Penelope went to the store to buy x pounds of broccoli for $1.10 per pound and y cans of soup for $2.50 each. in total, she spent less than $10. the inequality relating the purchases she made and the total purchase price is
1.10x + 2.50y < 10. which are reasonable solutions for this situation? check all that apply.

(–1, 4)
(0, 2)
(3, 2.5)
(2, 4)
(0.5, 3.78)
(1.5, 3)

(0,2)

(1.5,3)

Step-by-step explanation:

To determine the reasonable solutions for this inequality we simply substitute the values of x and y into the equation and check whether the condition on the right hand side is met. If the condition is met then the set of values are reasonable solutions otherwise if the condition is violated the set of values are not reasonable solutions;

The first alternative cannot be a reasonable solution since the quantity of broccoli cannot be negative.

For the second alternative, we substitute x with 0 and y with 2 in the left hand side;

1.10(0)+2.50(2) = 5. 5 is less than 10 hence (0,2) is a reasonable solution.

For the third alternative, we substitute x with 3 and y with 2.5 in the left hand side;

1.10(3)+2.50(2.5) = 9.55. 9.55 is less than 10 hence (3,2.5) is a reasonable solution but it will not be applicable in this case since we can not purchase a fraction of a can.

For the fourth alternative, we substitute x with 2 and y with 4 in the left hand side;

1.10(2)+2.50(4) = 12.2. 12.2 is greater than 10 hence (2,4) is not a reasonable solution.

For the fifth alternative, we substitute x with 0.5 and y with 3.78 in the left hand side;

1.10(0.5)+2.50(3.78) = 10. 10 is not less than 10 hence (0.5,3.78) is not a reasonable solution.

For the sixth alternative, we substitute x with 1.5 and y with 3 in the left hand side;

1.10(1.5)+2.50(3) = 9.15. 9.15 is less than 10 hence (1.5,3) is a reasonable solution.

50- 250 is correct i did that one too

step-by-step explanation: