Answer:
The value at the end of year 2 is $4400.
Step-by-step explanation:
The best approach here is to determine the expression for the line depreciation and then calculate the depreciation value at x = 2 years.
A line is given by

where m is the slope and b the bias (aka y-intercept). You can determine both directly from what is given. The slope is change in y divided by change in x. We know that over 5 years the car loses (500-7000)=-6500 in value. So, the slope is m=-6500/5 (note the negative sign). At time 0, the y-intercept is 7000, since that is the initial value (at year 0). So our line function is fully identified:

and gives you the value of the car in any given year. To answer the question, we now plug in 2 as value of x:

Answer:
Step-by-step explanation:
Outlier
X+y=4 :x=4-y,yR
x-y=6:x=6+y,yR
Answer:
Last equation given in the list of possible answers:
5 ( 1.5 + 1.5 + x ) = 25
Step-by-step explanation:
We need to include in the total addition of miles ridden during the week:
a) 1.5 miles to the school
b) 1.5 miles from school back home
c) x miles for the evening ride
so for the miles ridden per day we have: "1.5 +1.5 + x"
Now, since per week she does 5 days like this, then we need to multiply the expression above by 5 in order to total the number of miles she rides weekly (25 miles)
5 ( 1.5 + 1.5 + x ) = 25
And we can use this equation to find the amount "x" that Rin rides in the evening.