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2 years ago

Which inequality's graph will have a dashed boundary line?

Mathematics

1 answer:

Mnenie [13.5K]2 years ago

8 0

Dashed lines are having < or >

answer is A

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What does comp goku scale to?<br><br>

hammer [34]

Answer:

Goku (Super Saiyan 3)

Super Saiyan 3 Goku at full power has a battle power of 24,000,000,000, 2 billion above Kid Buu's level.

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2 years ago

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If f(x) = x2 - 2x and g(x) = 6x + 4 for which value of x does (t +g)(x)=0

Vinil7 [7]

Answer:

-2

Assumption:

Find the value of x such that $(f+g)(x)=0$ .

Step-by-step explanation:

$(f+g)(x)=0$

$f(x)+g(x)=0$

$(x^2-2x)+(6x+4)=0$

Combine like terms:

$x^2+4x+4=0$

This is not too bad too factor on the left hand side since 2(2)=4 and 2+2=4.

$(x+2)(x+2)=0$

$(x+2)^2=0$

So we need to solve:

$x+2=0$

Subtract 2 on both sides:

$x=-2$

Let's check:

$(f+g)(-2)$

$f(-2)+g(-2)$

$((-2)^2-2(-2))+(6(-2)+4)$

$(4+4)+(-12+4)$

$(8)+(-8)$

$0$

0 was the desired output of $(f+g)(x)$ .

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2 years ago

In a class of 32 boys, 16 passed the

vodka [1.7K]

Answer:

3/8

Step-by-step explanation:

12/32 failled and if you simplify this it means that about 3/8 of the class failled

8 0

2 years ago

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PLEASE ANSWERRR!!!!!

lawyer [7]

Answer:

I don't know how to do this. I'm so sorry that I couldn't help!!

Step-by-step explanation:

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2 years ago

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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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3 years ago