What is the ratio of A’s to B’s? ABBBAAA

Mathematics

1 answer:

kari74 [83]3 years ago

3 0

The ratio is 4/3
there's 4 As and 3 Bs so compared it's 4/3

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Jason bought pecans in bulk at the grocery store. He paid $19.20 for

murzikaleks [220]

Answer:

0.59 cents

Step-by-step explanation:

To find how much he paid for 1 pound, you divide $19.20 by 2 = $9.60

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8 0

3 years ago

Find all integers $n$ for which $\frac{n^2+n+1}{n-1}$ is an integer.

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$\dfrac{n^2+n+1}{n-1}=\dfrac{(n-1)^2+3(n-1)+3}{n-1}=n-2+\dfrac3{n-1}$

The remainder term $\dfrac3{n-1}$ will never vanish, and will always be rational as long as the denominator is not 1, 3, or -3.

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$n-1=3\implies n=4$
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3 years ago

Find the half range Fourier sine series of the function

worty [1.4K]

The full range is $-\pi$ (length $2L=2\pi$ ), so the half range is $L=\pi$ . The half range sine series would then be given by

$f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx$

where

$b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx$

Essentially, this is the same as finding the Fourier series for the function

$\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0$

Integrating by parts yields

$b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n$

So the half range sine series for this function is simply

$f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n$