Mathematics: Post your doubts here!

[Iridescent]

pleasure!

 

[Osama$]

plz help me
Calculate the determinant of the matrice
B=(-3 2) C=(-5 3)
( -4 4) (-2 -6)
Calculate :2CB

Plz help me

 

[Iridescent]

Determinant of B=(3*4)-(2*-4)
=(12)-(-8)
=-4
Determinant of C=(-5*-6)-(3*-2)
=(30)-(-6)
=36
2CB=2(-5 3) * (-3 2)
(-2 -6) (-4 4)
=2(15-12 -10+12)
(6+24 12-24)
=2(3 2)
(30 -12)
2CB =(6 -4)
(60 -24)

 

[Osama$]

can anyone plz solve Q7 h

 

[CaptainDanger]

can anyone plz solve Q7 h
View attachment 3849

Answer?

 

[Osama$]

Answer -576

 

[Serah]

when its an acute triangle and two sides are visible only without any angle AOB......!!
1) find angle O?
2) Find de third side?
we can find de third side by using sine rule bt we need an angke 4 dat??
how to find de answer for 1 question plzz!!

 

[ChrisChin]

when its an acute triangle and two sides are visible only without any angle AOB......!!
1) find angle O?
2) Find de third side?
we can find de third side by using sine rule bt we need an angke 4 dat??
how to find de answer for 1 question plzz!!

I'd like to know the answer to this one, too. If only the length of two sides OA and OB are known, the triangle is like the two hands of a clock -- it can form any triangle with and acute angle AOB...

 

[Spirit95]

when its an acute triangle and two sides are visible only without any angle AOB......!!
1) find angle O?
2) Find de third side?
we can find de third side by using sine rule bt we need an angke 4 dat??
how to find de answer for 1 question plzz!!

Have you got a link to the question? The diagram might help us.

 

[Serah]

i would appreciate ur help....
here z de link.......question number 3 pg 4
http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s09_qp_4.pdf

 

[CaptainDanger]

i would appreciate ur help....
here z de link.......question number 3 pg 4
http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s09_qp_4.pdf

In a) part use the area formula
Area = (1/2)PQ x PR sin P.

b) Use cosine rule...

cosP=PQ2 + PR2-QR2/2(PQ)(PR)

 

[Iridescent]

Cosine rule actually is a2=b2+c2-(2bcCosA)

 

[Spirit95]

i would appreciate ur help....
here z de link.......question number 3 pg 4
http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s09_qp_4.pdf

Alright, here is the solution for the first part:
0.5*QP*PR*sin(P)=48
0.5*10*14*sin(P)=48
sin(P)=48/70
P=sin^-1(48/70)
P=43.29 a.k.a 43.3

For the second part:
QR^2= 296-(280sin(43.3))
Then find the square root.

 

[C5FlyingSquad]

Hello guys,
I would like some help on number sequences, those ones for paper 4. An example is October/November 2008. Syllabus code is 0580. I also heard that there are some formulas. I don't really know. Some help would be appreciated!

 

[CaptainDanger]

Hello guys,
I would like some help on number sequences, those ones for paper 4. An example is October/November 2008. Syllabus code is 0580. I also heard that there are some formulas. I don't really know. Some help would be appreciated!

Check these :
http://www.astarmathsandphysics.com..._rules_or_nth_terms_for_simple_sequences.html
http://www.astarmathsandphysics.com...les_or_nth_terms_for_quadratic_sequences.html
http://www.scribd.com/doc/15913808/Sequences-number-revision-notes-from-GCSE-Maths-Tutor
http://www.gcsemathstutor.com/topicnotes-number-seq.php

 

[XPFMember]

Aoa!

Check the post below:

Look ....Sequence seems to be hard but actually if u use a little bit of ur brain u will surely find it easy!!Believe me!!Just look at the question ....the first few parts are easy and the next part which u will find hard ...will be based on these easy parts that is a combination of those answers or there will be some relation which varies according to the question!!
Eg.if u r given a table with first four columns like n=1 ....to ....n=4 and then u wil have a column for n
this table will have info abt three sequences and u will have to find the nth term!!now the first two will be simple one ...like some addition or multiplication which u can do by the rules or formulas u have...however for finding the third one u will have a problem...but if u look carefully u will find that there may be some relation with the first two sequences...may be multiplying the first two gives the third one or adding them may give the third one ...so what ever it is u will find nth term in the same way like suppose the first two are multiplied togethr to give the third one so to find the nth term for the third one u multiply the nth terms of the first two!!That simple....

Hope this helps !!
Jazak ALlah Khair!!

 

[ChrisChin]

i would appreciate ur help....
here z de link.......question number 3 pg 4
http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s09_qp_4.pdf

Aha! Now we have the missing bit of information -- the area means the triangle can have only one height.

a) Either side of the triangle x sin P = height of the triangle, the other side being the length. This is simplified by the formula for area

A = 1/2 * base * height = 1/2 * r * q * sin P [ r = length of PQ ; q = length of PR ]
.: sin P = 2A/rq

Substituting and solving for P, and since P < 90; we get P=43.29° or P=43.3°, to one decimal place.

b) using P=43.3°, we can now solve for p [ length of QR ] using the cos rule

p² = q² + r² - 2*q*r*cos P

Substituting and solving, we get p = 9.6 cm

Serah, I think your mistake was not taking all the given information into account. You can overcome this problem by reading through the question carefully and listing down all the information, together with all formulas that are related to this sort of problem, at one side.

In this question, both the diagram and the area were information you either considered unimportant, or else didn't know how to use. What you can do is:

1) write down the formulas that apply
2) use a hilighter pen to mark the variables you are given.
3) the formula you should use is the one with the least number of unmarked variables
4) if you have more than one unmarked variable, check to see if you have been given a piece of information that you have overlooked, or one that can be used to determine a variable you need.

Cheers.

 

[emma_rodge]

hey, can anyone please help me with these probability problems?

1) A die has its six faces marked 0,1,1,1,6,6. Two of these dice are thrown together and the total score is recorded. Draw a tree diagram.
(a) How many different totals are possible?
(b) what is the probability of obtaining a total of 7?

2) Two dice, each with four faces marked 1,2,3 and 4 are thrown together.
(a) what is the most likely total score on the faces pointing downwards?
(b) What is the probability of obtaining this score on three successive throws of the two dice?

thanks so much

 

[soumayya]

CB=(-5 3) (-3 2) = ( 3 2 )
(-2 -6) (-4 4) (30 -28)

2CB= ( 6 4)
(60 -56)

/2CB/ =(6* -56) - (4 * 240 ) =-576

 

[soumayya]

1) A die has its six faces marked 0,1,1,1,6,6. Two of these dice are thrown together and the total score is recorded. Draw a tree diagram.
(a) How many different totals are possible?
(b) what is the probability of obtaining a total of 7?

Totals = 0, 1, 1, 2, 6, 6, 7, 7, 12
No. of totals =9
P(obtaining a total of 7) =2/9