The volume generated by rotating the given region
about OC is
<h3>
Washer method</h3>
Because the given region (
) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.
solution
We first find the value of x and y









![v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy](https://tex.z-dn.net/?f=v%3D%20%5Cpi%20%5Cint%5Climits%5E2_o%3D%20%5B%5Cfrac%7By%5E%7B2%7D%20%7D%7B4%7D%20-%20%5Cfrac%7By%5E%7B8%7D%20%7D%7B2%5E%7B8%7D%20%7D%7D%20%20%5D%20dy)
![v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]](https://tex.z-dn.net/?f=v%3D%20%5Cpi%20%5B%5Cint%5Climits%5E2_o%20%7B%5Cfrac%7By%5E%7B2%7D%20%7D%7B4%7D%20%7D%20%5C%2C%20dy%20-%20%5Cint%5Climits%5E2_o%20%7B%5Cfrac%7By%7D%7B2%5E%7B8%7D%20%7D%20%5E%7B8%7D%20%7D%20%5C%2C%20dy%20%5D)
![v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi](https://tex.z-dn.net/?f=v%3D%5Cpi%20%5B%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7By%5E%7B3%7D%20%7D%7B3%7D%20%20%5Cint%5Climits%5E2_0%20-%20%5Cfrac%7B1%7D%7B2%5E%7B8%7D%20%7D%20%20%5Cfrac%7By%5E%7Bg%7D%20%7D%7Bg%7D%20%5Cint%5Climits%5E2_o%5C%5Cv%3D%20%5Cpi%20%5B%5Cfrac%7B1%7D%7B12%7D%20%282%5E%7B3%7D%20-0%29-%5Cfrac%7B1%7D%7B2%5E%7B8%7D%2A9%20%7D%20%282%5E%7Bg%7D%20-0%29%5D%5C%5Cv%3D%20%5Cpi%20%5B%5Cfrac%7B2%7D%7B3%7D%20-%5Cfrac%7B2%7D%7Bg%7D%20%5D%5C%5Cv%3D%20%5Cfrac%7B4%7D%7Bg%7D%20%5Cpi)
A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095
4 + 4 + 4 - 1 = 11? Was it supposed to be an equation? I am confused.
Minus 12x both sides
0=5x^2-12x+9
use quadratic formula
for 0=ax^2+bx+c
x=
given
5x^2-12x+9
a=5
b=-12
c=9
remember: i=√-1
Answer:
720 MPH
Step-by-step explanation:
2700 miles / 3.75 hours = 720 MPH
Answer:
I) |xz| ≈ 28.6 km
II) |yz| ≈ 34.8 km
Step-by-step explanation:
Let's assume that the position of ship due south of x is z (aà pictor representation of the question is attached)
|xy| = 20 km, |xz| = ?, |yz| = ?, θ(y) = 55°
Using Trigonometric ratio - SOHCAHTOA
I) Tan θ = |xz| ÷ |xy| ⇒ Tan 55° = |xz| ÷ 20
|xz| = 20 * Tan 55 = 20 * 1.428
|xz| = 28.56 km
|xz| ≈ <u>28.6 km</u>
<u />
II) Cos θ = |xy| ÷ |yz| ⇒ Cos 55° = 20 ÷ |yz|
|yz| * Cos 55° = 20 ⇒ |yz| = 20 ÷ Cos 55°
|yz| = 20 ÷ 0.574 = 34.84 km
|yz| ≈ <u>34.8 km</u>