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Maths, Addmaths and Statistics: Post your doubts here!

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But we have K with AE.
Can you please solve this further?
AE= 6p+kp-q
BC= 2p+q.
I know I'm very annoying. :p
Yes very annoying lol. Value hi galat btai thi.

pLgCKWW.png


Rule is to eliminate all variables as K is a numeric value. -5q can be eliminated by +5q
 
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Can anyone explain explicitly about these stupid nearest cm , nearest kg and nearest 10s and all that? What the hell is that? I don't get a thing in that!
http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_w08_qp_1.pdf
Question 11 part b. anyone? It's related to the nearest kg. again.
Whatever it is nearest to, divide that by 2 to get the bounds.
So the bound values of the box are 0.1/2 = ±.05 and for the container, they're 0.5/2 = ±0.25
You have to find the greatest possible mass, so use the + values of both bounds (each box is 1.55 kg and the container is 6.25 kg)
 
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Whatever it is nearest to, divide that by 2 to get the bounds.
So the bound values of the box are 0.1/2 = ±.05 and for the container, they're 0.5/2 = ±0.25
You have to find the greatest possible mass, so use the + values of both bounds (each box is 1.55 kg and the container is 6.25 kg)
So if it's nearest 10's of kg, degree whatever we'll divide 10 by two? And if it's 100 so we'll divide it by 2 as well.? When do we do 1/2?
 
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To find y:
cqXAchY.png


Rotate the object until you get the same shape. The red line has a new position. There is a angle between new and old position of the line:
mvm8cdL.png


The object has rotational symmetry of order 3 so you get that angle 3 times and in those 3 times the line has rotated 360 degrees so each angle is 360/3 = 120.

y = 360-120-53-40 = 147.

Read n-fold rotational symmetry
 
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In the locus questions when they usually ask us to define a region by giving three restrains on the region,they ask in condition for instance tht the region should be less than 3 cm from a point .then are we to draw a dotted line tht encloses the region(cos region is before 3 cm and the boundary is the line we r drawing) or shud we draw a solid line?
Also in estimation like this one it is fairly easy to get 39 *3/6 yielding 19.5 but the MS states 20 as the correct answer I mean we can do simple calculations such as this is one without further estm and this will give us a more closer answer.Is ther any sort of rule like to est all the relevant quantites to 1 or 2(which seems more apt) sig fig so 38.89=40 and wud give 20 as an answer
http://papers.xtremepapers.com/CIE/...Calculator Version) (4024)/4024_w13_qp_11.pdf
Q7 I am referring to in the abv context
 
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No...it is a bit tricky
here,s the deal
since OP is a common side side we just need to find the ratio of their heights right...
Since ba and op are paralle their HEIGHT component can be expressed as their ratio of lengths in general(if u wud like I will provide a detailed expl abt this)
so it simple becomes=(1/2 *OA*BA)/(1/2 *OA*OP)
so u will see it,s 1.5
 
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No...it is a bit tricky
here,s the deal
since OP is a common side side we just need to find the ratio of their heights right...
Since ba and op are paralle their HEIGHT component can be expressed as their ratio of lengths in general(if u wud like I will provide a detailed expl abt this)
so it simple becomes=(1/2 *OA*BA)/(1/2 *OA*OP)
so u will see it,s 1.5
i know that. but i dont understand where is their height?
 
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No...it is a bit tricky
here,s the deal
since OP is a common side side we just need to find the ratio of their heights right...
Since ba and op are paralle their HEIGHT component can be expressed as their ratio of lengths in general(if u wud like I will provide a detailed expl abt this)
so it simple becomes=(1/2 *OA*BA)/(1/2 *OA*OP)
so u will see it,s 1.5
For that area part, why don't we use the formula A1/A2= (l1/l2)^2 over here?
 
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