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antiseptic1488 [7]

2 years ago

9

how many different possible outcomes exist when each spinner shown below is spun one time? one is numbered from 1-3 and one is 1

-4.

2 answers:

Anit [1.1K]2 years ago

7 0

Answer:

there is 7 possible solutions

Ber [7]2 years ago

3 0

Answer: 12

Step-by-step explanation: I took the Quiz

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Free points, solve 4b+5b=8b-1+3b+15?

vazorg [7]

Answer: Hi!

First, we should combine like terms:

4b + 5b = 8b − 1 +3b + 15 =

9b = 11b + 14

Now, we should subtract 11b from both sides.

-2b = 14

Now, we should divide both sides by -2.

b = -7

Hope this helps!

6 0

2 years ago

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Paul can type 60 words per minute and Jennifer can type 80 words per minute. How does Paul’s typing speed compare to Jennifer’s

yan [13]

I'm not certain how exactly this question is supposed to be answered.

If it wants a ratio, Paul and Jennifer's speeds would be in a ratio of 3:4.

It also could be Paul types 20 words per minute slower than Jennifer.

It could be Paul types 33% slower than Jennifer.

Another answer could be Paul's typing speed is 3/4ths of Jennifer's typing speed.

Feel free to ask any further questions.

4 0

2 years ago

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Help! This is due today and I'm stuck on this problem.

Varvara68 [4.7K]

Answer:

280.6 $m^2$

Step-by-step explanation:

<u>find the length of each side:</u>

you're only given h = 9 m

the formula for h is $h=\frac{\sqrt{3} }{2} *a$ , where a is a side length of the hexagon

plug h in:

$9=\frac{\sqrt{3}}{2}*a\\\\a=9*\frac{2}{\sqrt{3}}\\\\$

a ≈ 10.392 m

<u>now find the area of a triangle:</u>

1/2 (10.392)(9) = 46.764 $m^2$

multiply that by 6 and you get 280.584 ≈ 280.6 $m^2$

you can also use the regular hexagon area formula instead of finding the individual area and multiplying by 6: $\frac{3\sqrt{3} }{2} *a^2$

7 0

2 years ago

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Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

3 0

2 years ago

If C is the center of the circle, find the measure of KM

stiv31 [10]

Your answer is 160. Sorry about that answer above I laughed reading it hahaha.

6 0

3 years ago

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