Answer:
Required series is:

Step-by-step explanation:
Given that
---(1)
We know that:
---(2)
Comparing (1) and (2)
---- (3)
Using power series expansion for 


![=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]](https://tex.z-dn.net/?f=%3D-%5Bc%2B%5Csum%5Climits%5E%7B%20%5Cinfty%7D_%7Bn%3D0%7D%20%28-1%29%5E%7Bn%7D%5Cfrac%7Bx%5E%7B2n%2B1%7D%7D%7B2n%2B1%7D%5D)


as

Hence,

Answer:
Step-by-step explanation:
You turn 5 3/5 into an improper fraction, to do that you do 5 x 5 = 25 then 25 + 3 = 28/5
so in a 7 day period she swims 5.6 kilometers. So you have to do 5.6/7 = 0.8
So it would be 0.8 kilometers in a day which is 4/5 kilometers so I don't know if the answers were just wrong?
Step-by-step explanation:
3:
h(x)=(-3x-15)/5
let y=(-3x-15)/5
interchanging role of x &y
x=(-3y-15)/5
5x+15=-3y
y=-(5x+15)/3
h-1(x)=-(5x+15)/3
not
equal to f(x)=(-3x-6)/4
<u>G</u><u>i</u><u>v</u><u>e</u><u>n</u><u> </u><u>func</u>tion <u>are</u><u> </u><u>not</u><u> </u><u>function</u><u> </u><u>of</u><u> </u><u>each </u><u>other</u><u> </u><u>.</u>
4:
g(x)=2/3x-2/3
let
y=2/3(x-1)
interchanging role of x &y
x=2/3(y-1)
3/2x+1=y
g-1(x)=3/2x+1
not equal to f(x)=½x+1
<u>Given function are not function of each other .</u>
3xy-5x+9y-45
Step-by-step explanation:
Step by Step Solution
STEP1:STEP2:Pulling out like terms
2.1 Pull out like factors :
3y - 15 = 3 • (y - 5)
Equation at the end of step2: (x • (3y - 5)) + 9 • (y - 5) STEP3:Equation at the end of step 3 x • (3y - 5) + 9 • (y - 5) STEP4:Trying to factor a multi variable polynomial
4.1 Split 3xy-5x+9y-45
4.1 Split 3xy-5x+9y-45
into two 2-term polynomials
-5x+3xy and +9y-45
This partition did not result in a factorization. We'll try another one:
3xy-5x and +9y-45
This partition did not result in a factorization. We'll try another one:
3xy+9y and -5x-45
This partition did not result in a factorization. We'll try another one:
3xy-45 and +9y-5x
This partition did not result in a factorization. We'll try another one:
-45+3xy and +9y-5x
This partition did not result in a factorization. We'll try