Answer:
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Step-by-step explanation:
The answer is <a is congruent to < d
because you already have <abe = <dbe
62 because 90 minus 15, minus 13, is equal to 62
Answer:
Using either method, we obtain: 
Step-by-step explanation:
a) By evaluating the integral:
![\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cint%5Climits%5Et_0%20%7B%5Csqrt%5B8%5D%7Bu%5E3%7D%20%7D%20%5C%2C%20du)
The integral itself can be evaluated by writing the root and exponent of the variable u as: ![\sqrt[8]{u^3} =u^{\frac{3}{8}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bu%5E3%7D%20%3Du%5E%7B%5Cfrac%7B3%7D%7B8%7D)
Then, an antiderivative of this is: 
which evaluated between the limits of integration gives:

and now the derivative of this expression with respect to "t" is:

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then

is continuous on [a,b], differentiable on (a,b) and 
Since this this function
is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

Answer:
Length: 6 m
Width: 4 m
Step-by-step explanation:
Area: 24 
Width: 2L - 8 m
Length: ?
Formula: A = W * L
Replace the variables:
24 = (2L - 8) * L
24 = 2
- 8L
0 = 2
- 8L - 24
or
2
- 8L - 24 = 0
Solve; if you don't know how to solve it using this method, let me know. It's the expression 
1. 
2. 
3. 
4. 
5. (L-6) (2L+4)
6. L-6 = 0 AND 2L+4 = 0
L - 6 =0
L = 6
---------------------------------
2L = -4
L = 
L = -2
Since the length cannot be negative, the only real value is 6.
Area: 24 
Width: 2L - 8 m
Length: 6 m
Calculate width:
W = 2L - 8
W = 2(6) - 8
W = 12 - 8
W = 4 m