Let g(t) denote the inner integral. By the fundamental theorem of calculus, the first derivative is

Then using the FTC again, differentiating g gives

Answer:
7/30
Step-by-step explanation:
P = 1/15 + 1/6 = (2+5)/30 = 7/30
Answer:
I-
Step-by-step explanation:
) 5)8%2=80)4)<em>88</em><em>=</em><em>5</em><em>=</em><em>5</em><em>=</em><em>88</em><em>=</em><em><u>9</u></em><em><u>=</u></em><em><u>5</u></em><em><u>)</u></em><em><u>5</u></em><em><u>=</u></em>
Given:
There are given that the parent functions as a cosine function:
Where,
The amplitude of the function is 9.
The vertical shift is 11 units down.
Explanation:
To find the cosine function, we need to see the standard form of the cosine function:

Where,
a is the amplitude of the function,
Now,
According to the question:
The amplitude of the function is 9, which means:

The vertical shift is 11 units down, which means:

For period:

Final answer:
Hence, the cosine function is shown below;
Answer:60
Step-by-step explanation:
5x4x3=60