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Mathematics 18 Online
OpenStudy (anonymous):

A child is given a push on a swing.Each time she swings back a little less, according to the equation d=2(0.85)^n , where d is the horizontal distance, in metre, from her position after n complete swings. After how many swings will the distance from rest be less than 15cm.

OpenStudy (anonymous):

mch

OpenStudy (anonymous):

what?

OpenStudy (johnweldon1993):

Well the equation we are given...is in meters.... \[\large d = 2(0.85)^n\] However what they want to know..is how long it will take to be less than 15 CM....so lets convert 15 centimeters to meters....we do that by dividing by 100 \[\large 15cm \times \frac{.01m}{1 cm}= .15m\] Now we just make our equation \[\large d = 2(0.85)^n \le .15\] and solve for 'n'

OpenStudy (johnweldon1993):

Do you know how to solve for 'n' @harz360 ?

OpenStudy (anonymous):

trial and error?

OpenStudy (anonymous):

anyone please help me

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