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adelina 88 [10]
3 years ago
11

What's the distance between 7,8 and 3,2

Mathematics
1 answer:
PolarNik [594]3 years ago
8 0
The anwser is 7.2
hope this helps

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Solve the system of equations algebraically.<br> 4x - 3y = 6<br> 6x - 5y = 6
muminat

Answer: x=6  y=6

Step-by-step explanation:

\displaystyle\\\left \{ {{4x-3y=6} \atop {6x-5y=6}} \right.\\

Since the right parts of the equation are equal, therefore the left parts of the equation are also equal:

4x-3y=6x-5y\\4x-3y+5y=6x-5y+5y\\4x+2y=6x\\4x+2y-4x=6x-4x\\2y=2x\\

Divide both parts of the equation by 2:

y=x\\Hence,\\4x-3x=6\\x=6\\Thus,\\x=y=6

3 0
1 year 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 1598726887465*65489
coldgirl [10]
The answer is 1.0469903e+17 Hope this helped 
6 0
3 years ago
The arithmetic sequence a; is defined by the formula:
Morgarella [4.7K]

Answer:

The sum of the first 650 terms of the given arithmetic sequence is 2,322,775

Step-by-step explanation:

The first term here is 4

while the nth term would be ai = a(i-1) + 11

Kindly note that i and 1 are subscript of a

Mathematically, the sum of n terms of an arithmetic sequence can be calculated using the formula

Sn = n/2[2a + (n-1)d)

Here, our n is 650, a is 4, d is the difference between two successive terms which is 11.

Plugging these values, we have

Sn = (650/2) (2(4) + (650-1)11)

Sn = 325(8 + 7,139)

Sn = 325(7,147)

Sn = 2,322,775

6 0
3 years ago
Marco Drew 20 pictures he drew 3/4 of them in art class how many pictures did Marco draw in the art class.
Zielflug [23.3K]
He drew 15 in the art class.                     
6 0
3 years ago
Read 2 more answers
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