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3 years ago

G(x)=x/x^2-49, what is g(x) when x=4?

Mathematics

1 answer:

Bingel [31]3 years ago

3 0

Answer:

x=4 is =0 that it

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what formulas could be used to determine the surface area of a cylinder sandwiched between one cone and half of a sphere?

skelet666 [1.2K]

Answer: Rectangle around the cylinder, Lateral surface of the cone, Half a sphere

To find this surface area, you have 3 different parts. First, you need the rectangular portion of the cylinder, because the circles will be on the inside of the composite shape. Then, you will need the lateral side of the cone, that's everything but the circular base. Again, the circle will be on the interior. Finally, you should find the surface area of half of the sphere.

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3 years ago

Look at the image below

boyakko [2]

Answer:

201.1 km²

Step-by-step explanation:

Surface area of the figure,

4πr²

= 4×π×4²

= 64π

= 201.1 km² (rounded to the nearest tenth)

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3 years ago

Which equations match the graph?<br><br><br> 5 STARS AND BRAINLYEST

Eva8 [605]

I believe that the answer is ‍♀️

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3 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

Convert 76,000 ml into L. Please show your work.

tester [92]

1000 ml = 1 L

76000 ml
76*1000 ml
76 L

8 0

3 years ago

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