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3 years ago

EXPLAIN YOUR ANSWER

1 answer:

7 0

Solve the equation by substitution method.
To get b-2-3b=4. Then you substitute b into simplest to get, b=-3. Repeat the same process with a to get, a=-5. So your answer would be (-5, -3)

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PLEASE HELP!!!!!!!!!

Naddik [55]

The answer is A. The player runs 7 meters forward and then runs 7 meters in the opposite direction. Hope it helped!

- Shadow.

4 0

3 years ago

A school administrator bought 6 desk sets. Each set consists of one table and one chair. He paid $48.50 for each chalr. If he pa

zlopas [31]

The answer is $62.00

Explanation: If you take $48.50 times 6 you get $291 now subtract $291 from $663 you will have an answer of $372 that is how much the total of tables came to now divide 372 by 6 for the number of tables and you will be left with 62

7 0

3 years ago

Domain:<br> Range:<br> Increasing:<br> Decreasing:

mihalych1998 [28]

Answer:

Domain- -2>x<2

Range- [4,∞)

Increasing-[∞,4]

Decreasing-[4,∞]

3 0

3 years ago

Which graph represents a linear function?

irakobra [83]

Answer:

On a coordinate plane, a line has a negative slope

Step-by-step explanation:

A linear function is a function whose graph is a straight line. For example, y = 3x - 2 represents a straight line on a coordinate plane and hence it represents a linear function.

5 0

3 years ago

Read 2 more answers

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