at t minutes past 2 p.m. the time needed by the minute hand of a clock to show 3 p.m. was found to be 3 minutes less than t2/4 minutes.find t
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Here according to the question,
t + (t)^2/4 -3 = 60
(4t + t^2 -12)/4= 60
4t + t^2 = 240 +12
t^2 + 4t -252 = 0
t^2 + 18t-14t -252=0
t (t + 18) -14 (t + 18) = 0
(t + 18)(t-14)=0
i.e t = -18( rejected since time cannot be negative)
So, t = 14.
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t + (t)^2/4 -3 = 60
(4t + t^2 -12)/4= 60
4t + t^2 = 240 +12
t^2 + 4t -252 = 0
t^2 + 18t-14t -252=0
t (t + 18) -14 (t + 18) = 0
(t + 18)(t-14)=0
i.e t = -18( rejected since time cannot be negative)
So, t = 14.
pls mark as brainliest...
yashagarwalla001:
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