What is 41 multiplied by 10 to the 0 power?
2 answers:
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Answer:
Collin: about $401 thousand
Cameron: about $689 thousand
Step-by-step explanation:
A situation in which doubling time is constant is a situation that can be modeled by an exponential function. Here, you're given an exponential function, though you're not told what the variables mean. That function is ...
In this context, P0 is the initial salary, t is years, and d is the doubling time in years. The function gives P(t), the salary after t years. In this problem, the value of t we're concerned with is the difference between age 22 and age 65, that is, 43 years.
In Collin's case, we have ...
P0 = 55,000, t = 43, d = 15
so his salary at retirement is ...
P(43) = $55,000(2^(43/15)) ≈ $401,157.89
In Cameron's case, we have ...
P0 = 35,000, t = 43, d = 10
so his salary at retirement is ...
P(43) = $35,000(2^(43/10)) ≈ $689,440.87
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Sometimes we like to see these equations in a form with "e" as the base of the exponential. That form is ...
If we compare this equation to the one above, we find the growth factors to be ...
2^(t/d) = e^(kt)
Factoring out the exponent of t, we find ...
(2^(1/d))^t = (e^k)^t
That is, ...
2^(1/d) = e^k . . . . . match the bases of the exponential terms
(1/d)ln(2) = k . . . . . take the natural log of both sides
So, in Collin's case, the equation for his salary growth is
k = ln(2)/15 ≈ 0.046210
P(t) = 55,000e^(0.046210t)
and in Cameron's case, ...
k = ln(2)/10 ≈ 0.069315
P(t) = 35,000e^(0.069315t)
At any time (min), the volume of solution in the tank is
If is the amount of salt in the tank at any time
, then the solution has a concentration of
.
The net rate of change of the amount of salt in the solution, , is the difference between the amount flowing in and the amount getting pumped out:
Dropping the units and simplifying, we get the linear ODE
Multiplying both sides by allows us to identify the left side as a derivative of a product:
Integrate and divide both sides by to get
The tanks starts off with 30 lb of salt, so and we can solve for
to get a particular solution of