First subtract 100v² from both sides to get:
C²L²=100c²-100v²
Then divide both sides by C²:
L²=(100c²-100v²)/C²
Then take the square root of both sides:
L=+ or - the square root of (100c²-100v²)/C²
Answer:
Step-by-step explanation:
Answer:
Step-by-step explanation:
Remark
Interesting detail.
Origami Paper is square -- perfectly.
So the properties listed will be related to a square.
Givens
Area = 35 in^2
Formulas
Area = s^2 where s is the length of 1 side.
Perimeter = 4s
Solution
<u>Area = s^2 = 35</u>
s^2 = 35 Take the square root of both sides.
√s^2 = √35
s = 5.91
<u>Perimeter = 4s</u>
s = 5.91
Perimeter = 4*5.91
Perimeter = 23.66
The way the question is worded, the answer you should submit is
Perimeter = 4 * √35
If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show
Let 
. Compute the inverse:
![f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}](https://tex.z-dn.net/?f=f%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%20%3D%20%5Csqrt%7B1%20%2B%20f%5E%7B-1%7D%28x%29%5E3%7D%20%3D%20x%20%5Cimplies%20f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7Bx%5E2-1%7D)
and we immediately notice that ![f^{-1}(x+1)=\sqrt[3]{x^2+2x}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%2B1%29%3D%5Csqrt%5B3%5D%7Bx%5E2%2B2x%7D)
.
So, we can write the given integral as
Splitting up terms and replacing 
in the first integral, we get
