Mathematics: Post your doubts here!

[Thought blocker]

Thank you!

Welcome

 

[Thought blocker]

guys i need help with question #9(b) on this paper...........
please help!!!!!!!!!!!!!!!!!!!!

For the one triangle BXC :
The area of segment formula u found out in a(ii) use that to find the area of segment BC with r = 4 and alpha = pi/6 = 8pi/3 - 4sqrt(3)
The area of triangle BXC = {1/2 * 2 * 2sqrt(3)/3 } * 2 = 4sqrt(3)/3
The shaded region BXC = area of triangle BXC - Area of segment BC = 16sqrt(3)/3 - 8pi/3

The shaded area BXC = shaded area AXC = shaded area AXB

Therefore total shaded area = 3 * shahded region BXC = 16sqrt(3) - 8pi

 

[aqibmansoor97]

If I got 35 in stats than? xD

Then B or also A BTW why ru giving again

 

[shingiechingz@8]

guys i need your help.....
P,Q and R are the angles of a triangle such that cosP=3/5 and cosQ=5/13
Without using a calculator find the value of cosR [2]

 

[M.Omar]

guys i need your help.....
P,Q and R are the angles of a triangle such that cosP=3/5 and cosQ=5/13
Without using a calculator find the value of cosR [2]

R=180-(P+Q) cos(R)=cos(180-(P+Q))= - cos( P+Q) (from identity cosx=-cos(180-x)) now simply find -cos(P+Q) by using the formula cos(A+B)=cosAcosB-sinAsinB
and sinx=(1-(cosx)^2)^(1/2)

 

[shingiechingz@8]

I got another question......
the variable complex number z is given by
z=2cosA+i(1-2sinA) whre A takes all values in the interval -pi<A<pi
prove that the real part of 1/(z+2-i) is constant for -pi<A<pi ............

 

[M.Omar]

I got another question......
the variable complex number z is given by
z=2cosA+i(1-sinA) whre A takes all values in the interval -pi<A<pi
prove that the real part of 1/(z+2-i) is constant for -pi<A<pi ............

are u sure the question is right ?

 

[Thought blocker]

I got another question......
the variable complex number z is given by
z=2cosA+i(1-sinA) whre A takes all values in the interval -pi<A<pi
prove that the real part of 1/(z+2-i) is constant for -pi<A<pi ............

1/4?
Substitute that z in that denominator so it will be :
1/(2cos(theta) - i(sinn(theta)))
then solve this as u usually solve your complex number problems when u have denominator in terms of i.
you will get 1/4 as the real part.

 

[bakhita]

okay I understand until we reach the critical values x =1 and x =-7 but I don't get this next part, the line method to get real roots. How do we use it?

 

[shingiechingz@8]

View attachment 60930
okay I understand until we reach the critical values x =1 and x =-7 but I don't get this next part, the line method to get real roots. How do we use it?

substitute any value when x<-7 we get a positive value which is not the solution (>0)
substitute any value when x>1 we get a positive value which is not the solution (>0)
sustitute values when -7<x<1 we get a negative value which is the solution....[<0]

 

[shingiechingz@8]

guys i need help with #8(ii) and (iii)

 

[Thought blocker]

guys i need help with #8(ii) and (iii)

Diagram not to scale. Just for an idea.
red line = cos(pi/4) * 3
blue line = sqrt(3^2 + 3^2) + 1

 

[Thought blocker]

View attachment 60930
okay I understand until we reach the critical values x =1 and x =-7 but I don't get this next part, the line method to get real roots. How do we use it?

They have used that to show the points satisfying in that region. The same is explained by :

substitute any value when x<-7 we get a positive value which is not the solution (>0)
substitute any value when x>1 we get a positive value which is not the solution (>0)
sustitute values when -7<x<1 we get a negative value which is the solution....[<0]

The easy way to remember this thing is :
if any inequality with < 0
then the roots will be sth like : less positive/ more negative value < x < more positive/ less negative value

if with > o
then, x > more positive/ less negative value and x < less positive/ more negative value

lets take ur example : x = 1 and x = -7 and inquality < 0
thius the answer will be -7 < x < 1

 

[Thought blocker]

You guys can vote the link in my signature, the first link in blue and do like it if i helped u guys It contains the best revision resource of PCM ur mbs wont be wasted

 

[shingiechingz@8]

They have used that to show the points satisfying in that region. The same is explained by :


The easy way to remember this thing is :
if any inequality with < 0
then the roots will be sth like : less positive/ more negative value < x < more positive/ less negative value

if with > o
then, x > more positive/ less negative value and x < less positive/ more negative value

lets take ur example : x = 1 and x = -7 and inquality < 0
thius the answer will be -7 < x < 1

you can also memorise this
if (x-a)(x-b)<0 then a<x<b if a<b
b<x<a if a>b
if (x-a)(x-b)>0 then x<a , x>b if a>b
x>a ,x<b if a<b

 

[Thought blocker]

http://dynamicpapers.com/wp-content/uploads/2015/11/9709_w15_qp_33.pdf

Q8(ii)
The formula is |a||b|.cos(x) = a.b then why we do |a||b|.sin(x) = a.b
Q8(iii)
I got that one vector (-3,11,-20)
how to find the other vector?

Doubts cleared. ^_^
Ty.

 

[shingiechingz@8]

need help with ques #7(b)

 

[Thought blocker]

need help with ques #7(b)

 

[Thought blocker]

need help with ques #7(b)

You already know one line is y = x
The other eqn of line can be found as i showed!
for the gradient use the formula the angle between two lines formula.

CAN YOU HELP ME OUT IN THIS : https://www.xtremepapers.com/commun...st-your-doubts-here.9599/page-884#post-969456 ?

 

[The Sarcastic Retard]

Yep that's an A for sure

ARE U SURE I WILL GET AN A?

Recently one of my friend said that total percentage of A levels is divided like this :
30% P1 and 20% S1 = 50% As level
30% P3 and 20% S2 = 50% A2 level
So overall A level grade will be depended on both, so as I scored 70% (b) in As level now I will have to get 90% (a) in my A2 level exam to score overall (A).
Is it like this or how did u said YES that's A for sure? :O Please temme asap, i wanna know how much hard work i should do xD