Pure 3 help.

[Deadly_king]

I am back now!

So you need to use the two equations. By elimination method, you eliminate one of the variables.

Eqn 1 : 2x - y - 3z = 7
Eqn 2 : x + 2y + 2z = 0
Step 1
Multiply Eqn  by 2 and you'll be getting 4x - 2y -6z =14 while Eqn 2 is still x +2y +2z = 0.

Add both equations and you'll note that -2y cancels 2y. You'll thus be getting 5x - 4z = 14
Make z subject of formula to get z = (5x - 14)/4

Now you need to find z in terms of y. So you should eliminate the values of x.
Step 2
Multiply Eqn 2 by -2 and you'll be getting -2x - 4y - 4z = 0 while Eqn 1 remains 2x - y -3z = 7

Add both equations and this time 2x will cancel -2x. Hence you'll obtain -5y - 7z = 7
Again make z subject of formula to get z = (-5y - 7)/7

Now you equate all the equations of z you obtained just like Requiem previously did.

Then you should make the coefficients of both x and y = 1

I guess by now, you'll be able to understand the workings of Requiem

[Deadly_king]

Nov 09 No 2, 6, 10

2.
3^(x+2) = 3^x + 3²
 
Divide everything by 3^x and you'll get : ( dividing implies powers are subtracted)
3² -1 = 3^(2-x)
Therefore 3^(2-x) = 9 - 1

Apply ln on both sides
ln 3(2-x) = ln 8
(2-x)ln 3 = ln 8

Hence x = 2 - (ln 8/ln3) = 0.107

Sorry but I need to go to tuition now. Will complete the other numbers once am back

[S.M.A.T]

Nov 09 No 2, 6

[S.M.A.T]

DK!! ...that was really helpful

+rep

[mousa]

God bless you all.  thanks for all reponses!!

[Deadly_king]

DK!! ...that was really helpful

+rep

That's alright br0

Thanks for completing the other numbers 

You're welcome mousa

[mousa]

Jun 09
3.
(i) Let pheta be x

Prove cosec 2x +cot 2x = cot x

cosec 2x + cot 2x = 1/(sin 2x) + (cos 2x)/(sin 2x)
Take sin 2x as common denominator to get (1 + cos 2x)/(sin 2x)
Use double angle formula sin 2x = 2sinx cosx and 1 + cos 2x = 2cos²x
You'll get (2cos²x)/(2sinx cosx)
Eliminate cosx from the numerator and denominator to obtain cosx / sinx = cot x

(ii) cosec 2x + cot 2x = 2
This also implies that cot x = 2 ----> 1/tanx = 2
therefore tanx = 1/2
Key angle x = tan^-1(1/2) = 26.6^o
tan is positive in first and third quadrant.
Hence x = pheta = 26.6^o and 206.6^o

8.
Let 100/x²(10-x) = A/x + B/x² + C/(10-x)
Take x²(10-x) as common denominator on the right hand side :
A(x²(10-x)) + B((10-x) + C(x²) = 100
Now to find C : take x=10 in the equation above -----> C=1
Now to find B : take x=0 ----> B 10
Now to find A : take x=1 ----> 9A + 9B + C = 100
Replace values of B and C calculated above to obtain A=1

Hence 100/x²(10-x) = 1/x + 10/x² + 1/(10-x)

(ii) Give dx/dt = 1/100*x²(10-x)
Send the terms in x to the left side and you'll obtain :
100/x²(10-x) dx = dt
Use the answer obtained in part (i) to replace 100/x²(10-x)
Integrate on both sides to get :
ln x - 10/x + ln(10-x) = t + c

Given when x=1, t=0 ----> replace in the equation to get the value of c
c = -10 - ln 9

Therefore ln x - 10/x - ln(10-x) = t -10 - ln 9
Make t subject of formula and use logarithm rules to obtain answer as :
t = ln (9x/(10-x)) - 10/x + 10


9
Given L lies in the plane ----> Vector 4i + 2j - k should fit in the equation of the plane.

Am sorry I don't know how to type it in column vectors. It's damn easier to explain in column vectors

(4i + 2j - k).(2i + bj + ck) = 1
From this dot product you'll obtain equation : 2b - c = -7

When L lies in the plane ---> It also implies that the direction vector of line is perpendicular to the normal vector of the plane. perpendicular implies dot product = 0.
(2i - j - 2k).(2 + bj + ck) = 0
From this you'll obtain : b + 2c = 4

Now you need to solve the two equations simultaneously to obtain b = -2 and c = 3

(ii) For PQ to be perpendicular to L ----> PQ.(2i - j - 2k) = 0
Take vector equation of PQ as (4+2t)i - tj - (5+2t)k
The dot product will indicate that t = -2

When t=-2 ----> point where L meets PQ perpendicularly is (4j + 3k)

Perpendicular distance : Square root of ( 0² +(4-2)² + (3-4)²)
Answer is square root of 5.

10
(i) Since M is a stationary point ---> dy/dx = 0 at M.

Use the product rule dy/dx = u.dv/dx + v.du/dx

Take u = x² ---> du/dx = 2x
Hence v = (1-x²)^-1/2 ---> dv/dx = -x(1-x²)^-1/2

therefore dy/dx = x2(-x(1-x²)^-1/2) + ((1-x²)^-1/2)(2x)
Solve dy/dx = 0 but x>0 (from range)

You'll b getting x as the positive square root of 2/3.

(ii) I'll take pheta as A
x = sin A ----> dx/dA = cos A
Hence dx = cos A dA

When x=0 ---> A = 0 and pie/2

Area = Integration of x²(1-x²)^1/2) dx
Substituting x=sinA and dx = cos A dA, you'll be getting :
Area = Integration of sin²A (1-sin²A)^1/2 cos A dA with limits 0 and pie/2

Replacing 1- sin²A by cos²A
Then sin²A cos²A = (sin²2A)/4

Area is shown to be 1/4 x integration of sin²2A dA with limits 0 and pie/2.

(iii) Replace sin²2A by (1-cos2A)/2 using double angle formula.

Then integrate normally

A = -1/8 * integration of (cos 2A -1) dA with limits 0 and pie/2

Solution will be pie/16

WOW, You NAILED it MAAN!! Thnxxxxxxxxxx

[Deadly_king]

WOW, You NAILED it MAAN!! Thnxxxxxxxxxx

Hehe.....anytime pal

I just hope you understand everything 

[Freaked12]

November 08
Question8
Part i)


note:in the second image k/4=0.05

[Freaked12]

part ii)

[Freaked12]

part iii)

[Freaked12]

I would like to add again

This is my tutor's work
I will in a month's time be able to master A2 maths.


And my tutor is the guy that got me a A in as maths so worry not

[Deadly_king]

I would like to add again

This is my tutor's work
I will in a month's time be able to master A2 maths.


And my tutor is the guy that got me a A in as maths so worry not

Hey that's alright dude. We trust you

Anyway thanks for your help

[astarmathsandphysics]

Those images are all black

[elemis]

Those images are all black

They were taken in poor light. Open them up and increase their brightness in MS Word.