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nignag [31]
3 years ago
7

Please help with 32 and 33, thank you.

Mathematics
1 answer:
PtichkaEL [24]3 years ago
4 0

Answer:

32. ∛((x-7)/4) = f^(-1)(x)

33. -10x - 9

Step-by-step explanation:

32. We want to switch f(x) and x, and then solve for f(x) to get the inverse.

x = 4f(x)³ + 7

subtract 7 from both sides

x -7 = 4f(x)³

divide both sides by 4

(x-7)/4 = f(x)³

cube root both sides

∛((x-7)/4) = f(x)

make f(x) f^(-1)(x) because this is now the inverse

∛((x-7)/4) = f^(-1)(x)

the second answer is correct

33. for composition, we can treat (f · g) (x) as attached picture (content filter!), so we plug g(x) into f(x). This results in

2(-5x-7) + 5

expand

-10x - 14 + 5

add

-10x - 9

the second answer is correct

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Answer:

About the x axis

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

About the y axis

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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Answer:

The inference that can be made using the dot plot is:

    The range of round 1 is greater than the round 2 range.

Step-by-step explanation:

<u>Round 1:</u>

Score                Frequency

  1                          0

  2                          2

  3                          3

  4                          2  

  5                          1

Hence, the minimum score of Round 1 is: 2

maximum score is: 5

Hence, Range=Maximum value-Minimum score

                      =5-2

                       =3

Similarly, <u>Round-2</u>

Score                Frequency

  1                          0

  2                         0

  3                          0

  4                          4  

  5                          4

Hence, the minimum score of Round 1 is: 4

maximum score is: 5

Hence, Range=Maximum value-Minimum score

                      =5-4

                       =1

The scores of round 2 are higher than round-1.

Since round 2 have a higher frequency for higher scores as compared to round-1.

Hence, Range of round 1 is greater than the range of Round-2.

 

6 0
3 years ago
Read 2 more answers
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