Respuesta :
Answer:
a. v = 40 000 (3/ 5)^d
b. v = 40 000 (3/5)^(4/10)
c. 0.95
Explanation:
The exponential growth is modelled by
[tex]v=A(b)^d[/tex]We know that points (0, 40 000) and (1, 24 000) lie on the curve. This means, the above equation must be satsifed for v = 40 000 and d = 0. Putting v = 40 000 and d = 0 into the above equation gives
[tex]40\; 000=Ab^0[/tex][tex]40\; 000=A[/tex]Therefore, we have
[tex]v=40\; 000b^d[/tex]Similarly, from the second point (1, 24 000) we put v = 24 000 and d = 1 to get
[tex]24\; 000=40\; 000b^1[/tex][tex]24\; 000=40\; 000b^{}[/tex]dividing both sides by 40 000 gives
[tex]b=\frac{24\; 000}{40\; 000}[/tex][tex]b=\frac{3}{5}[/tex]Hence, our equation that models the situation is
[tex]\boxed{v=40\; 000(\frac{3}{5})^d\text{.}}[/tex]Part B.
Remember that the d in the equation we found in part A is decades. Since there are 10 years in a decade, we can write
t = 10d
or
d = t/10
Where t = number of years
Making the above substitution into our equation gives
[tex]v=40\; 000(\frac{3}{5})^{\frac{t}{10}}[/tex]Therefore, the car's value at t = 4 is
[tex]\boxed{v=40\; 000(\frac{3}{5})^{\frac{4}{10}}}[/tex]Part C:
The equation that gives the car's value after t years is
[tex]v=40\; 000(\frac{3}{5})^{\frac{t}{10}}[/tex]which using the exponent property that x^ab = (x^a)^b we can rewrite as
[tex]v=40\; 000\lbrack(\frac{3}{5})^{\frac{1}{10}}\rbrack^t[/tex]Since
[tex](\frac{3}{5})^{\frac{1}{10}}=0.95[/tex]Therefore, our equation becomes
[tex]v=40\; 000\lbrack0.95\rbrack^t[/tex]This tells us that the car's value is changing by a factor of 0.95 each year.
