When you graph an equation, it should be dependent variable against the independent variable. For this problem, the independent variable is the time, so this is along the x-axis. The dependent variable is d, so this is along the y-axis. Since the slope is Δy/Δx, then it is also equivalent to Δd/Δt. Therefore, the answer is B.
The answer to number seven is EDF!
Answer:x= 5
Step-by-step explanation:
5x-12=63
>5x=63+12=75
>x=75÷5
>x=15
Answer:

Step-by-step explanation:
percentage of graduates with loan = 62%
total sample = 50
Number of student in the sample with student loan
= (percentage of graduates with loan) x (total sample)
= 62% x 50
= 31
Proportion of student in the sample with student loan = 
You can use the trigonometric identity

.

The requirement that

eliminates -1/6 from being another solution.