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4 years ago

A-5-B when A=9 and B=-4

Mathematics

1 answer:

Fed [463]4 years ago

3 0

Answer:

Step-by-step explanation:

(Refer to picture)

Plug in the value of A and the value of B into the expression

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I will give brainliest please help me easy math question!!!!!!!!!!!

maria [59]

Answer:

1. -1/2b

2. 3a/b^2

3. 3b^2/1a^2

4. -1a^2b/2

Step-by-step explanation:

Simplified answers and then the variables cancel out each out as much as they can (That's the best way to explain it)

7 0

3 years ago

Read 2 more answers

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

What is the expected value of X?

Soloha48 [4]

Answer:

5.6

Step-by-step explanation:

fr fr

7 0

4 years ago

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Taber invested money in an account where interest is compounded every year. He made no withdrawals or deposits. The function A(t

ioda

Taber invested money in an account where interest is compounded every year. He made no withdrawals or deposits.

The function A(t) = 525(1 + 0.05)t represents the amount of money in the account after t years.

The function is explained as below

Amount = Principal * ( 1 + interest rate ) ^ time

Principal is the amount that is invested in the beginning

Therefore, Taber originally invested $525

3 0

3 years ago

Help I don’t know answer

alex41 [277]

Answer:33

Step-by-step explanation:dwaWDwawawdawd

3 0

3 years ago