IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: angell on May 17, 2009, 08:49:42 pm
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Q.4 C plz:(
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Please attatch the question maybe i can help you ???
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Q.4 C plz:(
hey plz help even i m stuck on this 1
plzzzz help
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Q.4 C plz:(
hey plz help even i m stuck on this 1
plzzzz help
Please attach the question soo i can help
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Q4c
5e
and complete Q 9 PLEASE:(
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Q4c
5e
and complete Q 9 PLEASE:(
so did u get Q8 a
could u plz explain it to me
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Q4c
5e
and complete Q 9 PLEASE:(
Ummm sorry for Q4c i am nt suree sorry
But for Q5e...is : first draw a right angle triangle from and it ends in C
Sin 60.2=OPP/HYP
sin 60.2=OPP/85 = 73.76km
Q9 is :
a) P =42-(12+17+8) =5
Q= (40-(20+8)=12
R= 1
b) i) 17
ii)12
c) i) 26
ii)57
d) i) 8/100= 2/25
ii) 45/100=9/20
e) 37/74 * 36/73 and then u can find the answer
Hope i helped u ..and gud luck ;)
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Q4 C
u have to add the 2 pentagons = 536 X 2 =1070
drop a line from ) to the center of AB
then get AB by >> sin 36 = (AB÷2)
15
= 17.63
the total = (17.63X 5)+ 1070 = 5477.5 cm2
if u dont understand anything from the solution tell me
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Q4c
5e
and complete Q 9 PLEASE:(
so did u get Q8 a
could u plz explain it to me
x=78(bcoz alternate angle)
y=180-36=144(bcoz opposite angle of cyclic quadratic)
z=180-78=102(bcoz opposite angle of cyclic quadratic)
oki??
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thanks a lot guys
the explanations were superb :)
Thanks once again
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ur welcome.....glad i could help :)
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:) a +rep 4 all u guys :)
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ur welcome.....glad i could help :)
can u add my rep plzz :)
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yup thankx!!! +Rep from me too
but 1 more doubt
may/june 2008
q10
d)i d)ii and d)iii
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check the topic i made about it ZARA answered it.....
https://studentforums.biz/index.php/topic,1460.0.html
add to that N08.....Q 10i solved it but i dont fully get it :S....
anyone
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yup thankx!!! +Rep from me too
but 1 more doubt
may/june 2008
q10
d)i d)ii and d)iii
i)(x+2n+2-x)-(x+2n-x-2)=4
ii)g=x+2n i=x+2n+2
iii)(x+2n+2)x-(x+2n)(x+2)=-4n
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check the topic i made about it ZARA answered it.....
https://studentforums.biz/index.php/topic,1460.0.html
add to that N08.....Q 10i solved it but i dont fully get it :S....
anyone
Is that ai) bi) or ci)
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yup thankx!!! +Rep from me too
but 1 more doubt
may/june 2008
q10
d)i d)ii and d)iii
i)(x+2n+2-x)-(x+2n-x-2)=4
ii)g=x+2n i=x+2n+2
iii)(x+2n+2)x-(x+2n)(x+2)=-4n
Thanks
coul u plz explain how did u get g and i
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Go down 2 rows means add 2n. Go right 2 colums means add 2.
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Go down 2 rows means add 2n. Go right 2 colums means add 2.
wow i din't know that Thanks a million
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check the topic i made about it ZARA answered it.....
https://studentforums.biz/index.php/topic,1460.0.html
add to that N08.....Q 10i solved it but i dont fully get it :S....
anyone
Is that ai) bi) or ci)
all of it plz!!
the weird thing is that i did solve it all.....but i dont fully get it, lucky shot when solving it i guess
plz explain it
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can u attach the paper? .. mayb i can help
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check the topic i made about it ZARA answered it.....
https://studentforums.biz/index.php/topic,1460.0.html
add to that N08.....Q 10i solved it but i dont fully get it :S....
anyone
Is that ai) bi) or ci)
all of it plz!!
the weird thing is that i did solve it all.....but i dont fully get it, lucky shot when solving it i guess
plz explain it
as in 10 a to 10 d
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Q 10
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can u attach the paper? .. mayb i can help
the link is attached to the 5 th post(by angell)
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1 + 2 + 3 + 4 + 5 + .................................. + n = n(n +1)
10
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(a) (i) Show that this formula is true for the sum of the first 8 natural numbers. [2]
(ii) Find the sum of the first 400 natural numbers. [1]
(b) (i) Show that 2 + 4 + 6 + 8 + ..................... + 2n = n(n + 1). [1]
(ii) Find the sum of the first 200 even numbers. [1]
(iii) Find the sum of the first 200 odd numbers. [1]
(c) (i) Use the formula at the beginning of the question to find the sum of the first 2n natural
numbers. [1]
(ii) Find a formula, in its simplest form, for
1 + 3 + 5 + 7 + 9 + ........................ + (2n – 1).
[2]
Show your working.
ai)1+2+3+4+5+6+7+8=6=8/2(8+1)
ii)400(400+1)/2=80200
bi)AMS=2* the sum of the first 200 odd numbers=2*200(201)/2=40200
ii)ANS = sum of first 400 numbers- sum of first200 even numbers=80200-40200=40000
ci)2n(2n+1)/2=2n^2+n
ii)ANS=sum of firast wn numbers - sum of first 2n even numbers=2n^2+n- 2* sum of first n numbers=2n^2+n-2n(n+1)/2=n^2
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wow astar such a long post
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im failing tmorrw:'(
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yet i couldnt get it but Thanks.....lets just hope a Q like that is not in the exam tom. :-\
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i g2g ppl.......Talk to you later
and gd luck tomorrow.....we rabena yostor!! :(
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Quite tricky. Leave it til last and do the best you can.
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could some 1 explain Q7 a ii
how do u get y and z
same paper ???
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Rflection in line x=1? Is it Nov 08
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Rflection in line x=1? Is it Nov 08
no its june o8
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OAB is isoceles since )A and OB are radii. therefore y=(180-130)/2=25
ST and Ob are parelell so by Z rule x=40
ABC and ADB sum to 180 so z=180-(25+40)=115
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thank u so much for the prompt replies!!:D
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OAB is isoceles since )A and OB are radii. therefore y=(180-130)/2=25
ST and Ob are parelell so by Z rule x=40
ABC and ADB sum to 180 so z=180-(25+40)=115
how to get y =25
n Thanks a lottttt for such instant replies
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like astar said .. AO and OB are equal bcoz they r radii .. so triangle AOB is isosceles ..
the sum of the angles is 180 .. so
angle OAB = angle OBA and OAB + OBA + 130 = 180 => OBA + OBA + 130 = 180
so 2y + 130 = 180 => y=25
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oh ya it so simple
Thanks a lot ppl
i needa go because in next six hours i have to reach ma school 4 exam
best of luck to every1
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ahh i need help, my exam is in 3 hours :-\
do u guys know q7 part d)
why is the answer 57?
shouldnt it be 180-(78 x 2) = 24?
because if its isoceles than i can use that method right?
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q7 doesnt have a d)
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LOL :D.........every one is nervous
my xam is in 3 hours and 17 minutes ! :'(
God be with me....
good luck everyone :)