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2 years ago

I want answer. I need t’s solution.And I want explanation too.

Mathematics

1 answer:

monitta2 years ago

5 0

we have the equation

$log(\frac{1,000}{253.95})=12tlog(1+\frac{0.08}{12})$

Solve for t

That means

Isolate the variable t

$t=\frac{log(\frac{1,000}{253.95})}{12log(1+\frac{0.08}{12})}$ $t=17.2$

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3 years ago

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car 2: linear

B) car 1: $f(x)=18500(0.94)^{x-1}$

car 2: $f(x)=-1000x+19500$

C) Yes, there will be a significant difference of $1,100.40 which is approx 10% of the value of the cars after 10 years.

Step-by-step explanation:

Part A

Car 1: exponential as the projected value is not decreasing by the same amount each year

Car 2: linear as the projected value decreases by the same amount each year ($1000)

Part B

Car 1

General form of exponential function: $y=ab^x$

a = initial value = 18500

b = growth factor = $\dfrac{17390}{18500}=0.94$

$\implies f(x)=18500(0.94)^{x-1}$

NB We have to change x to "x-1" since we are told in the table that after 1 year the car's value is 18500.

Car 2

General form of linear function: $y=mx+b$

$m=\dfrac{\textsf{change in} \ y}{\textsf{change in} \ x}=\dfrac{17500-18500}{2-1}=-1000$

$y-y_1=m(x-x_1)$

$\implies y-17500=-1000(x-2)$

$\implies f(x)=-1000x+19500$

Part C

Car 1 after 10 years: $\implies f(10)=18500(0.94)^{10-1}=\$10,600.40$

Car 2 after 10 years: $\implies f(10)=-1000(10)+19500=\$9,500$

Difference = 10600.40 - 9500 = 1100.40

Yes, there will be a significant difference of $1,100.40 which is approx 10% of the value of the cars after 10 years.

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