B) $22,000
C) $21,000
D) $20,000
Answer: 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
okay so you would do it like this
2x + (3 - x)
2x + 3 - x
x + 3
the answer would be x + 3
hope this helps!
Following PEMDAS, the parentheses comes first. Essentially, (3 - x) is times 1, which removes the parentheses. So 2x + 3 - x. Subtract the x first. This leaves x + 3.