# The exponential decay graph shows the expected depreciation for a new boat, selling for $3500, over 10 years a. write an exponential The exponential decay graph shows the expected depreciation for a new boat, selling for$3500, over 10 years a. write an exponential function for the graph. b. use the function in part a to find the value of the boat after 9.5 years.

The general equation for exponential growth and decay is:

A = Ao x (1+r)^t

where
t  = number of units of time.
A = the amount after "t" units of time.
Ao =  The initial amount

Hence the final equation after substituting :
A = 3500 x (1+r)^10

(a) (b) $245.27 Explanation: (a) From the below graph it is clear that the graph it passes through the points (0,3500) and (2,2000). The general form of an exponential function is where, a is the initial value and b is growth or decay factor. Initial value is 3500, it means a=3500. f(x)=2000 at x=2.     The exponential function for the graph is (b) We need to find the value of the boat after 9.5 years. Substitute x=9.5 in the above function.   Therefore, the value of the boat after 9.5 years is$245.27. y = 3500 e^(-k⋅ 9.5)

Step-by-step explanation:

The exponential decay graph shows the expected depreciation for a new boat, selling for $3500, over 10 years a. Write an exponential function for the graph. b. Use the function in part a to find the value of the boat after 9.5 years. Explanation: Exponential equation is given by y = 3500 ⋅ e^( k 9.5) whereby: y : value A : constant; k : rate of change t : time value In this when t =0 3500= A ⋅ e^ k 0 3500 = A after 10 years we have , after 9.5 years, the value of the boat is: y = 3500 e^(-k⋅ 9.5) k is the rate of change and it shows that it is negative because there is a depreciation in value. Note that the rate of change is not given in this case. For the answer to the question above, V(n) = a * b^n, where V(n) shows the value of boat after n years. V(0) = 3500 V(2) = 2000 n = 0 V(0) = a * b^0 = 3500 a = 3500 V(2) = a * b^2 2000 = 3500 * b^2 b = sqrt (2000/3500) b ≈ 0.76 V(n) = 3500 * 0.76^n We can check it for n = 1 which is close to 2500 in the graph: V(1) = 3500 * (0.76)^1 V(1) = 2660 And in the graph we have V(3) ≈ 1500, V(n) = 3500 * (0.76)^3 ≈ 1536 Now n = 9.5 V(9.5) = 3500 * (0.76)^(9.5) V(9.5) ≈ 258 Part a: the formular for the exponential decay of a function is given by where f(t) is the value of the function after time t, is the initial value of the function (i.e. the value of the function at time t = 0), k is the decay constant and t is the time passed.given the graph showing the expected depreciation for a new boat, selling for$3500, over 10 years.thus, .from the graph, the value of the boat after 2 years is $2,000, thus we have therefore, the exponential function for the graph is given by part b: the value of the boat after 9.5 years is given by therefore, the value of the boat after 9.5 years is$245.27

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