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Yuri [45]
3 years ago
13

Solve the system of equations. y=x-1 y=-x-5

Mathematics
2 answers:
koban [17]3 years ago
8 0

There is something wrong with the question. Check it again

lyudmila [28]3 years ago
7 0

Answer:

there is some mistake in question

Step-by-step explanation:

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In a talent show of singing and comedy acts, there are a total of 11 acts.
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跳べほねんさくらふじ貼るぇご 74 号 acts are in the show
7 0
3 years ago
Write the equation in slope intercept form. y + 6 = -3( x - 4)
Maslowich

Answer:

y = -3x + 6

Step-by-step explanation:

Distribute -

y + 6 = -3x + 12

Subtract 6 from both sides, final answer

y = -3x + 6

4 0
3 years ago
1. The state of Wyoming is approximately a
VikaD [51]

Answer:

4.86 people/sq.miles

Step-by-step explanation:

the area of the state of Wyoming =

360 × 280 = 100,800 sq. miles

so, the population density =

490,408 ÷ 100,800

= 4.86 people/sq.miles

7 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.

5 0
4 years ago
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