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gogolik [260]
3 years ago
9

PLZ HELP ME. I am learning slope.

Mathematics
2 answers:
sp2606 [1]3 years ago
7 0
The slope is -8. I hope you have an amazing day!!!
Rom4ik [11]3 years ago
4 0

Answer:

-0.125x + 12.5

Step-by-step explanation:

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Help fast please.....
anyanavicka [17]

Answer:  294

7x7=49

49x6=294

7 0
3 years ago
What is white light,
matrenka [14]

Answer:

The absence of any color

Step-by-step explanation:

7 0
3 years ago
If a= 1/3 5a minus 2/3 please help me
Yuki888 [10]

Exact Form: \frac{14}{3}

Decimal Form: 4.6 repeating

Mixed Number Form: 4\frac{2}{3}

4 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
Identify the missing symbol. 6/5 ? 1 1/5
pshichka [43]

Hello from MrBillDoesMath!

Answer:

=   (the equal sign)


Discussion:

Note sure if this is what they wanted but 6/5 = 1 1/5 so I'd guess the symbol is "=".  That is,

6/5 = 1 1/5


Thank you,

MrB

7 0
4 years ago
Read 2 more answers
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