hey, could someone help me? solve these problems???

Question

1. A population, P, is increasing exponentially. At time t=0 years, the population is 35,000. In 10 years, the population is 44m 400.

a) Find the growth rate, b, in P(t) = a(b)^t. (note: lil t on top)
b) using the value of b calculated in part a), write an equation that models the population, P, after t years.
c) using your equation, find the population after 25 years.

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2. u want to buy a new car, u have investigated the trade in value of ur current car. three months ago, the trade-in value was 3200$. now it is 3125$- what will be the trade in value of ur car 6 months from now if it is depreciating exponentially.

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3. the temperature of a cooling liquid over time can be modelled by the exponential function: T(x) 60(1/2) ^x/30 + 20. (** 20 is not a little number lol whatever they are called**). where T(x) is the temperature, in degrees Celsius, and x is the elapsed time, in minutes. determine the temperature after 2 hours.

Answer 1

Hello

First one.

Let p = a*b^t putting in the values 35000 = a*b^0 so a = 35000 now p = 35000b^t and putting in the second values gives 44000000 = 35000b^10 so we have

44000000/35000 = b ^10 or taking the tenth root we have b = 2.04 now

p = 35000*2.04^t

So after 25 years we have

p = 35000*2.04^25 = big number

Second one
a = p(1+r/12)^n bis the equation for finding the amount of an investment of p in n periods at rate of r per year so since we are using months we divide r by 12.
So in three months we use

3125 = 3200 (1 - r/12)^3
3125/3200 = (1 - r/12)^3

take the cube root of both sides and we have
0.992126 = 1 - r/12 subtract 1 giving us -0.0078743 = -r/12 now multiply both sides by 12 and we have
r = 0.094449211

Now let's put the values in to find the answer in 6 months from now

a = 3125*(1 - 0.09449211/12)^6 = 3125* 0.953675 = 2980


the third one
t(x) = 60*(.5)^(2*120/30) + 20 = 60 * .5^4 + 20 = 60*0.0625 + 20 = 3.75 + 20 = 23.75




Hope This Helps!