Respuesta :

CastleRook
If f(4) = 246.4 when r = 0.04 for the function f(t) = Pe^rt , then what is the approximate value of P?
Solution:
f(t)=Pe^rt
when r=0.04, the f(t)=246.4, thus the value of P will be:
246.4=Pe^0.04t
P=(246.4)/(0.04t)
where t=time
SociometricStar

Answer:

209.96

Step-by-step explanation:

The given function is [tex]f(t)=Pe^{rt}[/tex]

Given that f(4) = 246.4 when r = 0.04. Thus, we have

[tex]f(4)=Pe^{r(4)}\\\\246.4=Pe^{0.04\cdot4}[/tex]

Simplifying the exponent

[tex]246.4=Pe^{0.16}[/tex]

Now, divide both sides by [tex]e^{0.16}[/tex]

[tex]\frac{246.4}{e^{0.16}}=P\\\\P=209.97[/tex]

Hence, the approximate value of P is 209.96

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