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blsea [12.9K]
2 years ago
9

Which is a solution of the inequality?p + 4 – 2(p – 10) > 0

Mathematics
2 answers:
boyakko [2]2 years ago
5 0

Hi Laila


p+4 -2(p-10)>0

The first thing you need to do is simplify the both sides

So that will be

p-2p + 4+ 20 > 0

-p + 24 > 0

-p > 0-24

-p > -24

Now since we can not have the negative sign in front of P, we need to do something. What we need to do is divide both sides by -1

-p/-1 > -24/-1

p < 24

Remember to change the symbol.

I hope that's help and if you have any further questions please let me know :)

san4es73 [151]2 years ago
3 0

Answer:

I got 24 as being WRONG! ITS NOT 24!! the other answer is wrong! its dumb that it got verified, verified by who!


This is the REAL ANSWER: i don't know, just pick 26 or something. i looked and looked all over the internet and no one had any other answer other than 24! so yeah..... just guess and if you get it right please leave a comment to say the real answer. I already got rid of 1 for you. GOOD LUCK!  33.33% chance!

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

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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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