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VPU=38 (same segment-take VU as base line)help somebody in these circle
VPU=QPR=38 (vertically opposite angles)
Take QR as base, thus QPR=QTR=38 (same segment)
38 is the answer!
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VPU=38 (same segment-take VU as base line)help somebody in these circle
http://clickpapers.net/past papers/math-/4024_w10_qp_22.pdf
q.5 last part !! i don't know how to calculate the width and height of the bar pls if anyone can explain it plz ? :O
yes xactly every shape must hv 1 rotational order of symmetry. so 1 R.O.S is called zero R.O.S???Yes. that's what i was arguing about last night with a couple of guys, when they say no rotational symmetry it basically means it only has the one rotational symmetry which every shape has.
height=frequency density.
Frequency density= frequency/class interval
so for 90-95 .. the class interval = 95-90 = 5
FD = F/X
FD = 20/5
FD = 4
so the height of 90-95 = 4 and the width = class width = 5.
I guess you can take it for the rest yourself now.
ok c! lets calculate frequency density which is also called height. for less than or equal to 90, it is 16/10=1.6the answer is 1 and 10 :O
the answer is 1 and 10 :O
thnx 4 ur help, but why would BMQ and ANP be congruent? the question says APB and BQC....HarisLatif said:for Q9ciii ; as you have worked out that AN is 6 so BM is also 6 since BMQ and ANP are congruent. and if BM is 6 then BN will be 12 since B Is the centre of enlargment and scale factor is 2. so BN-BM=12-6=6. Hope you get it.
Question 3 a part ii
even i am not able to solve it.. lol
yes it is 38. i got it know. thankyou...HarisLatif is the answer for <QTR=38 please let me know quick !!??
what?? my answer is coming 1:2 and besides how could R be less than r?? according to 10:1 answer r is 10x as big as R!! that's not possib;e. there must b some mistake!!http://clickpapers.net/past papers/math-/4024_w11_qp_22.pdf
how do we get 10 : 1 in q.3 part a (ii) ?? plz explain any1 ?
yep and now i also know how to solve it. thanks brotherIs the answer 38 ?? if it is then I know how to solve it
post link.HEyy guys!! the nightmare question for me:
w11_4042 paper 22 question 8(c)(iii) ......... plz hury up and tell me how we can calculte calues for B and C!!
http://www.xtremepapers.com/papers/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_w10_qp_21.pdf Question 10 c part. which tranformation is represented by that matrix? shear or stretch?
Find the shaded area first equate it to the area of small circle and then find ratio. its easyhttp://clickpapers.net/past papers/math-/4024_w11_qp_22.pdf
how do we get 10 : 1 in q.3 part a (ii) ?? plz explain any1 ?
yeah but we first have to know this before finding the cordinates of its image as asked in the question. i don't know how to get themIts a combined transformation of some kind but it never asks to define so why wwaste time thinking abt it
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