Pat bought a boat for $37,000 in 2001. in 2006 the boat was worth $27,000. if the boat depreciation is linear, how much will the boat be worth in 2009?
a) $27,000
b) $22,000
c) $21,000
d) $20,000

C) $21,000

Step-by-step explanation:

Given : Pat bought a boat for $37,000 in 2001. In 2006 the boat was worth $27,000.

If the boat depreciation is linear, then the amount by which the value of boat depreciates must be constant.

Let x be the constant depreciation in the value of boat per year.

Then , the value of boat (in dollars) after n years from 2001 is given by :-

               (1)

For year 2006 , n=5 and V = 27000

Then ,

i.e.  

i.e.  

i.e.  

Thus , the constant amount of depreciation in the value of boat per year. = $2000

Now for year 2009 , put in n=8 and x= 2000 in (1), we get

Hence, the value of boat  in 2009 = $21,000

f(x) is going down by 4

step-by-step explanation:

(its either that or its going left by 4)

the answer is 2/54

step-by-step explanation: