Suppose

At the same time, we can write

Note that
(just reverse the sum). Let's pair the first terms of
and
, and the second, and the third, and so on:

Now, each grouped term in the sum on the right side adds to 151. There are 52 grouped terms on that same side (because there are 50 numbers in the range of integers 51-100, plus 50 and 101), which menas

But
, as we pointed out, so

It is (0,-1) hope that helps
Answer:
56 cm squared
Step-by-step explanation:
First things first: Cop one triangle off the rectangle, and attach it to the other one, so the shape looks like a L
Now I can actually solve this:
The left side is 8 cm (because y = 8 cm)
The top is 10 cm
The middle is 6 cm
The inside left is 3 cm
And the very bottom is 2 cm
First, we'll solve for the newly constructed rectangle: 3 x 2. That equals 6.
Next, solve for the longer rectangle: 10 x 5. That's 50.
Now, add the two areas, and we get 56. So the area of the whole thing is 56 cm squared.
(Please keep in mind that I could be wrong, so double check it for me, thanks!)
Answer:
Step-by-step explanation:
1.
To write the form of the partial fraction decomposition of the rational expression:
We have:

2.
Using partial fraction decomposition to find the definite integral of:

By using the long division method; we have:


<u> </u>

So;

By using partial fraction decomposition:


x + 20 = A(x + 2) + B(x - 10)
x + 20 = (A + B)x + (2A - 10B)
Now; we have to relate like terms on both sides; we have:
A + B = 1 ; 2A - 10 B = 20
By solvong the expressions above; we have:

Now;

Thus;

Now; the integral is:


3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.
Please check to the attached image below for the solution to question number 3.