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Math P2 Oct/Nov 2001 HELP !!!

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sanity_master:
well....i should've been gone now

so if u r not sure about ur answers...let twilight check it for u :)

By the way twilight.......Thanks......i answered the Q with the correct answer without knowing the exact way of explaining (gd way  ;) )



CYA L8r GUYs!!..........BYE

Q80BOY:

--- Quote from: Amber on May 05, 2009, 07:03:12 pm ---Q9) x=140 degress (70*2)
      y= 50 ( 90 - 40)
      z= 30 ( 180 - 140 /2 = 20 and 40+20 = 60
                                                            60+70+20+x = 180
                                                                           x = 30 )  
Q12) a) 60 and 300
b) i think its is 270 <x<359

Q19) d) because BC is double of AQ

Q20) a) 70
b) 15.3cm ( 25/sin70 = x/sin35)

Q22) a) d= 42 ( 72-30)
           e= 74 ( 180-30 = 150+136= 360-286 = 74)
           f= 64 (180- [74+42])
b) 252 ( 180-136 = 44+64 = 180-108 = 72+180 = 252)

Q24) no .. the area under the graph is the distance so u use the formula :

1/2 * b * h
1/2 * 4 * 20 = 40 :D

couldnt do the first one sorry .. hope i helped    

--- End quote ---

Thanks for the help amber, appreciate it !!  ;D + rep

Q80BOY:

--- Quote from: twilight on May 05, 2009, 07:08:11 pm ---Page 5, Question 12

(a)       cos x = 1/2
        cosine is positive in first and fourth quadrant
            x   =  cos-1 0.5    =   60o
     and  x   =           360 - 60            = 300o             =>     therefore   x = 60 or 300

(b)       cos x   is greater than 0 in these    0 < x < 90    and    270 < x < 360
             and it is less than 1/2  in this      60 < x < 300
  so combining both the required vales of x    are    270 < x < 300

--- End quote ---

Thanks, that's what i was looking for, +rep !!!  ;)

twilight:

--- Quote from: sanity_master on May 05, 2009, 07:20:37 pm ---well....i should've been gone now

so if u r not sure about ur answers...let twilight check it for u :)

By the way twilight.......Thanks......i answered the Q with the correct answer without knowing the exact way of explaining (gd way  ;) )



CYA L8r GUYs!!..........BYE

--- End quote ---

Thanks sanity master  ;)  and Q80BOY   ;)

sweetsh:

--- Quote from: Q80BOY on May 05, 2009, 06:19:33 pm ---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: 7

I really hope u can help me out here  :D

--- End quote ---

Page 8 Q19
d)triangle AOQ and traingle BOC are similar so OB/OA=OC/OQ
proprotion so AQis parallel to BC

Page 9 Q2a
a) angle KTP=70 as the small triangle has (20 +90+70) and the 7o angle is opposite to KTP so its 70
b)you use the sine rule
KT/sin35 =25/sin70
(sin70)(KT)=25sin35
KT=15.23m

Page11 q24
c)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

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