SOLUTION: The formula D(t)= 6e^-0.1t can be used to find the number of milligrams (mg) D of a certain drug that is in a patient’s bloodstream after t hours, assuming that 6 mg of the drug

Question 1149635: The formula D(t)= 6e^-0.1t can be used to find the number of milligrams (mg) D of a certain drug that is in a patient’s bloodstream after t hours, assuming that 6 mg of the drug is administered initially
when t = 0.
Use this data and sketch the graph of the exponential function below. Label the axes.
(Remember, you should have a starting point on the y-axis since time cannot be negative!)

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i think you mean D(t) = 6e^-t, not 6e^-0.
D(t) = 6e^-t is equivalent to D(t) = 6/e^t.
when t = 0, this becomes D(0) = 6/e^0 = 6/1 = 6.
note that any variable or constant raised to the power of 0 is equal to 1.
while e is a letter, it represents the scientific constant of 2.718281828.....
the graph of the function looks like this:

$$$

it starts off at 6 and rapidly gets smaller in the beginning, slowing it's descent as t gets larger.
the graph of D(t) = 6 / e^t will approach 0 but never become 0.
for example:
when t = 0, D(t) = 6 / e^0 = 6
when t = 10, D(t) = 6/e^10 = 2.723995786 ^ -4 = .0002732995786
when t = 20, D(tg) = 6/e^20 = 1.236692173 ^ -8 = .00000001236692175.
round to 3 decimal points and both t = 10 and t = 20 get you D(t) = 0.000.
on the graph, you can see that D(10) = 0 which is the same as .000 after you take off the trailing zeroes.
this is because the graph rounds to a maximum of 3 decimal digits.
when t = 1, D(t) = 2.207276647 which rounds to 3 decimal digits of 2.207.
when t = 5, D(t) = .040427682 which rounds to 3 decimal digits of .040 = .04 when you take off the trailing zero.
note that, for graphing purposes, i set y = D(t) and i set x = t.
D(t) = 6 / e^t therefore becomes y = 6 / e^x.

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