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Penelope went to the store to buy [tex]\( x \)[/tex] pounds of broccoli for \[tex]$1.10 per pound and \( y \) cans of soup for \$[/tex]2.50 each. In total, she spent less than \$10. The inequality relating her purchases is [tex]\( 1.10x + 2.50y \ \textless \ 10 \)[/tex].

Which are reasonable solutions for this situation? Check all that apply.

A. [tex]\((-1, 4)\)[/tex]

B. [tex]\((0, 2)\)[/tex]

C. [tex]\((3, 2.5)\)[/tex]

D. [tex]\((2, 4)\)[/tex]

E. [tex]\((0.5, 3.78)\)[/tex]

F. [tex]\((1.5, 3)\)[/tex]

Sagot :

GinnyAnswer
Consider the inequality representing the total amount spent:

[tex]\[ 1.10x + 2.50y < 10 \][/tex]

We need to check which given pairs [tex]\((x, y)\)[/tex] satisfy this inequality.

1. For the pair [tex]\((-1, 4)\)[/tex]:

[tex]\[ 1.10(-1) + 2.50(4) = -1.10 + 10 = 8.90 \][/tex]
Since [tex]\(8.90 < 10\)[/tex], [tex]\((-1, 4)\)[/tex] is a valid solution.

2. For the pair [tex]\((0, 2)\)[/tex]:

[tex]\[ 1.10(0) + 2.50(2) = 0 + 5 = 5.00 \][/tex]
Since [tex]\(5.00 < 10\)[/tex], [tex]\((0, 2)\)[/tex] is a valid solution.

3. For the pair [tex]\((3, 2.5)\)[/tex]:

[tex]\[ 1.10(3) + 2.50(2.5) = 3.30 + 6.25 = 9.55 \][/tex]
Since [tex]\(9.55 < 10\)[/tex], [tex]\((3, 2.5)\)[/tex] is a valid solution.

4. For the pair [tex]\((2, 4)\)[/tex]:

[tex]\[ 1.10(2) + 2.50(4) = 2.20 + 10 = 12.20 \][/tex]
Since [tex]\(12.20 \not< 10\)[/tex], [tex]\((2, 4)\)[/tex] is not a valid solution.

5. For the pair [tex]\((0.5, 3.78)\)[/tex]:

[tex]\[ 1.10(0.5) + 2.50(3.78) = 0.55 + 9.45 = 10.00 \][/tex]
Since [tex]\(10.00 \not< 10\)[/tex], [tex]\((0.5, 3.78)\)[/tex] is not a valid solution.

6. For the pair [tex]\((1.5, 3)\)[/tex]:

[tex]\[ 1.10(1.5) + 2.50(3) = 1.65 + 7.50 = 9.15 \][/tex]
Since [tex]\(9.15 < 10\)[/tex], [tex]\((1.5, 3)\)[/tex] is a valid solution.

So, the reasonable solutions are:

[tex]\[ (-1, 4), (0, 2), (3, 2.5), (1.5, 3) \][/tex]
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