Simple, no. This would not be enough.
This is because you mentioned that there are 20 students and each of them needed 2 plates. So, there would need to be at least 40 plates. Forks seem irrelevant in this question but if the teacher has 8 plates in 1 box and another 8 in the second box, that would sum up to 16 plates that are available. And the fact that the box doesn't even include the other items, should hint the lack of items available for the students.
This question seemed worded differently. But tried my best. :)
Please re ask
this is not a valid anwser or question please ask again in a simplified form
19 bears, 38 boxes of candy.
We can use a system of equations (b = # of bears, c = # of candy boxes):
<em>b + c = 57, 8b + 6c = 380</em>.
We can use substitution by isolating b in the first eq.: <em>b = 57 - c</em>, then substitution b with 57 - c in the second eq.: <em>8(57 - c) + 6c = 380</em>.
Then, we solve: 456 - 8c + 6c = 380, 76 = 2c, c = 38 candy boxes.
Finally we plug in 38 for c in the first eq.: b + 38 = 57, b = 19 bears.
Assuming the units on the graphs increase by 1, the points marked on the first graph are (-2, 1) and (2, -3), and the slope is -1/2; the points marked on the second graph are (-2, -2) and (3, 4), and the slope is 6/5; and the points marked on the third graph are (4, 0) and (1, -3), and the slope is -1.
Hope this helps.