At t minutes past 5 pm , the time needed by the minute hand of a clock to show 6 pm was found to be 25 minutes less than 3t^2 / 4 minutes . Find t .
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Answers
Answer:
t minutes past 2, the time needed to show 3 is 60-t.
t minutes past 2, the time needed to show 3 is 60-t.As per the question,
t minutes past 2, the time needed to show 3 is 60-t.As per the question,60−t=4t2−3
t minutes past 2, the time needed to show 3 is 60-t.As per the question,60−t=4t2−3240−4t=t2−12
t minutes past 2, the time needed to show 3 is 60-t.As per the question,60−t=4t2−3240−4t=t2−12t2+4t−252=0
t minutes past 2, the time needed to show 3 is 60-t.As per the question,60−t=4t2−3240−4t=t2−12t2+4t−252=0t2+18t−14t−252=0
t minutes past 2, the time needed to show 3 is 60-t.As per the question,60−t=4t2−3240−4t=t2−12t2+4t−252=0t2+18t−14t−252=0t(t+18)−14(t+18)=0
t minutes past 2, the time needed to show 3 is 60-t.As per the question,60−t=4t2−3240−4t=t2−12t2+4t−252=0t2+18t−14t−252=0t(t+18)−14(t+18)=0(t+18)(t−14)=0
t minutes past 2, the time needed to show 3 is 60-t.As per the question,60−t=4t2−3240−4t=t2−12t2+4t−252=0t2+18t−14t−252=0t(t+18)−14(t+18)=0(t+18)(t−14)=0t=−18,14
t minutes past 2, the time needed to show 3 is 60-t.As per the question,60−t=4t2−3240−4t=t2−12t2+4t−252=0t2+18t−14t−252=0t(t+18)−14(t+18)=0(t+18)(t−14)=0t=−18,14Ignoring the negative value
Answer:
24.9 , -32.9
Step-by-step explanation:
60-t = 3t^2/4 - 25
60-t = 3t^2-100/4
(60-t)4 = 3t^2-100
2400-4t = 3t^2-100
3t^2-100-2400+4t = 0
3t^2+4t-2500 = 0
D=b^2-4ac
D=(4)^2-4x3x2500
D=16-30000
D=29984
Roots: - b (+-) √b^2-4ac/2a
-4(+-)√29984/6
-4(+-)173.15/6
-4+173.15/6 , -4-173.15/6
24.85 , -32.85