Mathematics: Post your doubts here!

[Konstantino Nikolas]

You saved me, I also did silly mistake..
I hope I will score full marks, only this was my problem, rest other is as easy as piece of a cake. Moreover its as easy to say as binomial expansion, its just one silly mistakes can cost u many marks.. xD
I will tell u my marks afternoon.

Savior B|
piece of cake? really bro? cool ... i make mistakes in vectors and stuff ... prolly cuz i've only done 2 to 3 papers so far
Yeah sure do

 

[The Sarcastic Retard]

Savior B|
piece of cake? really bro? cool ... i make mistakes in vectors and stuff ... prolly cuz i've only done 2 to 3 papers so far
Yeah sure do

Paper was really damn easy, I think I made a mistake in binomial expansion xD That's it. As expected... completed paper in just 45 mins..
This was the question, can u please show me the working of it?

 

[The Sarcastic Retard]

Savior B|
piece of cake? really bro? cool ... i make mistakes in vectors and stuff ... prolly cuz i've only done 2 to 3 papers so far
Yeah sure do

Vectors are really easy to solve, just jot down all the concepts in one piece of a paper, and revise it everytime u come through that paper. And then see where u are making the mistakes in vector questions and work on it. Yes I also use to do mistakes till my 3 to 5 papers. But now, I have made my weak points to be strong points. Similarly do this with any conceptual questions. ^_^

 

[staying alive]

Repost: can someone help me with the last two problems? its a little urgent, would be much appreciated

 

[Konstantino Nikolas]

Paper was really damn easy, I think I made a mistake in binomial expansion xD That's it. As expected... completed paper in just 45 mins..
This was the question, can u please show me the working of it?

So yeah ... not as easy as just saying 'binomial expansion'
The 2nd part?

Vectors are really easy to solve, just jot down all the concepts in one piece of a paper, and revise it everytime u come through that paper. And then see where u are making the mistakes in vector questions and work on it. Yes I also use to do mistakes till my 3 to 5 papers. But now, I have made my weak points to be strong points. Similarly do this with any conceptual questions. ^_^

uhuh ... thanks man! will try to do but imma be lazy :/

 

[Konstantino Nikolas]

Repost: can someone help me with the last two problems? its a little urgent, would be much appreciated View attachment 59359

I think this is O level stuff? Not able to remember how to do this ... you can maybe post it in the IG threads here?

 

[nehaoscar]

This was the same type of question you had asked me to do binomial expansion in remember?
Well, I can't believe I'm getting it wrong again! So frustrating.
nehaoscar please once again?

Sorry for the late reply! I believe you answered this?
No silly mistakes this time!

 

[The Sarcastic Retard]

So yeah ... not as easy as just saying 'binomial expansion'
The 2nd part?


uhuh ... thanks man! will try to do but imma be lazy :/

Yeah
both (i) and (ii)

 

[Konstantino Nikolas]

Sorry for the late reply! I believe you answered this?
No silly mistakes this time!

yea that one done ... but now plz do this one

Retard bruh ... I ain't got the energy to try that and electric fields together so.

 

[nehaoscar]

yea that one done ... but now plz do this one

Retard bruh ... I ain't got the energy to try that and electric fields together so.

The Sarcastic Retard
Here you go

 

[Rizwan Javed]

Just the last two , thanks a lot ! View attachment 59312

let <ADE = a
then,
<ADE = <DCA = a

also,
<DCA = <DBA = a (<s subtended by arc at circumference are equal)

<ADB = 90 (angle subtended by diameter at circumference is 90)

<ADE + <EDB = <ADB
a + <EDB = 90
<EDB = 90 - a

now consider triangle DEB,
the sum of interior angles in a triangle is 180, so

<DEB + <DBA + <EDB = 180
<DEB + a + 90 - a = 180
<DEB = 90 (proved)

(ii) <ACB = 90 (angle subtended by diameter at circumference is 90)

in a cyclic quadrilateral, the sum of two angles facing each other is 180.

So,
<ACB = <FCB

<FCB + <FEB = 90+90 = 180

in a quadrilateral, the sum of all interior angles is 360, so

<FEB + <FCB + <CFE + <EBC = 360
180 + <CFE + <EBC = 360 ( <FCB + <FEB = 180 )
<CFE + <EBC = 180

Hence shown that FEBC is a cyclic quadrilateral.

 

[Konstantino Nikolas]

The Sarcastic Retard
Here you go
View attachment 59363
View attachment 59364

Great!

 

[staying alive]

let <ADE = a
then,
<ADE = <DCA = a

also,
<DCA = <DBA = a (<s subtended by arc at circumference are equal)

<ADB = 90 (angle subtended by diameter at circumference is 90)

<ADE + <EDB = <ADB
a + <EDB = 90
<EDB = 90 - a

now consider triangle DEB,
the sum of interior angles in a triangle is 180, so

<DEB + <DBA + <EDB = 180
<DEB + a + 90 - a = 180
<DEB = 90 (proved)

(ii) <ACB = 90 (angle subtended by diameter at circumference is 90)

in a cyclic quadrilateral, the sum of two angles facing each other is 180.

So,
<ACB = <FCB

<FCB + <FEB = 90+90 = 180

in a quadrilateral, the sum of all interior angles is 360, so

<FEB + <FCB + <CFE + <EBC = 360
180 + <CFE + <EBC = 360 ( <FCB + <FEB = 180 )
<CFE + <EBC = 180

Hence shown that FEBC is a cyclic quadrilateral.

thanks a lot! and the last problem if it isn't too much trouble....

 

[The Sarcastic Retard]

The Sarcastic Retard
Here you go
View attachment 59363
View attachment 59364

Thanks. ^_^

 

[phanttasy]

need help on this ! thanks part (ii)

 

[Konstantino Nikolas]

need help on this ! thanks part (ii)

 

[i_try9621]

How to solve 9 b) ?

 

[Rizwan Javed]

View attachment 59418 How to solve 9 b) ?

the common difference, r = 2cosθ / √3

-1 < 2cosθ / √3 < 1
multiply with √3, you'll get:

-√3 < 2 cosθ < √3

divide by 2, you 'll get:

-√3/2 < cosθ < √3/2

now taking cosine inverse,

π/6 < θ < 5π/6 (Ans.)

 

[Pixie Twinkles]

A book of mass 0.8 kg is placed on a rough plane inclined at an angle of 60 degrees to the horizontal. The book is just prevented from sliding down the plane by a force of magnitude 5N acting parallel to the line of greatest slope to the plane. Find the coefficient of friction between the book and the plane.

 

[Anum96]

A book of mass 0.8 kg is placed on a rough plane inclined at an angle of 60 degrees to the horizontal. The book is just prevented from sliding down the plane by a force of magnitude 5N acting parallel to the line of greatest slope to the plane. Find the coefficient of friction between the book and the plane.

Component of weight of book is F=mgsin60
normal to plane; R=mgcos60
Use abolute values;
F =0.8*9.8*sqrt(3)/2 = 6.79 N
R = 0.8*9.8*1/2 = 3.92 N.

Ff(Force of friction) = uR = 3.92u, where u is the coefficient of friction, and the applied force of 5N. ie
F = R + 5
6.79 = 3.92u + 5
1.79 = 3.92u
u = 0.456