## anonymous one year ago A proportional relationship between the number of pounds of cabbage (x) and the price in dollars (y) is graphed, and the ordered pair (5, 2) is on the graphed line. Part A: What is the price of 1 pound of cabbage? Show your work, including the proportion you used to determine the price. Part B: What does the ordered pair (10, 4) on the graph represent? Explain in words.

1. anonymous

@Z4K4R1Y4

2. Z4K4R1Y4

you'll need to post the graph.

3. anonymous

There is no graph....

4. anonymous

@Z4K4R1Y4

5. anonymous

@Michele_Laino @hartnn @Hero

6. Z4K4R1Y4

(5,2) means 5 pounds of cabbage costs $2. so to get 1 pound you divide the weight of cabbage by 5. then to get the price of 1 pound you divide the price by 5. 7. anonymous Ok I don't get it 8. Z4K4R1Y4 |dw:1432752532427:dw| 9. anonymous Can u just give me the answer 10. anonymous in essay form 11. anonymous Sorry but I'm horrible at math 12. Z4K4R1Y4 5 pounds divided by 5 = 1 pound.$2 divided by 5 = \$0.4

13. Michele_Laino

we can write the subsequent proportion: $\Large 2:5 = p:1$ p is the unit price

14. anonymous

Im sorry i still don't get it

15. Michele_Laino

Hint: if 5 pounds of cabbage cost 2 dollars, then 1 pound of cabbage costs 2/5 dollars

16. anonymous

ok

17. Michele_Laino

so we can write this formula: $p = \frac{2}{5}$

18. anonymous

ok

19. Michele_Laino

now, if we apply the fundamental property to this proportion: $2:5 = p:1$ we get: $5p = 2$

20. anonymous

ok

21. Michele_Laino

next I divide both sides of that last formula, by 5, so I can write: |dw:1432753316992:dw|

22. anonymous

ok

23. Michele_Laino

which is the same equation which I wrote before, at the beginning. So we can conclude, that the proportion: $2:5 = p:1$ is the right proportion

24. anonymous

ok So now how would i write it in essay form

25. Michele_Laino

a possible statement is: "since 5 pounds of cabbage cost 2 dollars, then 1 pound of cabbage costs 2/5 of dollars, or 40 cent"

26. anonymous

ok I'm sorry i am not in a school mood to day can you help with one more question

27. Michele_Laino

ok!