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egoroff_w [7]
3 years ago
5

The sum of 2 number is 42 and when you switch the order the difference is 5

Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0
So, x + y = 42 and y - x = 5;
Then, y = x + 5;
Finally, x + x + 5 = 42;
2x + 5 = 42;
2x = 37;
x = 18.5;
y = 18.5 + 5;
y = 23.5;
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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.)

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1. The numbers ______ and _____ are respectively the additive and multiplicative identities of rational numbers
miss Akunina [59]

Answer:  0 and 1, in that order

The numbers <u>   0   </u> and <u>   1   </u> are respectively the additive and multiplicative identities of rational numbers.

===========================================================

Explanation:

The additive identity is 0 because adding 0 to any number leads to the original number. For instance, 7+0 = 7. In general we can say x+0 = x or we could also say 0+x = x.

The multiplicative identity is 1 because multiplying 1 with anything leads to that original number. Example: 1*5 = 5 or 9*1 = 1. The general template is x*1 = x which is the same as saying 1*x = x.

These ideas not only apply to rational numbers, but to real numbers as well.

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It should be, B) 7.25.
4 0
3 years ago
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What is the missing number. Explain how you determined your answer. <br><br> 5–8–14–26–50–?—194
katrin2010 [14]

Answer:

5-8 is +3

8-14 is +3*2

14-26 is 3*4

26-50 is 3*8

which means 50 to the next number is 3*16, which is 48

48+50 is 98

Step-by-step explanation:

4 0
2 years ago
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