Maths, Addmaths and Statistics: Post your doubts here!

[muzirox]

This is a june '89 paper's question

AB=[8 -6] (it's a column vector) and CD=3/2 AB

a) Calculate |AB|
b) Express CD as a column vector
c) Given that A is the point (6,9) find the coordinates of the point B.

Kindly help me in part 'c'

Fr part c:
We can write coordinates ov A as column vector that is OA=[6 9] so now we can formulate an equation i.e:
AB=OB-OA
[8 -6]+[6 9]=OB
[8+6 -6+9]=
[14 3]=OB , now OB can b written as coordinates of B which is (14,3)
Here is the concept that coordinates of any point can b written as column vector by adding O (for origin) which shows the
distance frm origin to that particular point
Hope u gt the concept

 

[Spidey21]

Thanks @spidey21 for the equation but i gave this a little thinking myself and was able to get the answer

np !!

 

[screamingskull]

Okay its official that I suck in the chapter vectors so one more question:

For those who have classified Maths D pastpapers this is question 5 on chapter 7

In the diagram OA=3x, OB=2x+y and OB=2y
a) Express CO in terms of y
b) The point D is such that OD=h(OA). Show that CD=3hx-2y
c) Express CB in terms of x and y
d) Given also that CD=k(CB), write down an equations in terms of x,y,h and k
e) Find the value of h and k

Please help me in part d and e. They are very confusing!!

 

[Tharindu]

on certificate , space is left blank ! so you will have no result .
appear and pray to Allah , he will help you

Then other subjects will be there right? DDDD. So is it like they put French and keep a blank or...do they even mention that subject?

 

[Spidey21]

Okay its official that I suck in the chapter vectors so one more question:

For those who have classified Maths D pastpapers this is question 5 on chapter 7

In the diagram OA=3x, OB=2x+y and OB=2y
a) Express CO in terms of y
b) The point D is such that OD=h(OA). Show that CD=3hx-2y
c) Express CB in terms of x and y
d) Given also that CD=k(CB), write down an equations in terms of x,y,h and k
e) Find the value of h and k

Please help me in part d and e. They are very confusing!!

there is some problem in your question..please recheck ..there cant be 2 OBs...one of them has to be OC..please correct it so i can solve it for you...

 

[Spidey21]

Okay its official that I suck in the chapter vectors so one more question:

For those who have classified Maths D pastpapers this is question 5 on chapter 7

In the diagram OA=3x, OB=2x+y and OB=2y
a) Express CO in terms of y
b) The point D is such that OD=h(OA). Show that CD=3hx-2y
c) Express CB in terms of x and y
d) Given also that CD=k(CB), write down an equations in terms of x,y,h and k
e) Find the value of h and k

Please help me in part d and e. They are very confusing!!

assuming OC=2y


CD= K(CB)
CB= OB -OC
CB= 2x +y - 2y
CB= 2x-y

CD= 2kx -ky
CD= 3hx-2y (as seen in b)

2kx-ky= 3hx-2y
2k=3h --------------1

k=2 -------------------2

eq . 1 ----> 2(2)=3h

h=4/3...

please tell me if i am right or right!! ..

 

[screamingskull]

Yes the answer that you have calculated but how is 2k=3h please elaborate

 

[Spidey21]

Yes the answer that you have calculated but how is 2k=3h please elaborate

well when you have 2 eqs of CD...they are obviously going to be equal so you put like this..
2kx-ky=3hx-2y...

right now you see here that the coefficient of X on both sides will be equal and so will it be with Y..
so its going to be
2k=3h-----------------1


and
-k=-2
k=2---------------2

hope you get it...

 

[ashiqbal]

Yes the answer that you have calculated but how is 2k=3h please elaborate

for any vector which is defined in two ways,
CD= 2kx -ky
CD= 3hx-2y (as seen in b)
the coeffecient of x (or sometimes i) is equal to coeffecient of x(or i) and the coeffecient of y(sometimes j) is equal to coeffecient of y.

 

[screamingskull]

Thanks again I got it now

 

[Spidey21]

Thanks again I got it now

np mate...

 

[Mayedah]

http://www.xtremepapers.com/papers/...(Calculator Version) (4024)/4024_s07_qp_1.pdf
Question no14 part b .! Can anyone solve this ?

 

[Spidey21]

http://www.xtremepapers.com/papers/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_s07_qp_1.pdf
Question no14 part b .! Can anyone solve this ?

its easy...
find the area of big circle with radius 2x and minus the area of the smaller circle with the radius x from it...
hope this helps!!

 

[Tharindu]

Can anyone help me in add maths 2005 paper 1... Question 6... The last part which functions have an inverse

 

[Spidey21]

Can anyone help me in add maths 2005 paper 1... Question 6... The last part which functions have an inverse

put in the values of x and youll have the maximum and minimum possible value of the function..

 

[Tharindu]

put in the values of x and youll have the maximum and minimum possible value of the function..

I mean they are asking which of the function f,g,h has an inverse

 

[Spidey21]

I mean they are asking which of the function f,g,h has an inverse

aaaaahh....well for a function to have an inverse...it should be a one to one function and when you remove the modulus youll get two values...so the one without the modulus will be having an inverse...none other...
hope this helps

http://answers.yahoo.com/question/index?qid=20070829165125AAF5QCH

check this link...much more vast...

 

[Tharindu]

aaaaahh....well for a function to have an inverse...it should be a one to one function and when you remove the modulus youll get two values...so the one without the modulus will be having an inverse...none other...
hope this helps

http://answers.yahoo.com/question/index?qid=20070829165125AAF5QCH

check this link...much more vast...

Thanks.......does all modulus functions have no inverse?

 

[Spidey21]

Thanks.......does all modulus functions have no inverse?

when there is a modulus there will be no inverse...cuz there will be two values then and it wont be a one - one function!!

 

[Mayedah]

its easy...
find the area of big circle with radius 2x and minus the area of the smaller circle with the radius x from it...
hope this helps!!

Thanks alot. I follwed the method and was able to solve this part