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Anuta_ua [19.1K]
3 years ago
6

Find the distance between 2 3/5 and -1 on a number line

Mathematics
1 answer:
True [87]3 years ago
7 0

Answer:

The distance between these two is 3 3/5

Step-by-step explanation:

In order to find the distance between two points, we subtract one from the other. Then if the result is negative, we take the absolute value.

2 3/5 - -1

2 3/5 + 1

3 3/5

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For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:
Oksi-84 [34.3K]
The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

where

b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}

Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

while for odd n, you have

a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}

So the half-range cosine series expansion would be

f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t
f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

Attached are plots of the first few terms of each series overlaid onto plots of f(t). In the half-range sine series (right), I use n=10 terms, and in the half-range cosine series (left), I use k=2 or n=2(2)-1=3 terms. (It's a bit more difficult to distinguish f(t) from the latter because the cosine series converges so much faster.)

5 0
3 years ago
What is the slope line through (-9,-6) and (3,-9)?
chubhunter [2.5K]

Answer:

-1/4

Step-by-step explanation:

We can find the slope of the line given two points

m = (y2-y1)/(x2-x1)

   = (-9- -6)/(3- -9)

   = (-9+6)/(3+9)

    =-3/12

    = -1/4

8 0
3 years ago
What is an appropriate scale to graph a credit of $35 and a debit of $40?Explain your reasoning
miss Akunina [59]
A bar graph since I'm only going up by fives it's easy to show as a bar graph
4 0
3 years ago
The length of a rectangle is 5 less than twice the width. If the perimeter of the rectangle is 146 cm, find the dimensions of th
Marta_Voda [28]

Length = 2x - 5

Width = x

Perimeter = 146

P = 2L + 2W

146 = 2(2x - 5) + 2x

146 = 4x - 10 + 2x

146 = 6x - 10

146 + 10 = 6x

156 = 6x

156/6 = x

26 = x

Length = 2(26) - 5

Length = 52 - 5

Length = 47 cm

Width = x

Width = 26 cm

Did you follow?

4 0
3 years ago
Can someone please help me I can’t afford to get anymore wrong
yuradex [85]
You answer would be B. Vehicle B needs fewer repairs then Vehicle A on average and its repair rate is more consistant
3 0
3 years ago
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