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Mathematics: Post your doubts here!

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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
 
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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`15doubt2.png
 
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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!doubt5.png
 
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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 :(
 

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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 (y)
 
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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
 
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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 (y)
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?
 
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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! :)
 
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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}
 
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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.
 
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