a baseball card sold for $205 in 1979 and was sold again in 1985 for $488. assume that the \na) find the…

a baseball card sold for $205 in 1979 and was sold again in 1985 for $488. assume that the \na) find the value k of the exponential growth rate. assume ( v_0 = 205 ).\n( k = 0.145 )\n(round to the nearest thousandth.)\nb) find the exponential growth function in terms of t, where t is the number of years since 197\n( v(t)=205 e^{0.145 t} )\nc) estimate the value of the baseball card in 2013.\n$ square )\n(round to the nearest dollar.)

a baseball card sold for $205 in 1979 and was sold again in 1985 for $488. assume that the \na) find the value k of the exponential growth rate. assume ( v_0 = 205 ).\n( k = 0.145 )\n(round to the nearest thousandth.)\nb) find the exponential growth function in terms of t, where t is the number of years since 197\n( v(t)=205 e^{0.145 t} )\nc) estimate the value of the baseball card in 2013.\n$ square )\n(round to the nearest dollar.)

Answer

Explanation:

Step1: Calculate the number of years from 1979 to 2013

The number of years (t = 2013 - 1979=34)

Step2: Substitute (t = 34) into the growth function (V(t)=205e^{0.145t})

[ \begin{align*} V(34)&=205e^{0.145\times34}\ &=205e^{4.93}\ \end{align*} ] We know that (e^{4.93}\approx138.87) (using a calculator for the exponential function)

Step3: Calculate the value of (V(34))

[ \begin{align*} V(34)&=205\times138.87\ &=205\times(138 + 0.87)\ &=205\times138+205\times0.87\ &=28290+178.35\ &=28468.35 \end{align*} ]

Answer:

(28468)