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3 years ago

Finding domain f(x)=3x+1/x-5

1 answer:

4 0

The function is undefinded if the denominator, x-5, equals 0.
So $x-5 \neq 0 \\ 
x \neq 5$ .
The domain is $x \neq 5$ .
In interval notation, this is (-infinity, 5) u (5, infinity).

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PLEASE HELP I AM DESPERATE I WILL GIVE THANKS, 5 STARS, AND THE BRAINLIEST!!!

Amanda [17]

g(-2) + g(2) = 22

Step-by-step explanation:

For x = -2, g(x) = x^2 - 3x

g(-2) = (-2)^2 - 3(-2) = 4 + 6 = 10

For x = 2, g(x) = 12, therefore

g(-2) + g(2) = 10 + 12 = 22

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2 years ago

A mixed number greater than 8.45? my answers are

zalisa [80]

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2 years ago

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Rufina [12.5K]

Ones that never intersect

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2 years ago

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A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

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3 years ago

Here are the heights (in inches) of 12 students in a seminar.

Gala2k [10]

Answer: 75%

Step-by-step explanation:

Height given :

69, 62, 73, 67, 63, 65, 73, 71, 70, 66, 59, 75

The total number of height given = 12

The height less than 73 are 69 , 62, 67 ,63 , 65 , 71 , 70 , 66 , 59

The total number of heights less than 73 = 9

Therefore : the percentage of these students who are shorter than 73 inches = total number of students shorter than 73/ total number of height x 100

= 9/12 x 100

= 75%

5 0

2 years ago