Answer:
b = 14
Step-by-step explanation:
Since both angles are diagonal each other, they are equal to each other.
5b = b + 56
-b -b (subtract b from both sides)
<u>4b</u> = <u>56</u>
4 4 (divide both sides by 4 to solve for b)
b = 14
See attached photo for solution
please give a thanks or brainliest if you found my answer helpful :)
0.250*140,000,000=35,000,000
40,000,000/35,000,000= 1.14
So about 1 time
2.8.1

By definition of the derivative,

We have

and

Combine these fractions into one with a common denominator:

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

3.1.1.
![f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3](https://tex.z-dn.net/?f=f%28x%29%20%3D%204x%5E5%20-%20%5Cdfrac1%7B4x%5E2%7D%20%2B%20%5Csqrt%5B3%5D%7Bx%7D%20-%20%5Cpi%5E2%20%2B%2010e%5E3)
Differentiate one term at a time:
• power rule


![\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%27%20%3D%20%5Cleft%28x%5E%7B1%2F3%7D%5Cright%29%27%20%3D%20%5Cdfrac13%20x%5E%7B-2%2F3%7D%20%3D%20%5Cdfrac1%7B3x%5E%7B2%2F3%7D%7D)
The last two terms are constant, so their derivatives are both zero.
So you end up with

Answer:
the factor of 25 is : 1, 5, 25