Mathematics: Post your doubts here!

[techgeek]

View attachment 61473
May/June 2007 p6 how to do this?

here you go

appearing in May/June?

 

[Aathif Shaffy]

here you goView attachment 61478

appearing in May/June?

Thanks a lot.
Yeah hopefully may/june

 

[Hussain Jerry.]

Complex Numbers, i want this topic worked solution

 

[Thought blocker]

Complex Numbers, i want this topic worked solution

just ask the question and get solution here

 

[liptonice]

can someone tell me how to solve this question


they ask for the range -90<x<90

and they ask to solve sin(x)+sin(180-x)=1/2

what laws and theories do i need to know apart from the quadrant rule and the trigonometry identity ?

 

[Thought blocker]

can someone tell me how to solve this question


they ask for the range -90<x<90

and they ask to solve sin(x)+sin(180-x)=1/2

what laws and theories do i need to know apart from the quadrant rule and the trigonometry identity ?

Can u post the picture of that question? I am not able to understand else link would work else just a paper varient and year with season.

 

[liptonice]

Can u post the picture of that question? I am not able to understand else link would work else just a paper varient and year with season.

No man i didnt write it, it was just a question that the teacher gave us...

 

[liptonice]

https://latex.codecogs.com/gif.latex?sinx&space;&plus;&space;sin(180-x)=&space;1/2

 

[Thought blocker]

can someone tell me how to solve this question


they ask for the range -90<x<90

and they ask to solve sin(x)+sin(180-x)=1/2

what laws and theories do i need to know apart from the quadrant rule and the trigonometry identity ?

We just have to solve for x right?

 

[liptonice]

We just have to solve for x right?

Yeah

 

[liptonice]

anyone....

 

[Thought blocker]

anyone....

Answer is 14.5 degress in the domain of given x.
Here is the steps : https://drive.google.com/open?id=0B3NebwGVUDjGMlQzSEpsOXFZVWs

 

[liptonice]

Answer is 14.5 degress in the domain of given x.
Here is the steps : https://drive.google.com/open?id=0B3NebwGVUDjGMlQzSEpsOXFZVWs

Thank you sooo much, now its very clear, but shouldnt there be two answers ?

 

[Elena Gilbirt]

Can someone explain why in mark schemes dey put < « can you explain what dey mean

 

[Thought blocker]

Thank you sooo much, now its very clear, but shouldnt there be two answers ?

In the given domain, only that is the possible answer.

 

[Thought blocker]

Can someone explain why in mark schemes dey put < « can you explain what dey mean

can u post the pic?

 

[nehaoscar]

Here are complete AS and A2 notes for maths 9709 (S1 and P3)!
The notes include each and every point of the syllabus

https://www.xtremepapers.com/community/threads/maths-notes.43468/
https://www.xtremepapers.com/community/threads/maths-notes.43468/#post-966387

Please like and share if you find them helpful
Check out my signature for more A level notes and revision resources!

 

[iremos]

Hi people!
I need help on these questions, only managed to complete question 3b and 3c, any help will be greatly appreciated!
Thanks in advance!

 

[Thought blocker]

Hi people!
I need help on these questions, only managed to complete question 3b and 3c, any help will be greatly appreciated!
Thanks in advance!

This isnt from 9709 course. Which board is it?
I may give them a try if you can provide answers of it.

 

[Thought blocker]

All doubts except 9b) are here : https://drive.google.com/open?id=0B3NebwGVUDjGV0pXRUUwcVcxSFU
9b)
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
madridtas_CR7 Here are the solution you asked me for march 2016 paper 12 question 1(a)(ii) and question 9.