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Setler79 [48]
3 years ago
13

Plz help me well mark brainliest if you are correct!.....

Mathematics
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

The first one is the answer to the question

You might be interested in
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
Plsss help me, ASAP, Thank you
ankoles [38]

Answer:

\frac{1}{16}  \frac{10}{9}

Step-by-step explanation:

a reciprocal is a number that when multiplied by a given number gives 1 as  product.

16 * x = 1

x = 1/16

9/10 * x = 1

x = 10/9

8 0
3 years ago
Read 2 more answers
Please help me with this question!
Westkost [7]

Answer:

Your Answer Is 10

Step-by-step explanation:

1st Factor the Numerator and Denominator then cancel the common factor.

Hope It Help You If Yes then Please make me the Brainiest.

Thank You

Parshv

7 0
3 years ago
Could someone help with the questions in the images below? I found them difficult
zubka84 [21]
Question 1a - You have got this correct, the median mark is 35
Question 1b- To work out the range, you must do the largest value subtract the smallest value. For your data, this would be 57 - 13 = 44

Question 2 - You have drawn the lines in the correct places, but you have not used a ruler. In order to get full marks on a question like this you must use a ruler.

Question 3 - Your lower quartile and median is correct, to find the upper quartile, we do 3(n/4). 'n' is the number of data points - in this case 120.
120/4 = 30
30 x 3 = 90
Go across the graph at 90 to see when the line is hit.
From what I can tell, the Upper Quartile is 3.

Next just find the maximum value.
Again, by the looks of it, the Maximum value is 8.5.

Now you have all the data needed for the box plot.

Min = 0
LQ = 0.8
Med = 2.1
UQ =  3
Max = 8.5

Using this information, draw a box plot like you did on the previous question.
<em>Again, I must stress that any line that you draw (unless purposely curved) must be drawn with a ruler; even on the graph at the top, during you working out. If you do not use a ruler, marks can be lost/taken away.</em>

Question 4a - For the median on a cumulative frequency chart, you must find the halfway point in the data. For this, we do n/2. In this case, n = 40
40/2 = 20
Draw a line (using  ruler) across at 20 until you reach the line.
Draw another one down to help you read the number.
The median looks like 34 seconds.
Question 4b - For this question, we already know the Min, Med and Max. Now we must work out the LQ and UQ.

Remember, LQ = n/4
In this case, n = 40
40/4 = 10
Look across at ten and you will find that the LQ = 16 seconds

To work out the UQ, we multiply the LQ place by 3 3(n/4).
3 x 10 = 30
30 on the graph takes us up to 45 seconds.

We now know that the:
Min = 9
LQ = 16
Med = 34
UQ = 45
Max = 57

With this information, draw a box plot like you have in the previous questions.

Question 5 - Good things to always compare on box plots are the Min/Max/Range and the Inter Quartile Range.
The boys had a lowest minimum and a higher maximum, this means that their range is larger, resulting in a large spread of data in comparison to the girls.
The Inter Quartile Range is the difference between the Upper Quartile and the Lower Quartile (how wide the box is).
The boy IQR = 45 - 16 = 29
The girls IQR = 34 - 23 = 11
Again, the girls have much more concise results; even though they did not get the quickest result, they were more like one another.

Question 5a - The median in a box plot is always the line down the centre of the box. In this case:
Median = 24 marks
Question 5b - The IQR is always the UQ - LQ
With this data, the IQR is the following:
36 - 17 = 19 marks

Hope this helps
6 0
3 years ago
Evaluate -15a (-3) for a= 2 and b=-3<br> A.) 5/24 B.) -5/24 C.) -1080 D.) -40/3
zloy xaker [14]
Hhjhhhu did I need ooo 40
4 0
3 years ago
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