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:
17.4 ml is equivalent to 1.74e-5 kl.
Hope this helps! ^_^
The nearest degree is b. 0 degrees
6x^2 - x - 40 =(<span>3x - 8)(2x + 5)
and
</span><span>5 + 2x = 2x + 5
</span><span>
so
</span>(6x^2 - x - 40)<span>÷ (5 + 2x)
= [</span>(3x - 8)(2x + 5)] ÷ (5 + 2x)
= 3x -8
answer
3x - 8