Mathematics, 13.06.2020 04:57 celeste961

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) measures the performance of someone learning a skill after a training time t, M is the maximum level of performance, and k is a positive constant. Solve this differential equation to find an expression for P(t). (Use P for P(t). Assume that P(0) = 0.)

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Answer from: scavalieri2421

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

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:

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

Answer from: Quest

I'll say b) log(- that if not let me know what i can do
Asap 25 pts + brainliest to right/best answer

Answer from: Quest

answer: e 9, 5.75

step-by-step explanation:

step 1

find the variation ab in the x coordinate

delta x=(x2-x1)=10-> 8 units

find the variation ab in the y coordinate

delta y=(y2-y1)=6-> 2 units

step 2

find the coordinates point c

x coordinate

cx=ax+delta x*2/> cx=2+8*(2/> cx=3

y coordinate

cy=ay+delta y*2/> cy=4+2*(2/> cy=4.25

the coordinates of point c are (3,4.25)

step 3

find the coordinates point d

x coordinate

dx=ax+delta x*4/> dx=2+8*(4/> dx=4

y coordinate

dy=ay+delta y*4/> dy=4+2*(4/> dy=4.5

the coordinates of point d are (4, 4.5)

step 4

find the coordinates point e

x coordinate

ex=ax+delta x*4/> ex=2+8*(6/> ex=5

y coordinate

ey=ay+delta y*4/> ey=4+2*(6/> ey=4.75

the coordinates of point e are (5, 4.75)

step 5

find the coordinates point f

x coordinate

fx=ax+delta x*4/> fx=2+8*(8/> fx=6

y coordinate

fy=ay+delta y*4/> fy=4+2*(8/> fy=5

the coordinates of point f are (6,5)

step 6

find the coordinates point g

x coordinate

gx=ax+delta x*4/> gx=2+8*(10/> gx=7

y coordinate

gy=ay+delta y*4/> gy=4+2*(10/> gy=5.25

the coordinates of point g are (7,5.25)

step 7

find the coordinates point h

x coordinate

hx=ax+delta x*4/> hx=2+8*(12/> hx=8

y coordinate

hy=ay+delta y*4/> hy=4+2*(12/> hy=5.5

the coordinates of point h are (8,5.5)

step 8

find the coordinates point i

x coordinate

ix=ax+delta x*4/> ix=2+8*(14/> ix=9

y coordinate

iy=ay+delta y*4/> iy=4+2*(14/> iy=5.75

the coordinates of point i are (9,5.75)

- r3ktforgood ☕

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