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nikklg [1K]
3 years ago
12

Help in 9 plzzz I will give u brainlist

Mathematics
1 answer:
marshall27 [118]3 years ago
4 0
6.9-6.0=0.9\\\\\begin{array}{ccc}6.0&-&100\%\\0.9&-&p\%\end{array}\ cross\ multiply\\\\6p=90\ \ \ |:6\\p=15\\\\\text{Answer: 15\%}
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Vikentia [17]
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Who’s joe ANSWER AND GET 35 points
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Step-by-step explanation:

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2 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}

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3 years ago
This question is about two positive numbers. Here are facts about these numbers:The numbers are consecutive even integers.The su
djyliett [7]

Now that we’ve learned how to solve word problems involving the sum of consecutive integers, let’s narrow it down and this time, focus on word problems that only involve finding the sum of consecutive even integers.

But before we start delving into word problems, it’s important that we have a good understanding of what even integers, as well as consecutive even integers, are.

Even Integers

We know that even numbers are integers that can be divided exactly or evenly by 22. Thus, the general form of the even integer nn, is n = 2kn=2k, where kk is also an integer.

In other words, since even numbers are the multiples of 22, we can represent an even integer nn by 2k2k, where kk is also an integer. So if we have the even integers 1010 and 1616,

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