a. Nahla can purchase school lunches for approximately 5.5 weeks with the money she deposited at the beginning of the quarter.
b. algebraically w = 5.56
a.
We notice that the intersection point appears to be slightly past the 5-week mark on the x-axis but not quite reaching the 6-week mark we can estimate that Nahla can purchase lunch for about 5.5 weeks.
b.
100 - 18w = 0
We can then solve for w and round the answer to the nearest hundredth.
we now add 18w to both sides:
100 = 18w
Divide both sides by 18:
w = 100 / 18
w = 5.555555556...
w = 5.56
Therefore, by solving the equation algebraically, we confirm that Nahla can purchase lunch for approximately 5.56 weeks (rounded to the nearest hundredth), which closely matches our estimated value of 5.5 weeks from the graph.