Mathematics: Post your doubts here!

[$~SauD~$]

ok from "3x^2 - 6ax - y^2 = 0", you can make y^2 the subject of the formula
so, "y^2 = 3x^2 - 6ax"
then substitute "y^2" into the equation of the curve and solve for "x"

Ohhhh How stupid of me
Last question
How to solve "4x^3 - 6ax^2 -6a^2x = 0"

 

[Kumkum]

Ohhhh How stupid of me
Last question
How to solve "4x^3 - 6ax^2 -6a^2x = 0"

lol, it happens
btw i think you should check your dy/dx again
shouldn't it be "3x^2 - 6ax + y^2 = 0" instead of 3x^2 - 6ax - y^2 = 0
sorry i should have said it earlier but i think your dy/dx is wrong

 

[$~SauD~$]

lol, it happens
btw i think you should check your dy/dx again
shouldn't it be "3x^2 - 6ax + y^2 = 0" instead of 3x^2 - 6ax - y^2 = 0
sorry i should have said it earlier but i think your dy/dx is wrong

Ni, actually I make loads of Silliest mistakes -.-
STUPID me again -.- I forgot to change the sign -.- I hate it when I try to take shortcuts
anyways, Thanks a lot Kumkum.. Got My answer too
btw, you giving any exam in oct/nov? cause.. I will bug you whenever i have maths doubts :3

 

[Kumkum]

Ohhhh How stupid of me
Last question
How to solve "4x^3 - 6ax^2 -6a^2x = 0"

if ur dy/dx is "3x^2 - 6ax + y^2 = 0"
then "y^2 = 6ax - 3x^2"

after substituting into the equation of the curve and simplifying u'd get:
6a^2x - 2x^3 = 0
x[(6a^2) - (2x^2)] = 0
so,
x = 0 or 2x^2 = 6a^2
x^2 = 3a^2 (taking square root both sides)
x = +/- (sq.root3)a
since "M" is on the positive x-axis , this means the x-coordinate = (sq.root 3)a

hope i've helped

 

[Kumkum]

Ni, actually I make loads of Silliest mistakes -.-
STUPID me again -.- I forgot to change the sign -.- I hate it when I try to take shortcuts
anyways, Thanks a lot Kumkum.. Got My answer too
btw, you giving any exam in oct/nov? cause.. I will bug you whenever i have maths doubts :3

it's ok...but be careful next time and ur welcome
yeah m giving maths, bio, phy nd chem
not a problem, i'll try my best to help you

 

[Khadija_1234]

How to solve the inequality: x-x^3 < 0
please help

 

[♫Prince Shah♫]

i dont knw anything abt quadratics can u teach me tat damn!

 

[JalalKaiser]

 

[JalalKaiser]

i dont knw anything abt quadratics can u teach me tat damn!

What you find attached is all there is to it, honestly!

 

[kaushar]

Ok. So we have
OA (1i, -2j, 2k) and OB (3i, pJ, qK)

for i) It says both are parallel. The clue is the "i" in both vectors. It goes from 1 to 3. Therefore "x3" OA to get to OB.

So
-2 x "3" = -6 = p
2 x "3" = 6 = q

for ii) it says 90 degrees which means the dot product =0
(1,-2, 2) and (3, p, 2p)
so:
(1 x 3) + (-2 x p) + (2 x 2p) = o
3 - 2p + 4p = 0
2p = -3
p=-3/2

finally iii)

AB means b-a (for e.g CB means b-c, AD means d-a etc etc)
so: (3, 1, 8) - (1, -2, 2) = (2, 3, 6)

To get the Unit vector first find the Magnitude of it which is Square root of (2^2 + 3^2 + 6^2)
Which is square root of 49 ==> 7

Lastly you gotta divide vector by this number.
Hope you understood!

yep i understood thnks a lot...

 

[kaushar]

thnks ...

 

[Dudu]

I'm not sure i understand it either by if you did the first part correctly you should have no trouble finding the answer.
You already found that OC = r x Sin θ and BC = r x Cos θ

r = 10 and θ = Pie/5

So: 10 x Sin (pie/5) = OC = 5.87
and: 10 x Cos (pie/5) = BC = 8.09

5.87 x 8.09 x 1/2 = 23.7

 

[salvatore]

There are two questions that are really troubling me.. they look quite simple though. Could someone please help me solve it?

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s10_qp_31.pdf
Qn no. 1

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_32.pdf
Qn no. 2

I'll appreciate your help.
Thanks!

 

[Ahmedm96]

in j12 p4 v1 no7 why didn't we consider rings wheight. in (I)

help pleaae

 

[Kumkum]

There are two questions that are really troubling me.. they look quite simple though. Could someone please help me solve it?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_31.pdf
Qn no. 1

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf
Qn no. 2

I'll appreciate your help.
Thanks!

M/J/10/P31:

1. |x + 3a| > 2|x - 2a| (to remove the modulus sign, u can square both sides)
(x + 3a) ^2 > (2x - 4a)^2
x^2 + 6a + 9a^2 > 4x^2 - 16ax + 16a^2
3x^2 - 22ax + 7a^2 < 0

so solving linearly, 3x^2 - 22ax + 7a^2 = 0 (factorising it)
3x^2 - 21ax - ax + 7a^2 = 0
(3x - a) (x - 7a) = 0
==> x = a/3 or x = 7a
now including the inequality sign you have : (a/3) < x < 7a

M/J/11/P32:

2.(i) log2[x + 5] = 5 - log2 [x]
log2[x + 5] + log2[x] = 5 <===here u use one of the laws of logarithm to simplify, when u add two logarithms, u multiply them since they have the same base(not sure if m clear enough)

so it will be,
log2[(x+5)*x] = 5 (u have to change this to exponential form, so 2 becomes the base and 5 the index)
x^2 + 5x = 2^5
x^2 + 5x - 32 = 0

(ii) u just solve the quadratic equation above

hope i helped

 

[JalalKaiser]

Q ] A (7,2) and C (1,4) are two vertices of a square ABCD.
(a) Find the equation of the diagonal BD.
(b) Find the coordinates of B and D.

First part answer is [ Y = 3X - 9 ]
Second part answer is (3,0) and (5,6)

HOW TO WE SOLVE THE SECOND PART

 

[JalalKaiser]

I'm not sure i understand it either by if you did the first part correctly you should have no trouble finding the answer.
You already found that OC = r x Sin θ and BC = r x Cos θ

r = 10 and θ = Pie/5

So: 10 x Sin (pie/5) = OC = 5.87
and: 10 x Cos (pie/5) = BC = 8.09

5.87 x 8.09 x 1/2 = 23.7

I had my calculator set to the degrees mode hahahahahhaha ooohh, important lesson learned in time thank Allah.

 

[salvatore]

M/J/10/P31:

1. |x + 3a| > 2|x - 2a| (to remove the modulus sign, u can square both sides)
(x + 3a) ^2 > (2x - 4a)^2
x^2 + 6a + 9a^2 > 4x^2 - 16ax + 16a^2
3x^2 - 22ax + 7a^2 < 0

so solving linearly, 3x^2 - 22ax + 7a^2 = 0 (factorising it)
3x^2 - 21ax - ax + 7a^2 = 0
(3x - a) (x - 7a) = 0
==> x = a/3 or x = 7a
now including the inequality sign you have : (a/3) < x < 7a

M/J/11/P32:

2.(i) log2[x + 5] = 5 - log2 [x]
log2[x + 5] + log2[x] = 5 <===here u use one of the laws of logarithm to simplify, when u add two logarithms, u multiply them since they have the same base(not sure if m clear enough)

so it will be,
log2[(x+5)*x] = 5 (u have to change this to exponential form, so 2 becomes the base and 5 the index)
x^2 + 5x = 2^5
x^2 + 5x - 32 = 0

(ii) u just solve the quadratic equation above

hope i helped

That was soo easy.. shame on me!

Thanks a lot

 

[JalalKaiser]

How to solve the inequality: x-x^3 < 0
please help

Take the x common. You'll get (x)(1 - x^2) < 0. That makes two roots, (x) < 0 and (1 - x^2) < 0. The first one's solved in itself, for the second one you just go about it as a regular equation; 1 < x^2 and then under-root both the sides and you'll get x > 1. I think I might be wrong here though but hopefully somebody can guide us both a bit here, or maybe you'll get it.

 

[Abhishek gohil]

can anyone give me all the formulas of the mechanics 1
plzzzzzz need help