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For accounting purposes, the value of assets (land, buildings, equipment) in a business depreciates at a set rate per year. The value, V(t), of $408,000 worth of assets after t years, which depreciate at 18% per year, is given by the formula V(t) = V0(b)t. What is the value of V0 and b, and when rounded to the nearest cent, what is the value of the assets after 8 years?

Respuesta :

squareruth04
V(t) = $408,000 - ( $408,000 x 18%)
       = $334,560 - ($334,560 x 18%)
       = $274,339.20 - ($274,339.20 x 18%)
       = $224,958.14 - ($224,958.14 x 18%)
       = $184,465.68 - ($184,465.68 x 18%)
       = $151,261.86 - ($151,261.86 x 18%)
       = $124,034.73 - ($124,034.73 x 18%)
       = $101,708.48 - ($101,708.48 x 18 %)
        = $83,400.95

The new carrying value of the asset on the current year is deducted with the depreciation rate to get the carrying value of the next year
carlosego
For this case we have an equation of the form:
 
[tex] V (t) = V0 * (b) ^ t [/tex] 
 Where,
 v0: initial value in assets
 b: depreciation rate
 t: time in years.
 Substituting values we have:
[tex] V (t) = 408000 * (0.82) ^ t [/tex]
 For year 8 we have:
[tex] V (t) = 408000 * (0.82) ^ 8 V (t) = 83400.94703[/tex]
 Rounding off we have:
 V (t) = 83401
 Answer:
 the value of V0 and b are:
 V0 = $ 408,000
 b = 0.82
 the value of the assets after 8 years is:V (t) = 83401 $

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