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2 years ago

6

Hays is standing outside on a sunny day. He is 6 ft tall and casts a 4 ft shadow. What is the distance from the top of Hays' hea

d to the end of the shadow?
Round to the nearest tenth, if necessary.

Question 1 options:

4.5 ft

5 ft

7.2 ft

10 ft

1 answer:

IceJOKER [234]2 years ago

5 0

7.2 will be the answer

we use phyagerous thereom

D^2 = 6^2 + 4^2
= sq. root of 52
= 7.2

hope it help\ed

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Answer:

49/12

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3 years ago

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Answer:

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2 years ago

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2 years ago

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2 years ago

Read 2 more answers

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