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Mathematics: Post your doubts here!

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10(1-x/10) – (10-x) – 1/100 (10-x) =0.05

plx sove this
Yo I started solving it but i just wanted to make sure your question is 10(1-x/10) – (10-x) – 1/100 (10-x) =0.05
Or 10((1-x)10)-(10-x)-(1/100)(1o-x) =0.05
Well this is kinda confusing can i have look at the original question cause we can't write it clearly over here!
 
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15b) use the ratios rule, square the ratio 1:5 and you will get 1:25 and 12:x, solve for x ie 300

21c) draw a tree diagram with three branches(RBG) and you will find 3 probabilities: RG(3/9*2/8),BG(2/9*4/8) and G and NOT G(ie blue+red(4/9*3+2/8)
that should give you 5/9
sorry but i didnot understand any of thm / :/
 
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guys!!! for sets how in the world are we suppose to know when we have to shade everything or only the circles? for ex A' U B do we shade the box or only B??? HELP PLX
 
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guys!!! for sets how in the world are we suppose to know when we have to shade everything or only the circles? for ex A' U B do we shade the box or only B??? HELP PLX
Rules for shading:

A U B : First shade all of A and then shade all of B. Some parts may be shaded twice but that doesn't matter.

A ∩ B : First shade all of A and then shade all of B. Then erase anything that hasn't been shaded twice.


So for A' U B, you first shade everything that is not in A (this includes the outside box area). Then shade all of B.
 
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Rules for shading:

A U B : First shade all of A and then shade all of B. Some parts may be shaded twice but that doesn't matter.

A ∩ B : First shade all of A and then shade all of B. Then erase anything that hasn't been shaded twice.


So for A' U B, you first shade everything that is not in A (this includes the outside box area). Then shade all of B.
ahhhh smart!!! ty a lot.. so one more question as u said for A' U B we shade the box too.. so how about A' ∩ B do u also shade the outside box?
 
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http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w11_qp_41.pdf
q4b(i) y will we minus 2.7 from 4.5.....while the radius of the cup is 4.5...so why shld we sbtract
q6b- y the answer is not 48.5?? y is it 131?
PLEASE HELP ASAP......
ok and q6 is a really good question.. but read it very carefully.. it says OBTUSE angle and when we get 48.5 thats acute so minus it from 180 to get the obtuse part.. usually in math when a word is written in bold.. u should stress on that part.. many make this mistake
 
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First of all,the answer to this:[ AA^-1 & A^-1 A ] Which one is the correct way to use and will the answer BE DIFF?

Here. Find P if Q=[2 -1 (4&0 below) ] and PQ= [ 6 -3 (8 & -2 below) ]



PS: I have another one tho. :/ Anyone here good at this?
 
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how to solve question 10 and 20 paper 2 oct/nov 2007
10)

First find the area of the big semicircle. This is:

pi*r^2 / 2 = pi*(12^2)/2 = ...

Then find the area of the small semicircle:

pi*r^2 / 2 = pi*(6^2)/2 = ...

And finally subtract these two answers.


20)

There are many isosceles triangles here: AOD, AOB, OBC and OCD. Can you fill in all the angles?

E.g. In AOB, the two unknown angles must be 45 and 45.
 
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ahhhh smart!!! ty a lot.. so one more question as u said for A' U B we shade the box too.. so how about A' ∩ B do u also shade the outside box?
A' ∩ B : First shade A'. This is everything not in A (including the part outside the circles).

Then shade all of B.

The part that has been shaded twice is the answer (erase everything else).

Here's a diagram to shows this:

KlL9ga9.jpg


The \ lines show A' and the / lines show B. Then the shaded area is the part that has been shaded twice.



If you call A', "not A" and ∩ "and", then it makes things easier. E.g. A' ∩ B is "B and not A". So this is everything in B that is not in A.
 
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