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Mathematics: Post your doubts here!

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having found the factorised form fro p(Z)....just replace all the Zs by Z^2 ,,......it shudnt be a problem now
 
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can someone please help with q10 part ii b
please
 

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sorry made a typing mistake... October/November 2010 Variant 3.Question 10, part ii b

having found the factorised form fro p(Z)....just replace all the Zs by Z^2 ,,......it shudnt be a problem now
Aoa wr wb!

yeah just as hassam said...like in the a part..u got

z = -2..

instead of z it's z^2 now
so z^2 = -2

similarly do for the other 2 roots...when u take the square root u obviously get..2 answer ;)
 
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Can someone help me?


Find the cartesian equations of the plane through (--1,1,2) and (2,0,-1) perpendicular to the plane x+y-2z=1.



use the two points to obtain a direrction vector in the plane....and usee the given plane equation's normal as another vector in the required plane....cross these two vectors and there u go ....


Can someone please write it? I'm still stuck. What does 'cross these vectors' mean please?
 
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how to get my self done wd p3 .. :S :(
aoa wr wb!
solve one paper ...try recent ones..check ur paper from the mark scheme and mark yourself....now make out which topics you werent really sure..revise them thoroughly...
go through your notes for all chapters...solve few more papers....mark urself..if you feel u r unsure about something revise that chapter again ;)
 
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Can someone help me?


Find the cartesian equations of the plane through (--1,1,2) and (2,0,-1) perpendicular to the plane x+y-2z=1.



use the two points to obtain a direrction vector in the plane....and usee the given plane equation's normal as another vector in the required plane....cross these two vectors and there u go ....


Can someone please write it? I'm still stuck. What does 'cross these vectors' mean please?
 
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Why did you reach to this?
This is not an equation, it is an inequality! :p

25cos(θ - 73.74) = 15
Simply solve this but remember the range is negative also so also take the value of negative (360 - x) in the 4th quadrant
 
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