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Juliette [100K]

3 years ago

10

I don't know how to do Pythagorean theorem

2 answers:

Darya [45]3 years ago

8 0

The Pythagorean theorem is

A sq x b sq = h sq

Sq means squared

H is the hypotenuse or longest side or the triangle

A and b are the shorter sides that make a 90 degree angle

romanna [79]3 years ago

6 0

One of the best ways to understand something in math is to watch videos of people explaining it on youtube. Here are a coulle videos, hopefully one of them helps :)

https://youtu.be/AA6RfgP-AHU

https://youtu.be/JDmglbduKvE

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2 years ago

What is 3n-22= -20

viva [34]

Answer:

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Step-by-step explanation:

3n - 22 = -20

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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In a previous exercise we formulated a model for learning in the form of the differential equation dP dt = k(M − P) where P(t) m

GalinKa [24]

Answer:

$\frac{dP}{M-P}= kdt$

And we can integrate both sides of the equation using the following substitution:

$u= M-P, du =-dP$

And replacing we got:

$\int -\frac{du}{u} = kt +C$

$-ln (u)= kt+c$

If we multiply both sides by -1 we got:

$ln (u ) = -kt -c$

$ln (M-P) = -kt -c$

And using exponential in both sides of the equation we got:

$M-P = e^{-kt} e^{-c}$

And solving for P we got:

$P(t) = M- e^{-kt}e^{-c}$

And replacing $P_o =e^{-c}$ we got:

$P(t) = M - P_o e^{-kt}$

We can use the condition $P(0)=0$ and we got:

$0 = M -P_o e^0$

And we see that $M = P_o$ and replacing we got:

$P= M(1- e^{-kt})$

Step-by-step explanation:

For this case we aasume the following differential equation:

$\frac{dP}{dt}= k(M-P)$

Is a separable differential equation so we can do the following procedure:

$\frac{dP}{M-P}= kdt$

And we can integrate both sides of the equation using the following substitution:

$u= M-P, du =-dP$

And replacing we got:

$\int -\frac{du}{u} = kt +C$

$-ln (u)= kt+c$

If we multiply both sides by -1 we got:

$ln (u ) = -kt -c$

$ln (M-P) = -kt -c$

And using exponential in both sides of the equation we got:

$M-P = e^{-kt} e^{-c}$

And solving for P we got:

$P(t) = M- e^{-kt}e^{-c}$

And replacing $P_o =e^{-c}$ we got:

$P(t) = M - P_o e^{-kt}$

We can use the condition $P(0)=0$ and we got:

$0 = M -P_o e^0$

And we see that $M = P_o$ and replacing we got:

$P= M(1- e^{-kt})$

8 0

3 years ago