SOLUTION: ok here is my super dooper hard question ok ready? a bus company has 4000 pasangers daily, each paying a fait of 2 dollars. for each 0.15$ increase the company looses about 40 pa

Question 65237: ok here is my super dooper hard question ok ready?
a bus company has 4000 pasangers daily, each paying a fait of 2 dollars. for each 0.15$ increase the company looses about 40 passangers. if the company needs to tske in $10450 per day to stay in buisness what fare should be changed?
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
THE EQUATION FOR THIS PROBLEM IS:
(4000-40X)(2+.15X)=10450
8000-80X+600X-6X^2=10450
6X^2-520X+2450=0
3X^2-260X+1225=0 USING THE QUADRATIC EQUATION TO SOLVE FOR X WE HAVE
X=(-B+-SQRT[B^2-4AC])/2A
X=(260+-SQRT[-260*-260-4*3*1225])/2*3
X=(260+-SQRT[67600-14700])/6
X=(260+-SQRT[52900])/6
X=(260+-230)/6
X=(260+230)/6
X=490/6
X=81.666666 SOLUTION
X=(260-230)/6
X=30/6
X=5 SOLUTION
PROOF
(4000-40*5)(2+.15*5)=10450
(4000-200)(2+.75)=10450
3800*2.75=10450
10450=10450
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MERRY CHRISTMAS

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