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Author Topic: Additional Math Help HERE ONLY...!  (Read 75550 times)

Offline Dibss

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Re: Additional Math Help HERE ONLY...!
« Reply #285 on: June 06, 2010, 03:58:03 pm »
October/Nov 2003 P2 Q8 iii) anyone?
I'm REALLY sorry I don't have the paper to attach :|
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Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #286 on: June 06, 2010, 04:47:20 pm »
Quote from: Dibss on June 06, 2010, 03:58:03 pm
October/Nov 2003 P2 Q8 iii) anyone?
I'm REALLY sorry I don't have the paper to attach :|

iii) Find no. of ways where there is an absence of rose bush of each color.

total ways  = 210

no. of ways in which there is no yellow rose bush = 8C6 = 28

no. of ways in which there is no pink bush = 7C6 = 7

Red bushes have to be dere in all selections as there are 5 of them

so 210 - 28 - 7 = 175
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Offline Dibss

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Re: Additional Math Help HERE ONLY...!
« Reply #287 on: June 06, 2010, 06:11:06 pm »
Quote from: A@di on June 06, 2010, 04:47:20 pm
iii) Find no. of ways where there is an absence of rose bush of each color.
That makes sense.
Thank you =]
+REP
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Offline Ghost Of Highbury

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Re: Additional Math Help HERE ONLY...!
« Reply #288 on: June 06, 2010, 08:38:13 pm »
No problem..and...Thansk for the rep :)
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Offline slvri

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Re: Additional Math Help HERE ONLY...!
« Reply #289 on: June 07, 2010, 08:12:24 am »
hi guys id be willing to solve your add math doubts as well seeing as im back
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #290 on: June 07, 2010, 09:19:16 am »
Hey guys, would you please help me to solve this question, it's from MJ 2003 paper 1.
Thanxx :D
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Offline slvri

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Re: Additional Math Help HERE ONLY...!
« Reply #291 on: June 07, 2010, 09:32:17 am »
y=kx-2
y2=4x-x2
substitute the first equation in the second
(kx-2)2=4x-x2
k2x2-4kx+4=4x-x2
k2x2+x2-4kx-4x+4=0
(k2+1)x2+(-4k-4)x+4=0
now the line meets the curve provided that for the above equation the discriminant or b2-4ac>=0
where a=k2+1,b=-4k-4 and c=4
b2-4ac>=0
(-4k-4)2-4(k2+1)(4)>=0
16k2+32k+16-16k2-16>=0
32k>=0
k>=0
i hope this helped, and is this the correct answer? :)
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #292 on: June 07, 2010, 09:48:01 am »
Quote from: slvri on June 07, 2010, 09:32:17 am
y=kx-2
y2=4x-x2
substitute the first equation in the second
(kx-2)2=4x-x2
k2x2-4kx+4=4x-x2
k2x2+x2-4kx-4x+4=0
(k2+1)x2+(-4k-4)x+4=0
now the line meets the curve provided that for the above equation the discriminant or b2-4ac>=0
where a=k2+1,b=-4k-4 and c=4
b2-4ac>=0
(-4k-4)2-4(k2+1)(4)>=0
16k2+32k+16-16k2-16>=0
32k>=0
k>=0
i hope this helped, and is this the correct answer? :)

Oh great, thanxx +rep. Yes this is the correct answer ;)
By the way are you taking Add Maths this year?
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #293 on: June 07, 2010, 11:05:59 am »
Can you guys please help me with Q8 of MJ 2004 P1?
Thanxx :D
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Offline astarmathsandphysics

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Re: Additional Math Help HERE ONLY...!
« Reply #294 on: June 07, 2010, 12:24:07 pm »
here
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Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #295 on: June 07, 2010, 02:11:31 pm »
Helloooo, :D
ummm i need help with Q2 please? :) i can find the first two values but... how do you find the rest? :o and how do you know how many 'values' can fit in within the range?  :-\
THANKS IN ADVANCE ;D
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Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #296 on: June 07, 2010, 02:48:47 pm »
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Offline slvri

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Re: Additional Math Help HERE ONLY...!
« Reply #297 on: June 07, 2010, 03:25:40 pm »
okay
3sin(x/2-1)=1
sin(x/2-1)=1/3
convert the given range for x into a range for (x/2-1) which is the basic angle here
0<x<6pi
0<x/2<3pi
-1<x/2-1<3pi-1
-1<x/2-1<8.425
now denote x/2-1 by a(alpha but i cant write that here so a it is)
sin a is positive in 1st and 2nd quadrants
a=sin-1(1/3)=0.3398
in the 1st quadrant a=0.3398
in 2nd a=pi-0.3398=2.802
but we can go till 8.425 as indicated in the range
so another value of a=2pi+0.3398=6.623
but no other value for 2nd quadrant cause it goes outside range(2.802+2pi=9.085>8.425
so a=0.3398,2.802,6.623
x/2-1=0.3398,2.802,6.623
x/2=1.3398,3.802,7.623
x=2.6796,7.604,15.246
x=2.68,7.60,15.2 correct to 3sf
is this correct?
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Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #298 on: June 07, 2010, 03:32:32 pm »
Quote from: slvri on June 07, 2010, 03:25:40 pm
okay
3sin(x/2-1)=1
sin(x/2-1)=1/3
convert the given range for x into a range for (x/2-1) which is the basic angle here
0<x<6pi
0<x/2<3pi
-1<x/2-1<3pi-1
-1<x/2-1<8.425
now denote x/2-1 by a(alpha but i cant write that here so a it is)
sin a is positive in 1st and 2nd quadrants
a=sin-1(1/3)=0.3398
in the 1st quadrant a=0.3398
in 2nd a=pi-0.3398=2.802
but we can go till 8.425 as indicated in the range
so another value of a=2pi+0.3398=6.623
but no other value for 2nd quadrant cause it goes outside range(2.802+2pi=9.085>8.425
so a=0.3398,2.802,6.623
x/2-1=0.3398,2.802,6.623
x/2=1.3398,3.802,7.623
x=2.6796,7.604,15.246
x=2.68,7.60,15.2 correct to 3sf
is this correct?

YEP its correct ;) Thank Youuuu :D !
btways if the range is from 0<x< 6pi radians..
isnt the max supposed to be = 18.85? :( why is it ...8.425?  :o :o :o :o
youre right ;D ..but why isnt  it 6pi = 18.85?  :-\ thanks!!!!!
« Last Edit: June 07, 2010, 03:34:39 pm by jellybeans »
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Offline slvri

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Re: Additional Math Help HERE ONLY...!
« Reply #299 on: June 07, 2010, 03:39:25 pm »
thats because theyve given you the range for x but the angle whos sine is being taken in the question is x/2-1, not x so whatever angle you find after taking inverse sine will be for x/2-1, not x itself. thats why you have to convert the given range for x to a range for x/2-1
that means for the limits given (0 and 6pi) you divide by 2 to get x/2 and then subtract 1 to get x/2-1 which is the required range
and by the way i gave add math in june 2009 and i gave a level math in june 2010
so yeah just felt like helping others out for the add math exam tomorrow :)
oh and HI ASTAR....remember me?
« Last Edit: June 07, 2010, 03:42:09 pm by slvri »
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