Step-by-step explanation:
hope this will help you..
Answer:
C
Step-by-step explanation:
An approximation of an integral is given by:

First, find Δx. Our a = 2 and b = 8:

The left endpoint is modeled with:

And the right endpoint is modeled with:

Since we are using a Left Riemann Sum, we will use the first equation.
Our function is:

Therefore:

By substitution:

Putting it all together:

Thus, our answer is C.
*Note: Not sure why they placed the exponent outside the cosine. Perhaps it was a typo. But C will most likely be the correct answer regardless.
Answer:
7/3
Step-by-step explanation:
It would be 7/4 divided by 3/4 which would just be 7/4 multiplied by 4/3. This is 28/12. Simplify:7/3
Answer:
2\9=1
2\3=1
Step-by-step explanation:
Answer:
Use quetexy i use it all the time
Step-by-step explanation: