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.

Answers

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) f(x)=3500(frac{2}{sqrt{7}})^x

(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

f(x)=ab^x

where, a is the initial value and b is growth or decay factor.

Initial value is 3500, it means a=3500.

f(x)=3500b^x

f(x)=2000 at x=2.

2000=3500b^2

frac{2000}{3500}=b^2

frac{4}{7}=b^2

sqrt{frac{4}{7}}=b

frac{2}{sqrt{7}}=b

The exponential function for the graph is

f(x)=3500(frac{2}{sqrt{7}})^x

(b)

We need to find the value of the boat after 9.5 years.

Substitute x=9.5 in the above function.

f(9.5)=3500(frac{2}{sqrt{7}})^{9.5}

f(9.5)=245.26598

f(9.5)approx 245.27

Therefore, the value of the boat after 9.5 years is $245.27.


The exponential decay graph shows the expected depreciation for a new boat, selling for $3500, over

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

y=3500e^{-k10}

, 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 byf(t)=f_0e^{-kt}where f(t) is the value of the function after time t, f_0 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, f_0.from the graph, the value of the boat after 2 years is $2,000, thus we havef(9.5)=3500e^{-0.2798times9.5} \ \ =3500e^{-2.658}=3500(0.0701) \ \ =245.27therefore, the exponential function for the graph is given byf(9.5)=3500e^{-0.2798times9.5} \ \ =3500e^{-2.658}=3500(0.0701) \ \ =245.27part b: the value of the boat after 9.5 years is given byf(9.5)=3500e^{-0.2798times9.5} \ \ =3500e^{-2.658}=3500(0.0701) \ \ =245.27therefore, the value of the boat after 9.5 years is $245.27

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