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Maths, Addmaths and Statistics: Post your doubts here!

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First find the relative velocity im pretty sure u guys know that :D
use the equation
V(L/S) = V(L) + (-V(S))
then use a scale of possibly 1 cm = 5km/h and plot a relative velocity line at a bearing of 270* which should if i remember it correctly about 9 cm then make an arrow showing that its moving towards the left !
Now connect the -V(s) line towards the left end of the relative velocity line u made before in such a way that u make 150* so the angle between the relative line and this new line is 60* now this line should be of 3 cm !
Last just join the remaining line to the figure and u got the diagram now use cosine rule for first part then sine rule and a bit of bearing knowledge in second part that all :D
but how do we know that the velocity of lifeboat relative to ship is 45 km/h ?? and is it necessary to use the scales?
 
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http://www.xtremepapers.com/papers/...(Calculator Version) (4024)/4024_s08_qp_2.pdf
question no. (4)(iv)(b), (7)(c), (9)(c)(i)&(ii).
Thnk u in advance. with method if possible plz.. :)
Question no 7 c
U have to do this by area to volume formula
(h/12)^3 = (1080/5000
Ull get the answer

Question no 9c(i)
The shotest distance is always the perpendicular distance which makes 90 degress so will will apply sin = perpendicular /hypotenuse and so multiply sin 65 x 2.8 !
 
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but how do we know that the velocity of lifeboat relative to ship is 45 km/h ?? and is it necessary to use the scales?

Exactly we can find that using the info given in the question abt the time it takes to cover the journey and the distance when the lifeboat sails that will give u the relative velocity for scales no its not a necessity but if u can make a scaled diagram it sometimes helps in understanding the bearing wala part and it also can allow u to check ur answer with the calculated answer so if u do the question u may know abt how successful u were :D
 
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Exactly we can find that using the info given in the question abt the time it takes to cover the journey and the distance when the lifeboat sails that will give u the relative velocity for scales no its not a necessity but if u can make a scaled diagram it sometimes helps in understanding the bearing wala part and it also can allow u to check ur answer with the calculated answer so if u do the question u may know abt how successful u were :D
yep got it now thanks a lot
 
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No problem !
i also want to ask one more thing, it's how you calculate area in geometry in add maths? i mean do we have to first get all the lengths by pythagorus way using all the cordinates that we have founded? it will be a very long process is there any other direct way?i don't understand the method using matrices as solved in pastpapers.
 
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i also want to ask one more thing, it's how you calculate area in geometry in add maths? i mean do we have to first get all the lengths by pythagorus way using all the cordinates that we have founded? it will be a very long process is there any other direct way?i don't understand the method using matrices as solved in pastpapers.

Heron's Formulae.
 
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The function f is defined for the domain –3 <_x<_ 3 by
f(x) = 9(x – 1/3)– 11.
(i) Find the range of f.
(ii) State the coordinates and nature of the turning point of
(a) the curve y = f(x),
(b) the curve y =  /f(x)/ . its dec 06 paper1 question can any one ans the b part please

http://www.xtremepapers.com/papers/CIE/Cambridge International O Level/Mathematics - Additional (4037)/4037_w06_qp_1.pdf
i think in that you just have to reflect the part along x-axis which lies below x-axis drawn in (a) part it will give you the curve for /f(x)/
 
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i also want to ask one more thing, it's how you calculate area in geometry in add maths? i mean do we have to first get all the lengths by pythagorus way using all the cordinates that we have founded? it will be a very long process is there any other direct way?i don't understand the method using matrices as solved in pastpapers.

Well yup man thats the biggest problem and thats one of the long reasons i will leave Coordinate geometry if it comes in the option part its ttooooooo lengthy !

Heron's Formulae.

Are talking abt that Modulus wala formula then man its not allowed and if u use it they give u like 1 mark out of the three check the MS they state it there ! :p
 
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Well yup man thats the biggest problem and thats one of the long reasons i will leave Coordinate geometry if it comes in the option part its ttooooooo lengthy !



Are talking abt that Modulus wala formula then man its not allowed and if u use it they give u like 1 mark out of the three check the MS they state it there ! :p
I'll still do that because i'm only good in Geometry :D
 
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Another question regarding relative velocity :( Question 10 last part, I don't understand this. And how can we get the direction as required in the first part? :unsure:
http://www.xtremepapers.com/papers/CIE/Cambridge International O Level/Mathematics - Additional (4037)/4037_s08_qp_1.pdf

For the last part u have to use t as the time where both of the objects meet at one point so u have to do this first find the position vector at time t by first multiplying t to velocity vector and then add it to the original position vector now equate both equations together as at t the position vector of both are same now u get t !
After finding t insert t into any of the position vectors at time t and here u have the position vector at the point of intersection !
 
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