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Mamont248 [21]
3 years ago
9

Multiples of 8 and 12

Mathematics
1 answer:
Wewaii [24]3 years ago
8 0
Some multiples are 24, 48,96
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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
Please help!! This is for a test!!
masya89 [10]
Cos=(adjacent/hypotenuse)
So the answer will be cos(16/20) or cos(4/5) both will give u the same answer
I hope it helps
6 0
3 years ago
Write the slope intercept form of a line going through
anyanavicka [17]
Perpendicular = opposite sign and reciprocal slope
Slope 2 turns into -1/2
Y = -1/2x + b
Plug in the point
-5 = -1/2(2) + b, b = -4
Solution: y = -1/2x - 4
5 0
3 years ago
I need help with number 4
Stella [2.4K]
The answer would be A 4.2x3.7
7 0
3 years ago
What is the y-value of the vertex of the function f(x)=−(x−3)(x+11)? −8 −4 33 49
alisha [4.7K]

Answer:

y - value of the vertex is 49.

Step-by-step explanation:

Given function is f(x) = -(x - 3)(x + 11)

f(x) = -(x² - 3x + 11x - 33)

     = -(x² + 8x - 33)

     = -(x² + 8x + 16 - 49)

     = -[(x + 4)² - 49]

     = -(x + 4)² + 49

Comparing this equation with the vertex form of a quadratic function,

f(x) = -(x - h)² + k

Where (h, k) is the vertex of the function.

Vertex of the parabola is (-4, 49)

Therefore, y-value of the vertex is 49.

4 0
3 years ago
Read 2 more answers
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