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3 years ago

pleassssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeehelp

Mathematics

2 answers:

kifflom [539]3 years ago

3 0

U kno how to simplify

Alexeev081 [22]3 years ago

3 0

SOLUTION:

= - ( 7 / 8 ) - ( - 5 / 6 )

= - ( 7 / 8 ) + ( 5 / 6 )

= - ( 21 / 24 ) + ( 20 / 24 )

= ( - 21 + 20 ) / 24

= - ( 1 / 24 )

Hope this helps! :)
Have a lovely day! <3

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You have 19 digs for your current volleyball season. There are 5 games left in the season. You want to break your previous recor

Charra [1.4K]

Answer:We must get more than 7 digs in the remaining 5 games to break record of 26.

Step-by-step explanation:

Given situation:- Let x represents the number of digs which we must get in the remaining 5 games to break your record ,and we have 19 digs for our current volleyball season.

To break previous record of 26 digs in a season, we have to get more than 26 i.e. 19 digs added to x must be greater than 26

⇒x+19 ≥ 26

⇒x≥26-19⇒x≥7

Therefore,we must get more than 7 digs in the remaining 5 games to break record of 26.

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3 years ago

Tax is paid on meals in a restaurant.

liberstina [14]

Answer:

17.52%

Step-by-step explanation:

65.35/11.45 = 0.1752

.1752 * 100

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2 years ago

Pg 436 Questions 10-13

Rzqust [24]

10. P- 30 ft , A- 8064
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2 years ago

Solve through substitution –3x–8y=20 y=5x+19 Point of intersection, please

Sedaia [141]

Answer:

(-4, -1)

Step-by-step explanation:

-3x - 8y = 20; y = 5x + 19

y = 5x + 19; -3x - 8y = 20

y = 5x + 19

-3x - 8y = 20

-3x - 8(5x + 19) = 20

-43x - 152 = 20

-43x - 152 + 152 = 20 + 152

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2 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

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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2 years ago