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In a beanbag toss game, Janelle scores 5 points for landing on a round target and 8 points for landing on a square target. She needs more than 50 points to win. Let [tex]\( x \)[/tex] represent the number of times Janelle lands on the round target and [tex]\( y \)[/tex] represent the number of times she lands on the square target.

Which inequality represents the situation?

A. [tex]\( 5x + 8y \ \textgreater \ 50 \)[/tex]

B. [tex]\( 5x + 8y \geq 50 \)[/tex]

C. [tex]\( 8x + 5y \ \textgreater \ 50 \)[/tex]

D. [tex]\( 8x + 5y \ \textless \ 50 \)[/tex]



Answer :

GinnyAnswer
To solve this problem, let's break it down step-by-step:

1. Identify the Points System:
- Janelle scores 5 points for each time she lands on a round target.
- She scores 8 points for each time she lands on a square target.

2. Define Variables:
- Let [tex]\( x \)[/tex] be the number of times Janelle lands on the round target.
- Let [tex]\( y \)[/tex] be the number of times Janelle lands on the square target.

3. Formulate the Total Points:
- The total points Janelle scores from landing on round targets is [tex]\( 5x \)[/tex].
- The total points she scores from landing on square targets is [tex]\( 8y \)[/tex].

4. Set Up the Inequality for Winning:
- Janelle needs more than 50 points to win. Hence, the total points from both targets must be greater than 50.
- This can be expressed as: [tex]\( 5x + 8y > 50 \)[/tex].

Therefore, the inequality that represents this situation is:
[tex]\[ 5x + 8y > 50 \][/tex]

The correct answer is:
[tex]\[ 5 x + 8 y > 50 \][/tex]

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