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Mathematics: Post your doubts here!
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papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s13_ms_32.pdf
Help me with question 10(ii) thankyou so much!!
Help me with question 10(ii) thankyou so much!!
9709 w10 qp33
10 The polynomial p(ß) is defined by
p(ß) = ß3 + mß2 + 24ß + 32,
where m is a constant. It is given that (ß + 2) is a factor of p(ß).
(i) Find the value of m. [2]
(ii) Hence, showing all your working, find
(a) the three roots of the equation p(ß) = 0, [5]
(b) the six roots of the equation p(ß2) = 0. [6]
(ii)how to solve?anyone can solve it??thanks
10 The polynomial p(ß) is defined by
p(ß) = ß3 + mß2 + 24ß + 32,
where m is a constant. It is given that (ß + 2) is a factor of p(ß).
(i) Find the value of m. [2]
(ii) Hence, showing all your working, find
(a) the three roots of the equation p(ß) = 0, [5]
(b) the six roots of the equation p(ß2) = 0. [6]
(ii)how to solve?anyone can solve it??thanks
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I hope you can do part (i) because that's just understanding what a complex conjugate is.
(ii)
|z - 10i| = 2|z - 4i|
You are told to square both sides. Question (i) tells you how to do this: |z - ki|^2 = (z - ki)(z* + ki).
(z - 10i)(z* + 10i) = 4(z - 4i)(z* + 4i)
Simplify to get 0 = zz* - 2iz* + 2iz - 12.
Meanwhile, |z - 2i|^2 = zz* - 2iz* + 2iz + 4. So the original equation becomes
|z - 2i|^2 - 16 = 0
| |z - 2i| | = 4
|z - 2i| = 4
(iii) The graph of |z| = k is a circle of radius k centered at the origin. What does subtracting 2i from z do to that graph?
Everything summarized. Hope you can understand. Am in a bit hurry.
If you don't get inbox me.
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Well RoOkaYya G is blocked. She forgot her password. So its just me to solve this piles of doubts. So I will take time.
guys help me in http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_3.pdf
question 7 part 2 !!!!!
question 7 part 2 !!!!!
Thank you..could you help with qn 7 part 2
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he normal to each plane is
<2, -1, -3> and <1, 2, 2> respectively.
To get the angle between the planes, consider the angle between the normals.
To get the equation of the line of intersection, consider the cross product of the normals. This will give you the direction numbers for the line of intersection.
That's all I can say.
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(ii)
|wz| = |w||z| = (1)(R) = R
arg|wz| = arg|w| + arg|z| = 2π/3 + θ
|z/w| = |z|/|w| = R/1 = R
arg|z/w| = arg|z| - arg|w| = θ - 2π/3
(iii)
Modulus of all three are same meaning their lengths are equal ( = R)
All angles subtended are π/3
(iv)
z = 4 +2i
The other two vertices are zw and z/w
zw = (4 + 2i)(-0.5 +i√3/2)
zw = - (2 + √3) + (2√3 -1)i
z/w = (4 + 2i)/(-0.5 +i√3/2)
Rationalise the denominator to get
z/w = - (√3 + 2) + (2√3 + 1)i
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its simple OP is actually the normal to the required plane.. coaz it says that AB is on the plane and OP is perpendicular to the plane.
simply use
r.n = a.n ( take OA or OB in place of a)
(x,y,z) x (2/3,5/3,7/3) = (1,2,2) x (2/3,5/3,7/3)
2x + 5y + 7z = 26
Not part 3
, I was asking about part 2..how do you get (2/3,5/3,7/3)?Thanks bro for the explanation I got it nowhe normal to each plane is
<2, -1, -3> and <1, 2, 2> respectively.
To get the angle between the planes, consider the angle between the normals.
To get the equation of the line of intersection, consider the cross product of the normals. This will give you the direction numbers for the line of intersection.
That's all I can say.
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I hope you can do part (i) because that's just understanding what a complex conjugate is.
(ii)
|z - 10i| = 2|z - 4i|
You are told to square both sides. Question (i) tells you how to do this: |z - ki|^2 = (z - ki)(z* + ki).
(z - 10i)(z* + 10i) = 4(z - 4i)(z* + 4i)
Simplify to get 0 = zz* - 2iz* + 2iz - 12.
Meanwhile, |z - 2i|^2 = zz* - 2iz* + 2iz + 4. So the original equation becomes
|z - 2i|^2 - 16 = 0
| |z - 2i| | = 4
|z - 2i| = 4
(iii) The graph of |z| = k is a circle of radius k centered at the origin. What does subtracting 2i from z do to that graph?
Everything summarized. Hope you can understand. Am in a bit hurry.
If you don't get inbox me.
I got it, Thank for your help =), can you help me this ? 32/0/n/11 question 7) iii)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_ms_32.pdf
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who can help me this ? 32/0/n/11 question 7) iii)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_ms_32.pdf
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_ms_32.pdf
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here you go http://sh.st/y4i3I
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Can you guys differentiate with respect to x ..
Q=
Sinx + Cosx
Sinx - Coxs
the answer is supposed to be 2 / Sin2x - 1
RoOkaYya G
Q=
Sinx + Cosx
Sinx - Coxs
the answer is supposed to be 2 / Sin2x - 1
RoOkaYya G
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Need help! :-( http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_33.pdf
Q.6 (ii). I know how to get the radius but I can't figure out how to get the center of the circle?
Q.6 (ii). I know how to get the radius but I can't figure out how to get the center of the circle?
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