Agroup of friends will be buying at most 10 items at the concession stand at the high school football game. they have no more than $25 to spend at the game. each hot food item is #3 and each cold food item is $2. the system of inequality graphed below represents this information. which combination of hot and cold food items could the group buy? 0 hot items and 11 cold items, 9 hot items and 1 cold item, 2 hot items and 8 cold items, 6 hot items and 4 cold items

0 hot items and 11 cold items

Step-by-step explanation:

He has 25 dollars to spend.

Hot: 3

Cold: 2

Let's eliminate some wrong answers.  

1. 0 hot items and 11 cold items: Possible, because 11 times 2 is 22.

2. 9 hot items and 1 cold item: Impossible, because 9 times 3 = 27 and that is already over 25 dollars. This one is eliminated.

3. 2 hot items and 8 cold items: 2 times 3 is 6. 8 times 2 is 16. 16 + 3 = 19. He buys 10 items here. Selection one bought 11 items though, so this one is eliminated.

4. 6 hot items and 4 cold items: 6 times 3 is 18, and 4 times 2 is 8. 18 + 8 = 26. He doesn't have enough money, so this one is eliminated.

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

this is not the answer this is just a question. is this middle school math?

step-by-step explanation: