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3 years ago

6

Please help !! ............

1 answer:

Sauron [17]3 years ago

4 0

I think the answer is C.... I’m not sure tho.

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Can someone pls help? ab to fail geometry. any help is nice.

Mamont248 [21]

By the law of cosines, the angle $C$ satisfies

$8.77^2=2^2+9^2-2\times2\times9\times\cos C$

which reduces to

$\cos C=0.2246\implies C=\arccos(0.2246)\approx1.34422\text{ rad}$

which converts to approximately 77 degrees (this can be done my multiplying by 180 and dividing by pi).

7 0

4 years ago

Help please help please

meriva

What does it say In the speaker

7 0

3 years ago

Read 2 more answers

In a survey of 400 seniors, x percent said that they plan on majoring in physics. One university has used this data to estimate

azamat

Answer:

2 percent

Step-by-step explanation:

We can use two separate fractions, one representing the amount of physics majors expected and the amount of seniors expected in physics majors by the university.

$\frac{400x}{66} = \frac{400}{3600}$

Now cross multiply.

$\frac{66*400}{400x * 3600}$

x will be removed.

$\frac{26400}{1320000}$

Simplifying this fraction will get you $\frac{1}{50}$ , which in percentage form, will be 2 percent.

8 0

2 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago

Find the median and mean of the data set below 19,12,44,15,44

vekshin1

Answer:

median =19

mean=26.8

Step-by-step explanation:

<u>-MEDIAN</u> is the middle number when arranged in ascending order

-MEAN is the sum (total) of all values set devided by the number of values on the list

4 0

3 years ago