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defon
3 years ago
13

What is the relationship between multiplying and factoring? You multiply numbers or expressions to produce a (select). You facto

r a product into the numbers or (select) that were multiplied to produce it.
Mathematics
1 answer:
ZanzabumX [31]3 years ago
6 0

Answer:

Step-by-step explanation:

1

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Write an equation of the line that passes through the points.<br><br> (2, 5), (0, 5)<br><br> Y =
vodomira [7]
Right off the bat we know that the y-intercept is going to be 5 because of the second co-ordinate.

Now for the slope, because the y co-ordinate doesn't change, it has no slope so the equation is simply: y=5

Hope this helps :)
7 0
3 years ago
For the function defined by f(t)=2-t, 0≤t&lt;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 measure of egf?​
BartSMP [9]

Answer:

∠EGF = 65°

Step-by-step explanation:

Since EF = EG the triangle is isosceles and the base angles are equal, that is

∠EGF = ∠GEF

∠EGF = \frac{180-50}{2} = \frac{130}{2} = 65°

7 0
3 years ago
Read 2 more answers
What slope of the points (9,7) and (3,3)
Ymorist [56]

Answer:

6/4

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is 1/2,0.55,5/7 in least to greatest?
Neporo4naja [7]

Answer:

1/2, 0.55, 5/7 is your answer

Step-by-step explanation:

Change each into decimal

1/2 = 0.50

0.55 = 0.55

5/7 = ~0.71

0.50 < 0.55 < 0.71

~

3 0
3 years ago
Read 2 more answers
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