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3 years ago

Round 26,984 to the nearest hundred

Mathematics

2 answers:

Klio2033 [76]3 years ago

8 0

27,000 is your answer

Basile [38]3 years ago

5 0

27000 because the nine would round up to another thousand to therefore leaving you with this answer

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Erick went bowling, and after 4 sets, his scores were: 120, 96, 85 and 105. His total after 5 sets were 525. What was his score

sp2606 [1]

The answer is 119 for the 5th set

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3 years ago

Find the approximate length of the diagonal of this rectangular prism

drek231 [11]

Is the answer around 24?

8 0

2 years ago

Read 2 more answers

marishachu [46]

Answer:

A. -4-3x>7

Step-by-step explanation:

-4-3(-5)>7

-4-(-15)>7

11>7

3 0

3 years ago

I NEED THE ANSWER IN THE NEXT MINUTE

OverLord2011 [107]

Answer:

508.68

*sorry if this is late

Step-by-step explanation:

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3 years ago

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