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Tanzania [10]
3 years ago
6

Find the slope VU Helppppppp pleaseee

Mathematics
2 answers:
Assoli18 [71]3 years ago
6 0
1/3 is the slope rise/rum
PilotLPTM [1.2K]3 years ago
4 0

Answer:

rise/run

1/3

Step-by-step explanation:

You might be interested in
Explain why solving the equation - 3n - 4 = 2 is different from solving the
Klio2033 [76]

Answer:

Step-by-step explanation:

First get the answer to -3n - 4 = 2

-3n - 4 + 4 = 2 + 4

-3n = 6

n = 6/-3

n = -2

That answer is the only one that is permitted. It is the only one that completely satisfies the equation.

Now when you do the inequality, look what happens.

-3n - 4 < 2              Add 4 to both sides.

-3n-4+ 4 < 2+4

-3n < 6                  Now there are a bunch of ways (2) to solve this.  

                            No matter which way you do it, the arrow will change.

-3n/-3 > 6/-3

n > - 2

That means that any number that is greater than - 2 will satisfy the inequality.

So n = 0 will work. Even n = - 1 will work. Anything bigger than -2 will work. The equation does not provide that kind of latitude.

7 0
3 years ago
Mike can read 15 pages of a book in 24 minutes.
pashok25 [27]
72 minutes

Take 24 divided by 15 = 1.6 minutes per page
Then take 1.6x45=72
8 0
3 years ago
Find the inverse laplace transform of: (2 s + 4) / (s - 3)^3
Serhud [2]

Answer:

e^{3t}(2t+5t^{2})

Step-by-step explanation:

L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=

Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3

Translation theorem:L^{1} [F(s-a)=L^{-1}[F(s)|_{s \to s-a}\\ L^{1} [F(s-a)=e^{at} f(t)

L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=e^{3t} L^{-1}[\frac{2(s+3)+4}{s^{3}} ]

Separate the fraction in a sum:

e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])

The formula for this is:

L^{-1}[\frac{n!}{s^{n+1}} ]=t^{n}

Modify the expression to match the formula.

e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ \frac{10}{2} L^{-1}[\frac{2}{s^{2+1}}])=e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])

Solve

e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])=e^{3t}(2t+5t^{2} )

6 0
3 years ago
The length of the rectangle is 12 units what is the width ​
sattari [20]

Answer:

it can be anything.

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Find the area of the rectangle in the picture
Volgvan

Answer:

2x^2-18

Step-by-step explanation:

2x to the power of 2 minus 18

7 0
2 years ago
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