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2 years ago

Plssssssssssssss help

Mathematics

1 answer:

3241004551 [841]2 years ago

6 0

What are the 7 & 9 for? Obviously, you had info from somewhere else for the 9 & 7 if they are THE ONLY NUMBERS here

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Graph

Rus_ich [418]

C is what I think/Know for sure

4 0

2 years ago

Can someone help me with this pls i’ll give brainliest

Rom4ik [11]

Answer: C

Step-by-step explanation:

1 1/8 = 1.125

3/4 = .75

4 0

3 years ago

Rewrite the expression 4+ the square root of 16-(4)(5) decided by 2 as a complex number in standard form a+bi

bija089 [108]

Answer: $4 + \sqrt{4i}$

Step-by-step explanation:

$4+ \sqrt{16-(4)(5)} = 4 + \sqrt{16-20} =4 + \sqrt{-4} = 4 + \sqrt{4i}$

if it matters then simply further...

4 + \sqrt{4i} = $4 + 4i^{1/2}$

8 0

3 years ago

Katie earns 15$ each hour. Round to total time katie worked to the nearest hour and estimate her pay by using mulitplacation

Cloud [144]

Answer:

How long did she work for though?

Step-by-step explanation:

3 0

1 year 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