Apool charges $4 each visit, or you can buy a membership for $100 for 3 months. write and solve an inequality to find how

Apool charges $4 each visit, or you can buy a membership for $100 for 3 months. write and solve an inequality to find how many times a person should use the pool so that a membership is less expensive than playing each time.

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  1. 26 times.

    Step-by-step explanation:

    If the pool charges $4 each visit, after x visits the amount paid would be 4x.

    Also, you can buy a membership that costs 100 dollars and it lets you go to the pool as many times as you want for 3 months.

    We need to find how many times a person should use the pool so the membership is less expensive than playing each time. To write this we would need that the 100 dollar membership was less than 4x

    [tex]100[/tex]

    Thus, x should be greater than 25 so the membership is less expensive. In other words, the person should use the pool at least 26 times in those three months so the membership is less expensive than playing each time

  2. 26 times.

    Step-by-step explanation:

    If the pool charges $4 each visit, after x visits the amount paid would be 4x.

    Also, you can buy a membership that costs 100 dollars and it lets you go to the pool as many times as you want for 3 months.

    We need to find how many times a person should use the pool so the membership is less expensive than playing each time. To write this we would need that the 100 dollar membership was less than 4x

    [tex]100[/tex]

    Thus, x should be greater than 25 so the membership is less expensive. In other words, the person should use the pool at least 26 times in those three months so the membership is less expensive than playing each time

  3. $100< $4x
    $100/$4<$4x/$4
    25<x

    $4 * 25 times = $100
    So using the pool 26 times would make the membership cheaper than paying each time

  4. $100< $4x
    $100/$4<$4x/$4
    25<x

    $4 * 25 times = $100
    So using the pool 26 times would make the membership cheaper than paying each time

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