The terms in the series are

The series converges, so we
. All the terms in this particular series are known to be positive, so
-th partial sum of the series,

Multiply both sides by
:

Subtracting this from
eliminates all the
terms between
and
, leaving us with

As
, the
term will converge to zero, and the value of the infinite series is

Now, we're given

By substitution, we have

so that

Then the sum of the first 8 terms is

and the difference between this and the infinite sum is

Both the function have same vertical stretch of 1.5 , Option B is the correct answer.
<h3>What is a Function ?</h3>
A function is a law that relates a dependent and an independent variable.
The functions given in the question is
f(x) = 3/2 |(x -7)| -2
g(x) = 1.5 ln(x-2)+7
For a function f(x)
f'(x) = a f(x) represents the stretching of a function if a> 1
here the value of a = 1.5 for both the functions , so they have same vertical stretch of 1.5 .
Therefore from the options given , Option B is the correct answer.
The complete question is attached as an Image.
To know more about Function
brainly.com/question/12431044
#SPJ1
4x - 26 = 3x + 4
-3x -3x
x - 26 = 4
+26 +26
x = 30
Since the 2 angles are opposite of each other you would set them equal to each other.
Answer:
a- about 3$
b-about 8$
c-about .04
Step-by-step explanation:
a-100.8/35=2.88
2.88 is around 3
b-49.2/6=8.2
8.2 is close to 8
c-.78/20=.039
.039 is the nearest to .04