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3 years ago

I only need the answer.TY

Mathematics

1 answer:

emmainna [20.7K]3 years ago

3 0

12a = 48!

I hope this helps you!

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jonny [76]

Answer:

1.8 m

Step-by-step explanation:

h(T)=-9.8T^2+20T+1

just replace the T with 2

h=-9.8(2^2)+20(2)+1

-9.8×4+40+1=1.8m

1.8m

8 0

3 years ago

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aksik [14]

Fifty six radical ten

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4 years ago

What is a solution to (x + 6)(x + 2) = 60?

Elden [556K]

Answer:

Ox=4

Step-by-step explanation:

expand he equation

x²+8x-48=0

using substitution method (of the options)

4 substituted as X eliminates all an results to zero

16+32-48=0

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4 0

3 years ago

Find the value of the variable and yz if y is between x and z xy=5p yz=p xy=25

fiasKO [112]

P is 5 , so yz=5 that’s the answer

they gave you that xy=25 and in the same time xy=5p , 5p should be 25 that’s why p is 5

7 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