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lutik1710 [3]
3 years ago
11

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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Answer please to the best of your ability
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
~Giving Brainiest~
aksik [14]
Fifty six radical ten
3 0
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
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