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tester [92]
3 years ago
13

Solve for x -8=32-4x

Mathematics
2 answers:
JulijaS [17]3 years ago
7 0

Hey there! In order to solve this problem you Isolate the variable by dividing each side by factors that don't contain the variable.  x = 10 so. That is your answer. I hope this helped! Sorry if I didn't explain the way you needed. Your fellow Brainly User, GalaxyGamingKitty.

marin [14]3 years ago
5 0

Answer: x= 10

First subtract 32 from both sides, giving you -40=-4x

Then, divide both sides by -4, giving you 10=x

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What set of transformations is performed on triangle ABC to form triangle A′B′C′? (4 points)
fenix001 [56]

Answer:

i am on this exact question and i am confused myself because it seems like the answer should be "180 degree rotation with a translation of 5 units to the right" but that is not an option :/ so my best guess is either A or D hope this helps at all

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Mike forgot to replace the cap on a bottle of room freshener. The room freshener began to evaporate at the rate of 15% per day.
aleksandrvk [35]
Given:
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x    y
1   34
2   28.9
3   24.57
4   20.88
5   17.75

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3 years ago
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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
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Dafna1 [17]

Answer:

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Step-by-step explanation:

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