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I did that, but I keep getting (cosx-1) instead of (1-cosx) in the numerator
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Mathematics: Post your doubts here!
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that
's awesome, I appreciate your help.
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Your 7 c)
https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-336#post-495659
7 b ii) This is a tricky question read these words of the question carefully.. "if the order in which Jessica chooses the 4 packets is taken into account" meaning it's permutation
8 Packets .. 1 Chocolate Biscuits(Bs) 1 Custard Cream (Cu)
4 Need to be chosen given Bs and Cu have to be chosen..
Bs Cu _ _
there were 8 total .. now 2 of those 8 have been chosen remaining = 6 packets .. and 2 need to be chosen.. and then there are 4 places in which these 4 packets can be arranged in...
So answer should be 6C2 * (cuz 2 of the remaining 6 items need to be chosen) 4! (cuz there are 4 places where each packet can go)
6C2*4! = 15 * 24 = 360 Ways
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No P3 here...
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Thanks!! though you've used add/sub formulas , which aren't a part of my sylllabus (i'm not appearing for P3 this year) ,so could you pls explain further :
how does tanπ -tanx / 1+tanπ tanx = -tanx/1 ?
thanks in advance!
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(Cosx-1)^2 = (1-cosx)^2 because minus sign taken common becomes plus due to the square.
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Tan x = k
If you have read about the quadrants ..
then Tan is positive in the first and 3rd quadrant.
Pi = 180 Degrees..
Pi - x means 2nd quadrant = -tan x = -k
ii) Tan(Pi/2 - x) = cot x = 1/tanx = 1/k
iii) Tan x = perp/base = k
= k/1
find the hyp. via pythogoras.. Hyp = √(k^2 + 1)
Sin = Perp/Hyp
Sin x = k/√(k^2+1)
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Well this part follows the previous part of sketching the graph. Also it is an easier method.
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he was sooo awesome... 90 percent of students in his class got A* 7 percent A and 3 percent others.
he made m,e go from E in school to A* in the cie o levels maths and A in add maths paper.
though in stats he is not as awsome as in pure maths
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you would find that the list of formula sheet. Then applying it would give u the answer btw its for p1 only.
I dint know how to do this, but searched through google and learned a useful method.
First we know the roots are
-√3 + i mod 2 arg 5/6pi
-√3 - i mod 2 arg -5/6pi
So according to De Moivre's theorem
therefore,
z^6 = r^6 (cos 6t + i sin6t)
2^6 (cos6(5/6pi) + isin 6(5/6pi))
64 (-1 + i (0))
-64
same thing with other root!
link.
http://www.cliffsnotes.com/study_guide/De-Moivres-Theorem.topicArticleId-11658,articleId-11634.html
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oh yeahhh i found those in the formula sheet! thanks again!
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Can anyone help me to give a detailed explanation for june 12 p 42 q7 thanks
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Can u help me to give a detailed explanation for june 12 p 42 q7 thanks
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I dint know how to do this, but searched through google and learned a useful method.
First we know the roots are
-√3 + i mod 2 arg 5/6pi
-√3 - i mod 2 arg -5/6pi
So accordingto De Moivre's theorem
therefore,
z^6 = r^6 (cos 6t + i sin6t)
2^6 (cos6(5/6pi) + isin 6(5/6pi))
64 (-1 + i (0))
-64
same thing with other root!
link.
http://www.cliffsnotes.com/study_guide/De-Moivres-Theorem.topicArticleId-11658,articleId-11634.html
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Hi, can someone please solve Question 5 http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
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here you go!
(i) divide that thingy into 2 triangles from the middle and u will have triangle AOX with angle 1/6pi.. so AX= 12tan1/6pi = 4√3
(ii) to get the area of the shaded region we need to find the area of the sector then subtract it from the whole thingy which is the 2 triangles..area of the sector will be= 1/2(12)^2(1/3pi)
area of sector = 24pi and the area of the 2 triangles will be = 1/2(4√3)(12) x 2 = 48√3
area of shaded region = 48√3 - 24pi
thats it, i hope you got it!
Okay, tangent means making 90 degrees with a point on circle.
so AX make 90 degree B A and BX make 90 at B and thus, line OX will bisect then angle 1/3pi (means cuts in half) giving 1/6pi
tan (1/6pi) = AX/12 (opposite over adjacent (radius))
1/√3 = AX/12
AX = 12/√3
(ii) Now the area can be found by finding area of the two triangles AOX and BOX (they are same) and find the sector area...and subtract
area of AOX = 12/√3 x 12 x 1/2 (1/2bh = area of triangle)
=24√3
then into 2 because 2 triangle =48√3
Now the sector 1/2 r^2 theta
1/2 x 12^2 x 1/3pi = 24pi
48√3 - 36pi =