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3 years ago

15

A company produces a cardboard box in the shape of a rectangular prism. The surface area of the box is represented by S=10x2, wh

ere x is the length of the base. Find the inverse by solving for x without switching the variables.

1 answer:

OLEGan [10]3 years ago

5 0

S = 10x^2

S/10 = x^2

+or-sqrt(S/10) = x

+or- sqrt(S10)/10 = x

so

_inverse = +or-sqrt(10x)/10

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What is the area of the circle below?<br> 10 ft

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Answer:

<h3>Circle</h3>

Solve for area

A=π r^2

r =Radius

<h2>You about this first: </h2>

The circumference of a circle is its perimeter or distance around it. It is denoted by C in math formulas and has units of distance, such as millimeters (mm), centimeters (cm), meters (m), or inches (in). It is related to the radius, diameter, and pi using the following equations: C = πd. C = 2πr.

Step-by-step explanation:

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Determine the domain and range of the function . f(x)=2 3√ 108^2x {x| all real numbers}; {y| y > 0} {x| all real numbers}; {y

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Answer:

Domain : {x | all real numbers} ; Range: {y | y > 0}

Step-by-step explanation:

The function can be written as :

$f(x)=\sqrt[\frac{2}{3}]{108^{2\cdot x}}\\\\\implies f(x)=(108)^{(\frac{3}{2})^{2\cdot x}}$

Now, since x is exponent so it can take any real values. So, its domain of f(x) is all real numbers

But value of f(x) can not be less than 1 because for x = 0 the value of f(x) is 1 and also for any values of x, the value of f(x) can never be less than 1

So, Range of f(x) is all real numbers greater than 0

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Domain : {x | all real numbers} ;

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3 years ago

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How do you find the volume of the solid generated by revolving the region bounded by the graphs

d1i1m1o1n [39]

Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

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$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.

We can find the area like this:

$A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy$

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= \pi \int_{0}^2 64-32x^2 +4x^4 dx$

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \frac{\pi}{2} \int_{0}^8 y dy$

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

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I need help with these math problems, i’m not really good at math. i’m gonna be asking more questions cause i really need help w

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Answer:

that's an isosceles triangle

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Marina CMI [18]

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