3 years ago

Does anyone know this ?? i’m super confused :(

Mathematics

1 answer:

LuckyWell [14K]3 years ago

8 0

Need some Some big brain so here take some

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RSB [31]

Answer:

12: 144

Step-by-step explanation:

As it fits all requirments.

5 0

2 years ago

Yes Or No Full Question Is Below Thanks

Dmitry [639]

No this is not a triangle. With these lengths

8 0

3 years ago

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Does anybody know anything about simplifying rational expressions? 4x/11x

babunello [35]

All you have to do is cancel the x 4/11

7 0

3 years ago

Solve for the letter d

Deffense [45]

Answer:

34 I am pretty sure

Step-by-step explanation:

34+55 well is 89 so yeah d=34

4 0

3 years ago

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Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note

Viktor [21]

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

By integrating both sides, we get

$\displaystyle-\ln(1-x)=C+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

If $x=0$ , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

$\displaystyle\ln(1-x)=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

We can shift the index to simplify the sum slightly.

$\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$ .

$\displaystyle x\ln(1-x)=-\sum_{n=1}^\infty\frac{x^{n+1}}n$

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

4 0

3 years ago