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3 years ago

Please help me. I'm very confused with this question. Thanks

Mathematics

2 answers:

Rudiy273 years ago

6 0

The answer is B, but they are very close in exact number.

PSYCHO15rus [73]3 years ago

4 0

I think its A but im not absolutely sure

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What does it mean about the data when a whisker is shorter?

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It means that the whisker is shorter in the data

Step-by-step explanation:

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3 years ago

Find the mode of the data.<br> 1, 2, 1, 3, 3, 4, 1<br> The mode is

drek231 [11]

Answer:

1 is the answer.

Step-by-step explanation:

The mode of a data set is the number that occurs most frequently in the set.

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3 years ago

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Use the Pythagorean Theorem to determine if a

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how can it be triangle with 0ft length

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3 years ago

Anyone who know how to do this answer i will give you 20 Brainly points and mark the Braninllest person who get it right.

notka56 [123]

Answer:

Answer is A

Step-by-step explanation:

(14x^4y^6)/(7x^8y^2)

because everything is one term, you can split into three terms

14/7, x^4 / x^8 , and y^6/y^2

multiplying these three terms will get us the starting term

14/7 = 2

x^4 / x^8

with division of exponents, you subtract the smaller exponent (4), from the big exponent (8), and leave it on the same side (bottom) as the big exponent

1/x^4

same thing with our y's

y^6/y^2

this time, the term stays on the top

y^4

take the simplified terms and multiply them together

(2) * (1/x^4) * (y^4) =

A: (2y^4) / (x^4)

7 0

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