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tia_tia [17]
2 years ago
6

Using elimination, solve-12x + 9y = 79x ÷ 12y = 6​

Mathematics
1 answer:
Anastaziya [24]2 years ago
7 0

Answer:

3x=24y

-12x+9y=7

let me know if this helps :3

Step-by-step explanation:

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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
Im not sure how to do this question
Pavel [41]
B. -9 is the most negative one, -2 is the least.
Hope this helps you.
3 0
2 years ago
Read 2 more answers
Two lighthouses flash their lights every 20 seconds and 30 seconds respectively. Given that they flash together at 8 p.m, when w
Free_Kalibri [48]
One minute later at 8:01pm. One lighthouse will flash 3 times (at 20, 40 and 60 seconds)...the other will flash twice (at 30 and 60 seconds). So they will both flash 60 seconds later, at 8:01
6 0
3 years ago
Let AA and BB be events such that P(A∩B)=1/73P(A∩B)=1/73, P(A~)=68/73P(A~)=68/73, P(B)=21/73P(B)=21/73. What is the probability
krok68 [10]

Answer:

P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}

Step-by-step explanation:

Let A and B events. We have defined the probabilities for some events:

P(A') =\frac{68}{73} , P(B) =\frac{21}{73} , P(A \cap B) =\frac{1}{73}

Where A' represent the complement for the event A

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: P(A)+P(A') =1

So for this case we can solve for P(A) like this:

P(A) = 1-P(A') = 1-\frac{68}{73}=\frac{5}{73}

And now we can find P(A \cup B) using the total probability rul given by:

P(A \cup B) = P(A)+P(B) -P(A \cap B)

And if we replace the values given we got:

P(A \cup B) = \frac{5}{73} +\frac{21}{73} -\frac{1}{73}=\frac{25}{73}

And that would be the final answer.

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2 years ago
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3 years ago
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