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3 years ago

15

when jerry drives 100 miles on the highway his car uses 4 gallons of gasoline how much gasoline would his car use if he drive 27

5 miles on the highway

1 answer:

Elanso [62]3 years ago

7 0

Answer:

11

Step-by-step explanation: Two one hundreds would be 8 gallons and 75 is 3/4 of one hundred making that 3 gallons

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The graph below shows the height of a tunnel f(x), in feet, depending on the distance from one side of the tunnel x, in feet:

Soloha48 [4]

<span>Part A: What do the x-intercepts and maximum value of the graph represent?

There are two x-intercepts: x = 0 and x = 36

Those values are the points at which the height of the tunnel is 0.

That means that the width of the tunnel at the ground level is 36 - 0 = 36.

The maximum value of the graph is 32, at x= 18. That, represents that at the center, x= 18, the tunnel has a height of 32.

What are the intervals where the function is increasing and decreasing, and what do they represent about the distance and height? (6 points)

The interval where the function is increasing is (0,18) and the interval where teh function is decreasing is (18,36)

That represents that the height starts to increase with the distance (from 0 to 18) reaches a maximum of 32, at x = 18, and starts to decrease with the distance until x = 36.

Part B: What is an approximate average rate of change of the graph from x = 5 to x = 15, and what does this rate represent? (4 points)

The average rate of change from x = 5 to x = 15 is the change in height divided by the change in distance

=> average rate of change = Δy / Δx

Δy ≈ 31 -14 = 17

Δx = 15 - 5 = 10

=> Δy / Δx ≈ 17 / 10 = 1.7

</span>

7 0

4 years ago

Find the value of y.

Kaylis [27]

Answer: 4.3

Step-by-step explanation:

11x2=22

22/3= 7.3

7.3-3= 4.3

4 0

3 years ago

Use the disk method or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs

JulsSmile [24]

Answer:

1) V = 12 π ㏑ 3

2) $\mathbf{V = \dfrac{328 \pi}{9}}$

Step-by-step explanation:

Given that:

the graphs of the equations about each given line is:

$y = \dfrac{6}{x^2}, y =0 , x=1 , x=3$

Using Shell method to determine the required volume,

where;

shell radius = x; &

height of the shell = $\dfrac{6}{x^2}$

∴

Volume V = $\int ^b_{x-1} \ 2 \pi ( x) ( \dfrac{6}{x^2}) \ dx$

$V = \int ^3_{x-1} \ 2 \pi ( x) ( \dfrac{6}{x^2}) \ dx$

$V = 12 \pi \int ^3_{x-1} \dfrac{1}{x} \ dx$

$V = 12 \pi ( In \ x ) ^3_{x-1}$

V = 12 π ( ㏑ 3 - ㏑ 1)

V = 12 π ( ㏑ 3 - 0)

V = 12 π ㏑ 3

2) Find the line y=6

Using the disk method here;

where,

Inner radius $r(x) = 6 - \dfrac{6}{x^2}$

outer radius R(x) = 6

Thus, the volume of the solid is as follows:

$V = \int ^3_{x-1} \begin {bmatrix} \pi (6)^2 - \pi ( 6 - \dfrac{6}{x^2})^2 \end {bmatrix} \ dx$

$V = \pi (6)^2 \int ^3_{x-1} \begin {bmatrix} 1 - \pi ( 1 - \dfrac{1}{x^2})^2 \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} 1 - ( 1 + \dfrac{1}{x^4}- \dfrac{2}{x^2}) \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} - \dfrac{1}{x^4}+ \dfrac{2}{x^2} \end {bmatrix} \ dx$

$V = 36 \pi \int ^3_{x-1} \begin {bmatrix} {-x^{-4}}+ 2x^{-2} \end {bmatrix} \ dx$

Recall that:

$\int x^n dx = \dfrac{x^n +1}{n+1}$

Then:

$V = 36 \pi ( -\dfrac{x^{-3}}{-3}+ \dfrac{2x^{-1}}{-1})^3_{x-1}$

$V = 36 \pi ( \dfrac{1}{3x^3}- \dfrac{2}{x})^3_{x-1}$

$V = 36 \pi \begin {bmatrix} ( \dfrac{1}{3(3)^3}- \dfrac{2}{3}) - ( \dfrac{1}{3(1)^3}- \dfrac{2}{1}) \end {bmatrix}$

$V = 36 \pi (\dfrac{82}{81})$

$\mathbf{V = \dfrac{328 \pi}{9}}$

The graph of equation for 1 and 2 is also attached in the file below.

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4 years ago

Write the equation of thr parabola in vertex form. vertex (1,1) point (2,-4)

scZoUnD [109]

Answer:

y = -5(x- 1)^2 + 1

Step-by-step explanation:

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3 years ago

WILL GIVE BRAINLIEST

Oksanka [162]

I took the quiz the answer is A,B,D,E
Hope I Helped!

4 0

3 years ago

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