D(5)= 50e^-2

How would I solve it from here?

1 answer

Uh......what...?

Similar Questions

Solve for x. 50e^-0.12x = 10

Top answer: 50e^-0.12x = 10 e^-0.12x = .2 take ln of both sides and use log rules -.12x ln e = ln .2, remember Read more.
A student is in a 20 week calc class. The function f(t)=50-50e^0.2t f(t) measures the students problem speed in t weeks-how many

Top answer: at the beginning --- > t = 0 f(0) = 50 - 50e^0 = 0 half-way through course --- > t=10 f(10) = 50 - Read more.
A student is in a 20 week calc class. The function f(t)=50-50e^0.2t f(t) measures the students problem speed in t weeks-how many

Top answer: Did you not look at and read my reply? http://www.jiskha.com/display.cgi?id=1342766530 Once you have Read more.
Let p(x)=90/9+50e ^ −x .What is p(5) ?

Top answer: To find p(5), we need to substitute x = 5 into the equation p(x) = 90 / (9 + 50e^(-x)). So, let's Read more.
solve each equation algebraically and check it by substituing into the orignall equation.50e^0.035x=200 3LN(x-3)+4=5 Method of

Top answer: 50e^(0.035x)=200 e^(.035x) = 4 take ln of both sides ln(e^(.035x)) = ln4 .035x lne = ln4, but lne = Read more.
50e-.367x=125

Top answer: Divide both sides by 50 and take log (to base e). Solve for x. Read more.
If y=5e^5x^2 find the double derivativeHow is it 50e^5x^2(10x^2+1)

Top answer: remember to use the product rule and the chain rule y = 5e^(5x^2) y' = 5(e^(5x^2)*10x) = 50xe^(5x^2) Read more.
in a parallel RLC circuit you haveL=100mH and i= inductor current i=50e^(-4000t)-10e^(-5000t) mA how do you find R and C??

Top answer: To find the values of R (resistance) and C (capacitance) in a parallel RLC circuit, you need Read more.
I am confused on how to write this function as an exponential function having a base 10:P(t)= 50e^(0.5t). The answer in my

Top answer: we want to change e^(.5t) to 10^(kt) let e^(.5t) = 10^(kt) log it log[e^(.5t)] = log [10^(kt)] Read more.
A 4.50e-2 M solution of unknown monoprotic weak base has a pH of 10.200 what is the value of kb for the base

Top answer: BOH ==> B^+ + OH^- Kb = (B^+)(OH^-)/(BOH) pH = 10.2; therefore, pOH = 3.8 pOH = -log(OH^-). Solve Read more.