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Quote from: Q80BOY on May 05, 2009, 06:19:33 pmIGCSE (CIE) Mathematics Paper 2 Oct/Nov 2001 (i've attached the question paper)Page 3, Question 5 (Done)Page 4, Question 9 (Done)Page 5, Question 12 (Done)Page 8, Question 19, Part (d)Page 9, Question 20, Part (a) and (b)Page 10, Question 22, Part (a) and (b)Page 11, Question 24, Part (c) < -- when i use the formula distance = speed * time i get 20 (ans in m/s is 40)Total Number of Questions: 7I really hope u can help me out here Page 8 Q19d)triangle AOQ and traingle BOC are similar so OB/OA=OC/OQproprotion so AQis parallel to BCPage 9 Q2aa) angle KTP=70 as the small triangle has (20 +90+70) and the 7o angle is opposite to KTP so its 70b)you use the sine ruleKT/sin35 =25/sin70(sin70)(KT)=25sin35KT=15.23mPage11 q24c)you dont have to use this formula! o my god dont you take physics?! You calculate the distance under the graph to calculate the distance..distance=0.5*4*20=40m
IGCSE (CIE) Mathematics Paper 2 Oct/Nov 2001 (i've attached the question paper)Page 3, Question 5 (Done)Page 4, Question 9 (Done)Page 5, Question 12 (Done)Page 8, Question 19, Part (d)Page 9, Question 20, Part (a) and (b)Page 10, Question 22, Part (a) and (b)Page 11, Question 24, Part (c) < -- when i use the formula distance = speed * time i get 20 (ans in m/s is 40)Total Number of Questions: 7I really hope u can help me out here
if u need further explaination of any particular question tell me
