Question 1073952: Gabby 's gym charges members an initial joining fee of $300 plus $50 per month. So members can calculate how much they have paid to the gym using the formula C= 50m+30, where m is the number of months they have been members. Will's workout charges $65 per month with no initial fee. So members of will's can calculate their charges using formula C=65, where m is the number of months they have been members.
1. Write the algebraic equation needed to indicate that the cost of Gabby's gym is equal to the cost of will's workout center.
2. Solve the equation from the previous equation
3. Write the algebraic inequality needed to indicate that the cost to belong to will's workout center is cheaper than the cost to belong to Gabby's gym.
4. Solve the inequality from the previous question
5. Based on these results, under what circumstances would you want to join will's workout center? Explain why?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Gabby 's gym charges members an initial joining fee of $300 plus $50 per month. So members can calculate how much they have paid to the gym using the formula
C= 50m+30, where m is the number of months they have been members. Will's workout charges $65 per month with no initial fee. So members of will's can calculate their charges using formula C=65m , where m is the number of months they have been members.
1. Write the algebraic equation needed to indicate that the cost of Gabby's gym is equal to the cost of will's workout center.
50m+30 = 65m
2. Solve the equation from the previous equation
15m = 30
m = 2 (# of months for equal cost)
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3. Write the algebraic inequality needed to indicate that the cost to belong to will's workout center is cheaper than the cost to belong to Gabby's gym.
65m < 50m+30
4. Solve the inequality from the previous question
15m < 30
m < 2 months
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5. Based on these results, under what circumstances would you want to join will's workout center? Explain why?
If you want the plan for few than 2 months.
Cheers,
Stan H.
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