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Mathematics 23 Online
OpenStudy (anonymous):

Annette drove to shawnee in 4 hours and drove back in 3 hours. what were her speeds if her speed coming back was 11 mph greater than her speed going?

OpenStudy (bahrom7893):

Speed = d*time Speed going = d*4 Speed returning = d*3 = speed going+11

OpenStudy (anonymous):

what would be the equation?

OpenStudy (bahrom7893):

wait lol speed is distance over time.. omg im an idiot

OpenStudy (bahrom7893):

s_g=d/4 s_r=d/3=s_g+11 d/3=(d/4)+11 4d=3d+132 d=132

OpenStudy (bahrom7893):

Thus: speed going: 132/4= 33 speed returning: (132/3)+11 = 55

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

u in smarter classes?

OpenStudy (bahrom7893):

define smarter classes

OpenStudy (anonymous):

u know smarter than the average person

OpenStudy (bahrom7893):

define an average person lol

OpenStudy (anonymous):

someone who is smarter than an ape

OpenStudy (bahrom7893):

oh then: <-- my pic says it all

OpenStudy (anonymous):

hahahahahahahahahahahaha

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