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Mathematics: Post your doubts here!

Haya Ahmed

Haya Ahmed

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Thought blocker said:
I dont know this method is correct or not :p
We had to show a point of reflection, hence we have x1,y1 as (-1,3) and we found a midpoint (3,9) by sim. equation,
So, as to find midpoint we do (x1+x2)/2, and (y1+y1)/2,
similarly,
We know have to find x2 and y2, we have x1 y1 and midpoint :)
so (x1,y1, + x2,y2)/2 = midpoints
=(-1,3 + x2,y2) = 2(3,9)
=(-1,3 + x2,y2) = (6,18)
x2=(6-(-1))=7
y2=(18-3)=15

so reflection on (7,15)


T H I S R E A L L Y W O R K S :)


Yes I nees work solution on question 8 :)
Click to expand...
How did you get 3,9 ?
 
Thought blocker

Thought blocker

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Haya Ahmed said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf

Q10b and Q7 and 9 (iii) (please explain the domain and range thingy cuz i'm confused with them :) )
Click to expand...
This 7 is easy, see my steps :) THAT REALLY WORKS.


Haya Ahmed said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_11.pdf
Click to expand...
Haya Ahmed said:
Q3
Click to expand...

Expand the brackets, and multiply it, then check the coff. of X^2, add them and as given in question keep it equal to 48 hence a = 3 :)
 
Namehere

Namehere

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I´ve got a question and would appreciate some help.

For Q9(i), if you are trying to find the normal to the plane using vector product, why if you do AC x AB or CA x BA you get a wrong answer, but if you use any other two vectors, you get the correct normal to the plane? The only difference for the normal to the plane I get using AC x AB or CA x BA is a minus sign for the k coordinate, instead of a positive one.

I would also like if someone could clarify me how to use vector product correctly, ie. what requisites should the vectors you are going to use for vector product have?

Thank you in advance.
 

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Namehere

Namehere

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Namehere said:
I´ve got a question and would appreciate some help.

For Q9(i), if you are trying to find the normal to the plane using vector product, why if you do AC x AB or CA x BA you get a wrong answer, but if you use any other two vectors, you get the correct normal to the plane? The only difference for the normal to the plane I get using AC x AB or CA x BA is a minus sign for the k coordinate, instead of a positive one.

I would also like if someone could clarify me how to use vector product correctly, ie. what requisites should the vectors you are going to use for vector product have?

Thank you in advance.
Click to expand...

Also, how do you do Q9(ii).
 
Talha Khatri

Talha Khatri

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Namehere said:
I´ve got a question and would appreciate some help.

For Q9(i), if you are trying to find the normal to the plane using vector product, why if you do AC x AB or CA x BA you get a wrong answer, but if you use any other two vectors, you get the correct normal to the plane? The only difference for the normal to the plane I get using AC x AB or CA x BA is a minus sign for the k coordinate, instead of a positive one.

I would also like if someone could clarify me how to use vector product correctly, ie. what requisites should the vectors you are going to use for vector product have?

Thank you in advance.
Click to expand...
Someone definitely has to help boxxy out :LOL:
 
syed1995

syed1995

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Haya Ahmed said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf
Q7 :3
Click to expand...

I gave this paper last year.. that question killed it for me :p

Haya Ahmed said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_12.pdf

Q10b and Q7 and 9 (iii) (please explain the domain and range thingy cuz i'm confused with them :) )
Click to expand...

Q10b

Well I used another logic here which made this question very easy for me. If you look closely.. You'll see The shaded region consists of a triangle with height OB and base OC. and the curve with limits (A,0)

From Part i you have Co-ordinates of B = (0,1) and equation of BC = y=-x/2+1
Find Co-ordinates of C.. as C lies on the x-axis.. there substitute y=0. 0=-x/2+1 // x/2=1 // x=2 Therefore C= (2,0)
Co-ordinates of A are .. as A lies on the x-axis y=0 .. sub y=0 in Curve Equation
0=√(1+4x)
0^2=1+4x
1+4x=0
x=-1/4

The length OB = 1 and the Length OC = 2 since O is origin (0,0).

Area of Triangle = 1/2*1*2 = 1unit^2

Now Area under the curve is
⌡y
⌡√(1+4x)
⌡(1+4x)^0.5
Integrating gives..
|[(1+4x)^1.5]/1.5 * 4|
Putting limits -0.25 and 0. we get Area under the curve as 1/6

Now Area of Shaded region = Area of Triangle + Area under the curve.. which equal 1+1/6 = 7/6 Unit^2 Answer.

9iii

Just Remember this.. Domain is values of x.. and range is values of y or f(x) whatever you like to call it.

And also memorize another important rule. f^-1(x), the inverse, is rewriting the equation in terms of y instead of x. like y=5/(1-3x) becomes x=(y-5)/3y

So technically as you have inversed the position of x and y.. the range (y values) of f(x) becomes the domain (x values) of f^-1(x) and the domain (x values) of f(x) becomes the range (y-values) of f^-1(x).
 
Haya Ahmed

Haya Ahmed

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syed1995 said:
I gave this paper last year.. that question killed it for me :p



Q10b

Well I used another logic here which made this question very easy for me. If you look closely.. You'll see The shaded region consists of a triangle with height OB and base OC. and the curve with limits (A,0)

From Part i you have Co-ordinates of B = (0,1) and equation of BC = y=-x/2+1
Find Co-ordinates of C.. as C lies on the x-axis.. there substitute y=0. 0=-x/2+1 // x/2=1 // x=2 Therefore C= (2,0)
Co-ordinates of A are .. as A lies on the x-axis y=0 .. sub y=0 in Curve Equation
0=√(1+4x)
0^2=1+4x
1+4x=0
x=-1/4

The length OB = 1 and the Length OC = 2 since O is origin (0,0).

Area of Triangle = 1/2*1*2 = 1unit^2

Now Area under the curve is
⌡y
⌡√(1+4x)
⌡(1+4x)^0.5
Integrating gives..
|[(1+4x)^1.5]/1.5 * 4|
Putting limits -0.25 and 0. we get Area under the curve as 1/6

Now Area of Shaded region = Area of Triangle + Area under the curve.. which equal 1+1/6 = 7/6 Unit^2 Answer.

9iii

Just Remember this.. Domain is values of x.. and range is values of y or f(x) whatever you like to call it.

And also memorize another important rule. f^-1(x), the inverse, is rewriting the equation in terms of y instead of x. like y=5/(1-3x) becomes x=(y-5)/3y

So technically as you have inversed the position of x and y.. the range (y values) of f(x) becomes the domain (x values) of f^-1(x) and the domain (x values) of f(x) becomes the range (y-values) of f^-1(x).
Click to expand...
In Q9 (iii) there is 0 included in the domain :( not exactly :)
and thanks for your effort .. appreciated ^_^
 
Ameena Eesa

Ameena Eesa

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Help me with these, how do they look tooo......

(a)what are angle properties of irregular polygons?
(c) angles in the same segment are equal look like?............
(d) angles in opposite segments are
supplementary; cyclic quadrilaterals..... what type of segments......... i cant get how they look.......................
 
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