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3 years ago

Solve the following equations : 2(x -1) + 3(x +1)=4(x+4)

1 answer:

3 0

This is how I'd work it:

2(x-1)+3(x+1)=4(x+4)
2x-2+3x+3=4x+16
5x+1=4x+16
<u>x=15</u>

To check, substitute x into the original equation and solve:

2(15-1)+3(15+1)=4(15+4)
28+48=76
76=76

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Is /62 rational or irrational

Fittoniya [83]

Answer:

irrational

Step-by-step explanation:

$\sqrt{62}$ is an irrational number because it is not a perfect square root

hope this helps... -_-

6 0

4 years ago

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Simply 3 -(4x -5) divide by 6

tamaranim1 [39]

3.1666666667 << round that and it should give you your answer

5 0

3 years ago

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angle A and angle B are supplementary. mA = (3x + 6), mB = (4x + 20). Find the measure of each angle.

Inga [223]

Answer:

Step-by-step explanation:

3x + 6 + 4x + 20 = 180

7x + 26 = 180

7x = 154

x = 22

mA= 3(22)+6 =66+6 = 72°

mB= 4(22)+20 = 88 + 20 = 108°

5 0

4 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

74 = 12 - 4x<br> Please help me

agasfer [191]

Answer:

Step-by-step explanation:

6 0

3 years ago