PLEASE!!!! I NEED HELP!!!!!!
Ron buys a lawnmower for $1,500. The salesperson says the value will depreciate about 30% per year over the next few years. However, his neighbor says it is likely to depreciate about $300 per year.

Which system could be used to determine when the two depreciation models will give the same value for the lawnmower?

y = 1,500(0.3)^x and y = 1,500 - 300x
y = 1,500(0.7)^x and y = 300 - 1,500x
y = 1,500(0.7)^x and y = 1,500 - 300x
y = 1,500(0.3)^x and y = x - 300

Respuesta :

your answer should be y = 1,500(0.7)^x and y = 1,500 - 300x (c)
hope this helped :) good day


Answer: [tex]y = 1,500(0.7)^x\ \text{and }y = 1,500 - 300x[/tex]

Step-by-step explanation:

Given : Ron buys a lawnmower for $1,500. The salesperson says the value will depreciate about 30% per year (r=0.3) over the next few years.

The exponential decay equation is given by :-

[tex]y=A(1-r)^x[/tex], where A is the initial amount , r is rate of decay and x is the time period.

Then , the equation of depreciation for Ron :-

[tex]y=1500(1-0.3)^x=1500(0.7)^x[/tex]

However, his neighbor says it is likely to depreciate about $300 per year, which is linear depreciation.

The linear equation is given by :-

[tex]y=ax+c[/tex], where 'c' is the initial amount and 'a' is the  rate of change.

Then equation of depreciation for his neighbor :-

[tex]y=-300x+1500=1500-300x[/tex]

Thus , the system could be used to determine when the two depreciation models will give the same value for the lawnmower:-

[tex]y = 1,500(0.7)^x\ \text{and }y = 1,500 - 300x[/tex]

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