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't' മിനുട്ടിൽ ഒരു കാർ സഞ്ചരിക്കുന്ന ദൂരം d = 4t2 – 3 ആണ് നൽകുന്നത്. രാവിലെ 9 മണിക്ക് കാർ സ്റ്റാർട്ട് ചെയ്താൽ, 9.02 am നും 9.03 am നും ഇടയിൽ കാർ സഞ്ചരിച്ച ദൂരം എത്രയാണ് ?

A33

B13

C30

D20

Answer:

D. 20

Read Explanation:

d = 4t2-3

t = 2 sec

d2 = 4x(2x2)-3

d2 = 16-3

d2 = 13

 

d = 4t2-3

t = 3

d3 = 4x(3x3) -3

d3 = 36-3

d3 = 33

 

d3-d2 = 33-13 =20

9.02 am നും 9.03 am നും ഇടയിൽ കാർ സഞ്ചരിച്ച ദൂരം =20


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