Mathematics: Post your doubts here!

[imanmalik]

just confuse if thats possible to learn the Probability & Statistics 1 in 3 month and test in May/June

Definitely possible. Statistics isn't hard at all. It's basically logic. So if you've got a good brain you're good to go!

 

[Iffat]

can sum1 plz help me asap
-Express 4x2-12x+3 in the form (ax+b)2+c where a, b and c are constants and a is greater than 0
-The line y=5x-3 is a tangent to the curve y=kx2-3x+5 at the point A. Find the cordinates of A

can sum1 plz REPLY ASAP!!!!

 

[imanmalik]

P is a point in an argand diagram corresponding to a complex number z which satisfies |4+z| - |4-z| = 6. Prove that |4+z|^2 - |4-z|^2 ≧ 48 and deduce that Re z ≧ 3.

The only trouble I'm having here is how do you get |4+z|+|4-z|≧ 8

 

[Ibtizam]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s12_qp_42.pdf q11 b explain me d answer

 

[soumayya]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s12_qp_42.pdf q11 b explain me d answer

Area = (22/360 x π r^2) + (22/360 x 2πrh)
= (22/360 x π x 12^2) +( 22/360 x 2 x π x 12 x 7)

 

[soumayya]

can sum1 plz REPLY ASAP!!!!

4x^2 -12x +3 = 4 (x^2 -3x ) +3
= 4 ( (x- 3/2)^2 - (3/2)^2 ) +3
= 4 ( (x- 3/2)^2 - 9/4 ) + 3
= 4 (x- 3/2)^2 - 9 + 3
= 4 (x- 3/2)^2 + 6

 

[soumayya]

can sum1 plz REPLY ASAP!!!!

y = 5x - 3 , y = kx^2 + 3x + 5

Equate both equations:
kx^2 + 3x + 5 = 5x - 3
kx^2 + 3x - 5x + 5 + 3 = 0
kx^2 - 2x + 8 = 0

Tangent ---- b^2 -4ac =0
(-2)^2 - 4(k)(8) = 0
4 - 32k = 0
32k = 4
k = 1/8

 

[Rahma Abdelrahman]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s12_qp_42.pdf q11 b explain me d answer

That was my exam

 

[rafayet]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf
I need help with Q.9(i). i got coor of M, but how can i get coor of D? can you please help me?

 

[Iffat]

4x^2 -12x +3 = 4 (x^2 -3x ) +3
= 4 ( (x- 3/2)^2 - (3/2)^2 ) +3
= 4 ( (x- 3/2)^2 - 9/4 ) + 3
= 4 (x- 3/2)^2 - 9 + 3
= 4 (x- 3/2)^2 + 6

but its not in the form (ax+b)^2+c, isnt the ans (2x-3)^2-6?

 

[Iffat]

y = 5x - 3 , y = kx^2 + 3x + 5

Equate both equations:
kx^2 + 3x + 5 = 5x - 3
kx^2 + 3x - 5x + 5 + 3 = 0
kx^2 - 2x + 8 = 0

Tangent ---- b^2 -4ac =0
(-2)^2 - 4(k)(8) = 0
4 - 32k = 0
32k = 4
k = 1/8

uhh... u made a mistake its supposed 2 b y=kx^2-3x+5 hence the value of k is 2. i get that part but now how do u find the cordinates?

 

[anonymous123]

P is a point in an argand diagram corresponding to a complex number z which satisfies |4+z| - |4-z| = 6. Prove that |4+z|^2 - |4-z|^2 ≧ 48 and deduce that Re z ≧ 3.

The only trouble I'm having here is how do you get |4+z|+|4-z|≧ 8

|4+z| - |4-z| = 6 ---------i
|4+z|^2 - |4-z|^2 ≧ 48-------------ii

if u notice, the second eq is in the form a^2 - b^2..so if we expand:
(|4+z|+|4-z|) (|4+z| - |4-z|) >=48
put (i) in the eq:
(|4+z| +|4-z|)(6) >=48
|4+z| +|4-z| >=8

 

[imanmalik]

|4+z| - |4-z| = 6 ---------i
|4+z|^2 - |4-z|^2 ≧ 48-------------ii

if u notice, the second eq is in the form a^2 - b^2..so if we expand:
(|4+z|+|4-z|) (|4+z| - |4-z|) >=48
put (i) in the eq:
(|4+z| +|4-z|)(6) >=48
|4+z| +|4-z| >=8

Yeah I worked backwards and got that. But that's by working backwards! If we didn't know the equation was (|4+z|+|4-z|) (|4+z| - |4-z|) >=48 how would we get |4+z|+|4-z|≧ 8 from |4+z| - |4-z| = 6 :/ ? That's why I'm confused

 

[anonymous123]

Yeah I worked backwards and got that. But that's by working backwards! If we didn't know the equation was (|4+z|+|4-z|) (|4+z| - |4-z|) >=48 how would we get |4+z|+|4-z|≧ 8 from |4+z| - |4-z| = 6 :/ ? That's why I'm confused

ok...if u don't want to do it backwards think about what you have to do in order to get the req inequality..we do know that its in a^2 - b^2 form so if we multiply (|4+z| + |4-z|) to both sides of (i) we should get some results..
=> |4+z|^2 - |4-z|^2 = 6 (|4+z| + |4-z|)
6 (|4+z| + |4-z|) >=48
|4+z| + |4-z| >= 8

now u cant complain even though its the same thing repeated..where did u get this question anyway...doesn't seem like its for A level

 

[华尔街小胖胖]

Definitely possible. Statistics isn't hard at all. It's basically logic. So if you've got a good brain you're good to go!

thanks for your advice

 

[imanmalik]

ok...if u don't want to do it backwards think about what you have to do in order to get the req inequality..we do know that its in a^2 - b^2 form so if we multiply (|4+z| + |4-z|) to both sides of (i) we should get some results..
=> |4+z|^2 - |4-z|^2 = 6 (|4+z| + |4-z|)
6 (|4+z| + |4-z|) >=48
|4+z| + |4-z| >= 8

now u cant complain even though its the same thing repeated..where did u get this question anyway...doesn't seem like its for A level

Thank you so much. Our maths book is damn hard. CIE not Edexcel! :/

 

[Yousif Mukkhtar]

Guys can you explain how to prove the identity from simplifying left side to right side? Q2
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w10_qp_12.pdf

 

[Ahmedraza73]

Can Anyone start The differentation and integeration tution For AS Level?
I want to learn it from the Begining

 

[french410]

Guys can you explain how to prove the identity from simplifying left side to right side? Q2
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_12.pdf

ok listen up: sin^2 means sin squared (^ means raised to the power of)

then we use common denominator

to obtain:

then we factor

out to obtain:

therefore:

hope that helped!!!!!

 

[Kumkum]

Can someone please help me with this:
Given that 'a' is a positive constant, solve the inequality
|x - 3a| > |x - a|