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like right nowI agree that this thread should be on topic as much as possible but sometimes I too get carried away sometimes.. haha
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like right nowI agree that this thread should be on topic as much as possible but sometimes I too get carried away sometimes.. haha
lol, thanksThis is CIE making life difficult for us...
To get the vectors we will find the unit vector and multiply by the magnitude..
OA = Unit Vector .. 3i − 4k/√25
15/5(3i-4k)
=3(3i-4k)
=9i-12k
Unit Vector of OB = 2i + 3j − 6k / √(2^2)+3^2+(-6^2)
Unit Vector of OB = 2i+3j-6k/7
OB = 14/7(2i+3j-6k)
OB = 2(2i+3j-6k)
OB = 4i+6j-12k
http://rechneronline.de/function-graphs/How did ya draw the graph?
u just find till wich term are negative...take their sum n add it to 525..as their sum will b negative will be subtracted frm original in the start..http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf
Someone please help me with question no. 10(a) iii..
MustafaMotani , A star , daredevil , PANDA-
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
Question No. 7.. whole question, I will try and solve it but you guys solve it as well.. so that I can tally my working.. since I don't exactly understand the question over here...
a)
an = a1 + (n-1)d
cos^2x = 1 +d
d= cos^2x -1
Sn = 10/2[2(1) + (10-1)(cos^2x - 1)]
Sn = 10/2[2(1) + 9(-sin^2x)]
Sn = 5[ 2 - 9sin^2x]
Sn = 10 - 45sin^2x
a = 10 ... b = 45
Thank you very much..tan inverse of gradient .. is the angle the line makes with x-axis , so when ever u need to find the angle of two lines intersecting ! just find their gradients and do tan^-1(gradient)
the one with larger angle minus the smaller angle so u get the angle ... this angle can be acute or obtuse ..
some time cie makes it thougher for us .. wen we get acute angle ! they might ask for an obtuse angle, so at that time you need to 180-(what ever angle u got) to get the obtuse angle
Yea I solved it.. the exact same way.. what about b part i ? I solved that as well but unsure of the working.. since mark scheme shows literally no working.
MustafaMotani , A star , daredevil , PANDA-
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
8 iv ..
How does one know when taking an inverse whether a + sign or a - sign would come before the √ equation?
like (x-2)^2 = y+4-k
x-2 = ±√(y+4-k)
How'd you know whether we will take + or - here??
Yea I solved it.. the exact same way.. what about b part i ? I solved that as well but unsure of the working.. since mark scheme shows literally no working.
MustafaMotani , A star , daredevil , PANDA-
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_11.pdf
8 iv ..
How does one know when taking an inverse whether a + sign or a - sign would come before the √ equation?
like (x-2)^2 = y+4-k
x-2 = ±√(y+4-k)
How'd you know whether we will take + or - here??
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