Asked by kathy m.
                Suppose that a certain municipality uses the equation y=3(x-50)+30 to determine the fine assessed a car which exceeds the 50 mph speed limit. Let x=speed of the car (in mph) and y=corresponding fine in dollars.
 
a)What is the fine for traveling at 55 mph?
 
b)If the fine is $126, what is the speed of the car?
 
c)If car A travels at 60 mph and car B travels at 90 mph, then car B is traveling 50% faster than car A. However the fine for car B is what percent higher than that of car A?
            
        a)What is the fine for traveling at 55 mph?
b)If the fine is $126, what is the speed of the car?
c)If car A travels at 60 mph and car B travels at 90 mph, then car B is traveling 50% faster than car A. However the fine for car B is what percent higher than that of car A?
Answers
                    Answered by
            kathy m.
            
    i just really need help with c. i know a is $45 and b is 102 mph but i don't get c
    
                    Answered by
            Reiny
            
    I disagree with your b) answer of 102
3(x-50) + 30 = 126
3(x-50) = 96
x-50 = 32
x = 32+50 = 82 mph
c) fine for A is 3(60-50) + 30 = $60
fine for B is 3(90-50) + 30 = $150
increase in speed = 30 mph
increase in fine = 150-60 = $90
percentage increase = 90/60 = 1.5 or 150% increase in the fine.
    
3(x-50) + 30 = 126
3(x-50) = 96
x-50 = 32
x = 32+50 = 82 mph
c) fine for A is 3(60-50) + 30 = $60
fine for B is 3(90-50) + 30 = $150
increase in speed = 30 mph
increase in fine = 150-60 = $90
percentage increase = 90/60 = 1.5 or 150% increase in the fine.
                    Answered by
            kathy m.
            
    yeah, i found out b was wrong after working it out again, but thanks so much for explaining c~
    
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