So.. in this case, the starting amount is the 500mg sample... and the rate of decay, negative rate, is 1%, and at the time, the elapsed days is 0, to t = 0, P = 400
Answer:
the constant rate of change is 20
Factoring the expression would be 6a(2a^2+4a-1)
Answer:
i think there was a typo bc u didnt put the number of zoo animals
Step-by-step explanation:
The correct question is
Which is the best approximation to a solution of the equation
e^(2x) = 2e^{x) + 3?
we have that
e^(2x) = 2e^{x) + 3-----------> e^(2x)- 2e^{x) - 3=0
the term
e^(2x)- 2e^{x)----------> (e^x)²-2e^(x)*(1)+1²-1²------> (e^x-1)²-1
then
e^(2x)- 2e^{x) - 3=0--------> (e^x-1)²-1-3=0------> (e^x-1)²=4
(e^x-1)=2--------> e^x=3
x*ln(e)=ln(3)---------> x=ln(3)
ln(3)=1.10
hence
x=1.10
the answer is x=1.10
