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UkoKoshka [18]
3 years ago
11

HELP FAST I'LL MARK YOU BRAINLIEST

Mathematics
1 answer:
Luda [366]3 years ago
7 0

Answer:

Parallel

Step-by-step explanation:

............

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Can Snap say your somewhere when your not with someone????
PtichkaEL [24]

Answer:

In my choice yes, it helped me I mean snap showed me where I am.

Step-by-step explanation:

if you want to know how then just be online and turn on location

5 0
2 years ago
Read 2 more answers
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
Which expression is equivalent to 4-10(9m-7)
saveliy_v [14]

Answer:

D

Step-by-step explanation:

4-90m+70

74-90m so

D

if my answer helps please mark as brainliest

7 0
2 years ago
A 6-yard piece of ribbon costs $36.72. What is the price per foot?
djyliett [7]

Answer:

$2.04

Step-by-step explanation:

Note that: 1 yard = 3 feet.

First, change the amount of yard (6) to feet:

6 x 3 = 18

6 yard = 18 feet.

Note: 18 feet = $36.72

Find the cost per foot (1 feet). Note the equal sign, what you do to one side, you do to the other. Divide 18 from both sides.

(18 ft)/18 = ($36.72)/18

1 ft = 36.72/18

1 ft = 2.04/1

1 ft = $2.04

~

4 0
3 years ago
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Mrs. Dugan plans to serve 100 barbecue sandwiches at the company picnic. How many packages of barbecue buns will she need if bun
Aleonysh [2.5K]

100 sandwiches, using bun packages of 8...

100/8=12.5 Since you cant buy a half a package, round up to 13 packages

100 sandwiches, using bun packages of 12...

100/12=8.33 Since you cant have a third of a package, round up to 9 packages
6 0
3 years ago
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