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cestrela7 [59]
2 years ago
10

300 x 25.4 SHOW YOUR WORK

Mathematics
1 answer:
Lana71 [14]2 years ago
6 0

Answer:

7620

Step-by-step explanation:

25x 3 = 75 so 25x 300 = 7500

4x 300 = 1200

0.4x300 = 120

7500+120 = 7620

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago
Can I get help with finding the Fourier cosine series of F(x) = x - x^2
trapecia [35]
Assuming you want the cosine series expansion over an arbitrary symmetric interval [-L,L], L\neq0, the cosine series is given by

f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx

You have

a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx
a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}
a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)
a_0=-\dfrac{2L^2}3

a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}

and so

a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}

So the cosine series for f(x) periodic over an interval [-L,L] is

f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx
4 0
3 years ago
Mac has 1/4 hour left of track practice and must be completed 8 laps around the track before he leaves which equation can be use
telo118 [61]

Answer:

(a) Unit = \frac{1}{4}/8

Step-by-step explanation:

Given

Time = \frac{1}{4}

Lap = 8

Required

Determine the time in each lap

The unit time in each lap is calculated by dividing the total time by the number of laps; i.e.;

Unit = \frac{Time}{Lap}

Substitute values for Time and Lap

Unit = \frac{1}{4}/8

5 0
2 years ago
If you had a bad day, Here is some free points.
OverLord2011 [107]

Answer:

thank you

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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