A bag of oranges costs \[tex]$1.30 per pound, and a pumpkin costs \$[/tex]5.20. The inequality [tex]1.30x + 5.20 \leq 18.20[/tex] models this situation, where [tex]x[/tex] is the number of pounds of oranges. Solve the inequality. How many pounds of oranges can Antoine buy?

A. [tex]x \geq 10[/tex]; Antoine can buy 10 pounds or more of oranges.
B. [tex]x \geq 8[/tex]; Antoine can buy 8 pounds or more of oranges.
C. [tex]x \leq 10[/tex]; Antoine can buy 10 pounds or less of oranges.
D. [tex]x \leq 8[/tex]; Antoine can buy 8 pounds or less of oranges.



Answer :

Let's solve the inequality step-by-step to determine how many pounds of oranges Antoine can buy.

Given inequality:
[tex]\[ 1.30x + 5.20 \leq 18.20 \][/tex]

1. Subtract [tex]\( 5.20 \)[/tex] from both sides:
[tex]\[ 1.30x + 5.20 - 5.20 \leq 18.20 - 5.20 \][/tex]
[tex]\[ 1.30x \leq 13.00 \][/tex]

2. Divide both sides by [tex]\( 1.30 \)[/tex]:
[tex]\[ \frac{1.30x}{1.30} \leq \frac{13.00}{1.30} \][/tex]
[tex]\[ x \leq 10 \][/tex]

Therefore, the inequality simplifies to [tex]\( x \leq 10 \)[/tex].

This means Antoine can buy 10 pounds or less of oranges.

So the correct answer is:
C. [tex]\( x \leq 10 \)[/tex]; Antoine can buy 10 pounds or less of oranges.

Answer:

C. x ≤ 10; Antoine can buy 10 pounds or less of oranges.

Step-by-step explanation:

1.30x + 5.20 ≤18.20

To solve this inequality, subtract 5.20 from each side.

1.30x + 5.20-5.20 ≤18.20-5.20

1.30x  ≤13.00

Divide each side by 1.30.

1.30x/1.3  ≤13.00/1.3

x  ≤10.00

Antoine can buy 10 lbs or less.