Mathematics: Post your doubts here!

[Awesome12]

the mark scheme says 154 for ACE.. Please check it, ON 2009,P22

AND also BCE and DBC can be co interior angles right? that means theey should be supplementart?? how can it be equal to 68 then? :S

no
/_ DBO and /_ COB are supplementary

 

[Awesome12]

also bro how is triangle OCA an isosceles triangle how can u find it and tell?

any line extending from the circumference of a circle towards the centre is equal to any other line extending from the circumference of a circle towards the centre
thus OC = OA

 

[Dubi Pro]

darn

any line extending from the circumference of a circle towards the centre is equal to any other line extending from the circumference of a circle towards the centre
thus OC = OA

thanks bro! so according to that. even that should apply in this question as well?

as the line AC also passes the centre O so any isoceles in this as well?

posted here? can u solve this as well? q`15

 

[Counter Logic Gamer]

bro got in a ditch!! problem in sequences when differences is varying....
Ex:20,17,13,8
7,9,12,16
bro hv xams cming so pps help asap!! if you have sequences notes pls send to [email protected]

 

[Dubi Pro]

guys the question number 7. will angle 37 be equal to angle c? as the rule applies co interior angles are supplementart or do they add up to 180? :S

PLEASE HELP GUYS!

 

[Awesome12]

guys the question number 7. will angle 37 be equal to angle c? as the rule applies co interior angles are supplementart or do they add up to 180? :S

PLEASE HELP GUYS!View attachment 37334

no there are 2 methods
(1) c + 90 + 37 = 180
c = 53

(2) 90 - c = 37
c = 53

 

[Dubi Pro]

no there are 2 methods
(1) c + 90 + 37 = 180
c = 53

(2) 90 - c = 37
c = 53

Bro. How? I mean see, 90-37= 53,

its an isoceles, so if thts 53, eve the 2nd angle c is 53

am i right?

 

[Awesome12]

Bro. How? I mean see, 90-37= 53,

its an isoceles, so if thts 53, eve the 2nd angle c is 53

am i right?

Yes you are right , that is another way of solving it

 

[Dubi Pro]

see bro lsn! the question i gave u now is quite similar to this circle theorEm.

see im really confused, not aasking you to solve the whole question,
BUT angle p and angle r SHOULDNT THEY BE CYCLIC QUADRILATERAL AND ADD UP TO 180"? darn! and the same thing in previous quest

1st ans :
96
48
97 how
35 how

 

[Phoenix Blood]

PLEASE HELP ME WITH THIS!

 

[Igcse stuff]

PLEASE HELP ME WITH THIS!

easy, just say since 4 terms = 30 and 3 terms = 14, then you can form a simultaneous equation.

[a(4)^3+b(4)^2+4/6]=30
[a(3)^3+b(3)^2+3/6]=14

64a+16b+4/6=30
27a+9b+3/6=14

64a+16b=30-4/6
27a+9b=14-3/6

64a+16b=29.3
27a+9b=13.5

now all you have to do is solve the simultaneous equation to find the values of a and b

 

[Igcse stuff]

see bro lsn! the question i gave u now is quite similar to this circle theorEm.

see im really confused, not aasking you to solve the whole question,
BUT angle p and angle r SHOULDNT THEY BE CYCLIC QUADRILATERAL AND ADD UP TO 180"? darn! and the same thing in previous quest

1st ans :
96
48
97 how
35 how

easy, all you have to do is use the circle properties

(iii) to find angle SPQ you have to do the following:

<QRP=<QSP=35 and <PRS=<PQS=48

therefore,

<SPQ= 180 - (<PQS + < QSP)
= 180 - ( 48 + 35)
= 180 - 83
= 97

(iv) use <QRP=<QSP=35

 

[Phoenix Blood]

easy, just say since 4 terms = 30 and 3 terms = 14, then you can form a simultaneous equation.

[a(4)^3+b(4)^2+4/6]=30
[a(3)^3+b(3)^2+3/6]=14

64a+16b+4/6=30
27a+9b+3/6=14

64a+16b=30-4/6
27a+9b=14-3/6

64a+16b=29.3
27a+9b=13.5

now all you have to do is solve the simultaneous equation to find the values of a and b

Thanks so much
But the answer i am getting is -o.5 and -1/3
Whereas in the marking scheme it is only 0.5 and 1/3
What's it?

 

[Phoenix Blood]

Hey, can you tell me how to solve Q.6(b)? Which sine rule do we use?

 

[Phoenix Blood]

Are the sine, cos, and tan rules (soh,cah,toa) only applicable to right angled triangles?

 

[Dubi Pro]

easy, all you have to do is use the circle properties

(iii) to find angle SPQ you have to do the following:

<QRP=<QSP=35 and <PRS=<PQS=48

therefore,

<SPQ= 180 - (<PQS + < QSP)
= 180 - ( 48 + 35)
= 180 - 83
= 97

(iv) use <QRP=<QSP=35

thanks!

 

[chocolatelover]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w06_qp_4.pdf
I need help in Q4 (d)(ii). I have drawn the graph and everything, but how do i use it to get the answer.

 

[Dubi Pro]

Help in the 1st quest please..

 

[***amd***]

Help in the 1st quest please..

Q9
b)
i) it is written 'Union' of the 2 sets, not intersection. so the answer will be 8.
ii) P' intersection Q' = {elements of epsilon except the elements of P} intersection {elements of epsilon except that of Q)
= {1, 2, 3, 4, 5} int. {4, 5, 6, 7, 8)
= {4, 5}

 

[Dubi Pro]

Q9
b)
i) it is written 'Union' of the 2 sets, not intersection. so the answer will be 8.
ii) P' intersection Q' = {elements of epsilon except the elements of P} intersection {elements of epsilon except that of Q)
= {1, 2, 3, 4, 5} int. {4, 5, 6, 7, 8)
= {4, 5}

but the answer for bi is 6 not 8


for second on its correct.