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ExtremelyBoredUser

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For C:

the equation would be P=1000e^(1/10*ln(2.5)t

so you would simply to substitute t =23 since 1995 is start and 2018 is what you're finding (2018-1995) = 23 yrs into the equation to find P:

P = 1000e^(23/10*ln(2.5))
= 8227.388777
therefore 8230

For e:

You use the value you get in d and then you would input that t value (t=25) into the differential equation dP/dt which would be k(1000e^(kt)) or kP and then you would just calculate the value of the rate of change.

dP/dt = k(1000e^25k) = 905

My maths might be a bit dodgy since its 11 here so I think someone here else could format this more nicely.
 

stressedadfff

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For C:

the equation would be P=1000e^(1/10*ln(2.5)t

so you would simply to substitute t =23 since 1995 is start and 2018 is what you're finding (2018-1995) = 23 yrs into the equation to find P:

P = 1000e^(23/10*ln(2.5))
= 8227.388777
therefore 8230

For e:

You use the value you get in d and then you would input that t value (t=25) into the differential equation dP/dt which would be k(1000e^(kt)) or kP and then you would just calculate the value of the rate of change.

dP/dt = k(1000e^25k) = 905

My maths might be a bit dodgy since its 11 here so I think someone here else could format this more nicely.
thank you the answers correct but i dont understand its 1/10 and not just 10 as t?
 

ExtremelyBoredUser

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View attachment 31227
could i also get a solution for b please
So B asks you to find k. Half of the salt left would mean that S at t = 3 would equal initial value of S divided by 2. To find initial value of S, make t = 0;

S(initial) = 20e^0 = 20.

Therefore the value of S after 3 mins would be 20/2 which would be 10. Then substitute the values into the equation, t = 3 and S = 10;

10 = 20e^-3k
1/2 = e^-3k

(taking natural log of both sides)

ln(1/2) = -3k and 1/2 = 2^-1 and using log laws we can rearrange ln(1/2) -> ln(2^-1) -> -ln(2)

Then -ln(2) = -3k

k = (1/3)*ln(2)
 

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