (800 - 80x - 40x + 4x²)x
V = 4x³ - 120x² + 800x
dV/dx= 12x² - 240x + 800 = 0
3x² - 60x + 200 = 0
x = [60 +/- sqrt(60² - 4(3)(200))]/6
x = [30 +/- 10sqrt(3)]/3
x = 10 +/- 10sqrt(3)/3
Approximately,
15.7735 and 4.2265
d²V/dx² = 24x - 240
For max, d²V/dx² < 0
So, x = 4.2265
V = (20 - 2x)(40 - 2x)x
Plug in x = 10 - 10sqrt(3)/3
V = 1539.600718
Answer:

Step-by-step explanation:
y=mx+c
where m is the gradient and c is the y-intercept of the line
Let's find the gradient of the line first.

Using the above formula,

Subst. the gradient into m:

Now, find the value of c by substituting a coordinate into the above equation.
when x=-1, y=3,

Therefore the equation of the line is
