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DaniilM [7]
3 years ago
6

Solve the compound inequality |3x-9|≤15 and |2x-3|≥5. Give answer in interval notation.

Mathematics
1 answer:
zavuch27 [327]3 years ago
8 0

Answer:

Step-by-step explanation:

|3x-9|≤15

-15≤3x-9≤15

-15+9≤3x≤15+9

-6≤3x≤24

-2≤x≤8

x∈[-2,8]

|2x-3|≥5

2x-3≤-5

2x≤-2

x≤-1

and

2x-3≥5

2x≥8

x≥4

solution is x∈[-2,-1]∪[4,8]

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Answer:

The Answer is Letter C!!!

Step-by-step explanation:

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3 years ago
Determine whether the equation is exact. If it​ is, then solve it. e Superscript t Baseline (7 y minus 3 t )dt plus (2 plus 7 e
umka21 [38]

Answer:

F(t,y)=(2+7e^t)y+3(1-t)e^t +C

Step-by-step explanation:

You have the following differential equation:

e^t(7y-3t)dt+(2+7e^t)dy=0

This equation can be written as:

Mdt+Ndy=0

where

M=e^t(7y-3t)\\\\N=(2+7e^t)

If the differential equation is exact, it is necessary the following:

\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}

Then, you evaluate the partial derivatives:

\frac{\partial M}{\partial y}=\frac{\partial}{\partial t}e^t(7y-3t)\\\\\frac{\partial M}{\partial t}=7e^t\\\\\frac{\partial N}{\partial t}=\frac{\partial}{\partial t}(2+7e^t)\\\\\frac{\partial N}{\partial t}=7e^t\\\\\frac{\partial M}{\partial t} = \frac{\partial N}{\partial t}

The partial derivatives are equal, then, the differential equation is exact.

In order to obtain the solution of the equation you first integrate M or N:

F(t,y)=\int N \partial y = (2 +7e^t)y+g(t)        (1)

Next, you derive the last equation respect to t:

\frac{\partial F(t,y)}{\partial t}=7ye^t+g'(t)

however, the last derivative must be equal to M. From there you can calculate g(t):

\frac{\partial F(t,y)}{\partial t}=M=(7y-3t)e^t=7ye^t+g'(t)\\\\g'(t)=-3te^t\\\\g(t)=-3\int te^tdt=-3[te^t-\int e^tdt]=-3[te^t-e^t]

Hence, by replacing g(t) in the expression (1) for F(t,y) you obtain:

F(t,y)=(2+7e^t)y+3(1-t)e^t +C

where C is the constant of integration

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