2 years ago

Please help me out??????

1 answer:

6 0

Answer:

Your answer should be 54!

You might be interested in

frez [133]

Answer:

-4 5\12

Step-by-step explanation:

Identify the Least Common Denominator, 12, then just simply evaluate.

8 0

3 years ago

enyata [817]

Let $f(x)=\tan^{-1}x$ . Then $f'(x)=\frac1{1+x^2}$ . Note that $f(0)=0$ .

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n\ge0}x^n$

This means that for $|-x^2|=|x|^2$ , or $-1$ , we have

$\displaystyle f'(x)=\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n\ge0}(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}$

Integrate the series to get

$f(x)=f(0)+\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

$\implies\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

6 0

2 years ago

ladessa [460]

-129 is the anwser to ur problem

3 0

2 years ago

lapo4ka [179]

Yes I work this out so yea you divide idk

7 0

2 years ago

tester [92]

The correct answer is d15

7 0

2 years ago

Please help me out??????​