the declining - balance method of depreciation is an accounting method businesses use to deduct most of the…

the declining - balance method of depreciation is an accounting method businesses use to deduct most of the cost of new equipment during the first few years of purchase. unlike other methods, the declining - balance formula does not consider salvage value. for the value of a crane, the function ( v(t)=375000(1 - 0.25)^{t} ), where ( v ) is value and ( t ) is time in years, can be used to find the value of the crane for the first 9 years of use.\na. what is the value if the crane after 4 years and 9 months? the value of the crane will be $.\nb. state the domain of the function.\n( leq tleq )\nc. state the range of this function.\n( leq v(t)leq )

the declining - balance method of depreciation is an accounting method businesses use to deduct most of the cost of new equipment during the first few years of purchase. unlike other methods, the declining - balance formula does not consider salvage value. for the value of a crane, the function ( v(t)=375000(1 - 0.25)^{t} ), where ( v ) is value and ( t ) is time in years, can be used to find the value of the crane for the first 9 years of use.\na. what is the value if the crane after 4 years and 9 months? the value of the crane will be $.\nb. state the domain of the function.\n( leq tleq )\nc. state the range of this function.\n( leq v(t)leq )

Answer

Explanation:

Step1: Convert time to years

9 months = $\frac{9}{12}=\frac{3}{4} = 0.75$ years. So $t=4 + 0.75=4.75$ years.

Step2: Substitute $t = 4.75$ into the formula $V(t)=375000(1 - 0.25)^{t}$

$V(4.75)=375000\times(0.75)^{4.75}$. First, $(0.75)^{4.75}=e^{4.75\ln(0.75)}$. $\ln(0.75)\approx - 0.2877$, then $4.75\times(-0.2877)\approx - 1.366$. $e^{-1.366}\approx0.255$. So $V(4.75)=375000\times0.255 = 95625$.

Step3: Determine the domain

Since the function is used for the first 9 years of use, $0\leq t\leq9$.

Step4: Determine the range

When $t = 0$, $V(0)=375000(1 - 0.25)^{0}=375000$. As $t$ increases, $V(t)$ decreases. When $t = 9$, $V(9)=375000\times(0.75)^{9}$. $(0.75)^{9}=e^{9\ln(0.75)}$, $\ln(0.75)\approx - 0.2877$, $9\times(-0.2877)\approx - 2.589$, $e^{-2.589}\approx0.075$. $V(9)=375000\times0.075 = 28125$. So $28125\leq V(t)\leq375000$.

Answer:

a. $95625$; b. $0\leq t\leq9$; c. $28125\leq V(t)\leq375000$