Author Topic: Additional Math Help HERE ONLY...!  (Read 68009 times)

Offline xlane

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Re: Additional Math Help HERE ONLY...!
« Reply #315 on: June 08, 2010, 01:26:00 am »
The value at angle pie/4 and at -pie/4 is equal from y =3 ..therefore equatn of axis of symmetry y=3..i.e C=3

period or range for 1 cycle is 180..use 180/b...so b=1
amplitude i.e height from axis of symmetry in either directn so 5-3=2, 3-1=2..therefore a=2

Alpha

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Re: Additional Math Help HERE ONLY...!
« Reply #316 on: June 08, 2010, 01:28:00 am »
I need help with this question, how can you find a?
Thanxx :D

a is the amplitude.

You get it just by comparing to the y = tan x curve.

tan x passes through the origin, it's passing at 3. So there has been a translation of 3. Therefore, c =3.

tan pie/4 = 1. But on the graph it's 5. You take 5 - 3 to get the amplitude. (y = 3 is the axis of curve here)
a = 2

Offline xlane

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Re: Additional Math Help HERE ONLY...!
« Reply #317 on: June 08, 2010, 01:28:42 am »
The value at angle pie/4 and at -pie/4 is equal from y =3 ..therefore equatn of axis of symmetry y=3..i.e C=3

period or range for 1 cycle is 180..use 180/b...so b=1

amplitude i.e height from axis of symmetry in either directn so 5-3=2, 3-1=2..therefore a=2

Alpha

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Re: Additional Math Help HERE ONLY...!
« Reply #318 on: June 08, 2010, 01:29:28 am »
The value at angle pie/4 and at -pie/4 is equal from y =3 ..therefore equatn of axis of symmetry y=3..i.e C=3

period or range for 1 cycle is 180..use 180/b...so b=1
amplitude i.e height from axis of symmetry in either directn so 5-3=2, 3-1=2..therefore a=2


 :) Just.  :D

Offline xlane

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Re: Additional Math Help HERE ONLY...!
« Reply #319 on: June 08, 2010, 01:34:30 am »
Lol yeah..."just" tryna help....:D:D:D...tan curve usualy never comes..so i put in da explanation 2

Alpha

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Re: Additional Math Help HERE ONLY...!
« Reply #320 on: June 08, 2010, 01:42:39 am »
Lol yeah..."just" tryna help....:D:D:D...tan curve usualy never comes..so i put in da explanation 2

 Yah, but nah. I meant just after or before me?  :D Both ;)

Offline xlane

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Re: Additional Math Help HERE ONLY...!
« Reply #321 on: June 08, 2010, 01:58:50 am »
Lol..oh dat...m usin cell..it got posted again..:P

Alpha

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Re: Additional Math Help HERE ONLY...!
« Reply #322 on: June 08, 2010, 02:05:07 am »
Ah no worries.  :)
I was getting bored... So, came in Maths thread.

Offline BlackBunny103

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Re: Additional Math Help HERE ONLY...!
« Reply #323 on: June 08, 2010, 02:14:52 am »
Thanxx a lot both of you  :D

Alpha

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Re: Additional Math Help HERE ONLY...!
« Reply #324 on: June 08, 2010, 02:15:48 am »
Thanxx a lot both of you  :D

Most welcome.  :)
From both of us.

Offline xlane

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Re: Additional Math Help HERE ONLY...!
« Reply #325 on: June 08, 2010, 02:25:52 am »
Thanks alpha :D...ur are most welcum frm both of us :P

Alpha

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Re: Additional Math Help HERE ONLY...!
« Reply #326 on: June 08, 2010, 02:27:34 am »
Thanks alpha :D...ur are most welcum frm both of us :P

LOL.  :P
Am honoured.  :D

Offline xlane

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Re: Additional Math Help HERE ONLY...!
« Reply #327 on: June 08, 2010, 02:30:11 am »
lol...r u nw :P @ alpha

Offline jellybeans

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Re: Additional Math Help HERE ONLY...!
« Reply #328 on: June 08, 2010, 02:43:04 am »
:o Heeeeeeelp.

if you rearrange for 6i) ;
isnt it 2t2 - 9t - 5 = 0
(x-5)(2x+1) = 0
x = 5 or x = -1/2

why does the MS say (2x+1) => x = 1/2

??? help pleaaaase thankssss

! and for Q7iii) when are we supposed to use the "> or equal to " and ">" ? .... :( !!! THANKS
« Last Edit: June 08, 2010, 02:49:47 am by jellybeans »
You said I must eat so many lemons cause I am so bitter.

Offline xlane

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Re: Additional Math Help HERE ONLY...!
« Reply #329 on: June 08, 2010, 02:46:48 am »
the mark scheme is wrng..i just solvd it nd it x=5 n x=-1/2...