Hello,
r=5(1+cos t)
r'=5(-sin t)
r²+r'²= 25[(1+cos t)²+(-sin t)²]=50(1-cos t)=50 sin² (t/2)
Between 0 and π, sin x>0 ==>|sin x|=sin x
![l= 2*5* \int\limits^{\pi}_0{sin( \frac{t}{2} )} \, dt= 5[-cos (t/2)]_0^{\pi}\\\\ =5(0+1)=5](https://tex.z-dn.net/?f=l%3D%202%2A5%2A%20%5Cint%5Climits%5E%7B%5Cpi%7D_0%7Bsin%28%20%5Cfrac%7Bt%7D%7B2%7D%20%29%7D%20%5C%2C%20dt%3D%205%5B-cos%20%28t%2F2%29%5D_0%5E%7B%5Cpi%7D%5C%5C%5C%5C%0A%3D5%280%2B1%29%3D5)
Here is the method but i may have make some mistakes.
A = b * h
b = 8 2/3
h = 5 3/4
A = 8 2/3 * 5 3/4.....turn to improper fractions
A = 26/3 * 23/4
A = 598/12
A = 49 5/6 ft^2 <====
Answer: The volume is 288π .
Step-by-step explanation:
V= 4/3 *n*r^3
V = 4/3*n * 6^3
v=4/3 *n * 216
v= 288n
The answer is 3/5 because u need to find the missing side in order to find sin
Answer:
<em>i: </em>x=-2, x=1
<em>ii: </em>x=-1/2
Step-by-step explanation:
Quadratic form:
You solve <em>i </em>by using FOIL (First, Outside, Inside, Last) because it is a multiplication problem.
<em>"first"</em> would be 
, which would equal 
<em>"outside"</em> would be 
, which would equal 
<em>"inside"</em> would be 
, which would equal 
<em>"last" </em>would be 
, which would equal 
Now you need to combine the terms so that they are one after the other
Combine like terms, and you should get:
i Solution
<em>You need to get the variable by itself.</em>
<em>Subtract two from both sides</em>
<em>Add one to both sides.</em>
ii Solution
<em>Add all the terms.</em>
