Mathematics: Post your doubts here!

[Suhayb.B]

Part (c)
(I've never faced any trouble in this kind of questions. .. don't knw why i can't solve this one )

Is the answer
0 1
1 0
?

 

[NOneed2speedd]

Is the answer
0 1
1 0
?

0 1
-1 0

 

[Suhayb.B]

0 1
-1 0

Yes sorry I meant that lol, I just forgot to put the negative sign. I 'll show you how to do it.

 

[misscute406]

Part (c)
(I've never faced any trouble in this kind of questions. .. don't knw why i can't solve this one )

that's pretty easy man!! Sorry about my handwriting
https://scontent-sin.xx.fbcdn.net/h...=eb18deff8b1643c9008036718f296c04&oe=55DB095D

 

[misscute406]

Part (c)
(I've never faced any trouble in this kind of questions. .. don't knw why i can't solve this one )

sorry i didn't read the question!! forget the last part!! rotation 90 degrees anticlockwise

 

[loaie Amgad]

M/J 2010 Paper 42 Question 3(d). What is the method to do it?

thats a practical question and you need to stay calm while solving just look carefully
16/243 = 2/6 * (4/6)to the power of x
shift 2/6 to the other side , to divide so
16/81 = (2/3)to the power of x , when you equate indices . You will find that x is 4
this means that probability of other than 2 comes 4 times and 2 comes next
so n=x+1 , n=4+1 , n=5......
Hope that helps....

 

[misscute406]

someone help me out in this....
0580/41/m/j/14-question 5 (f)
(f) The lighthouse stands on an island of area 1.5cm2
Work out the actual area of the island on the scale drawing.

 

[masterex567]

someone help me out in this....
0580/41/m/j/14-question 5 (f)
(f) The lighthouse stands on an island of area 1.5cm2
Work out the actual area of the island on the scale drawing.

the scale is 2 cm : 3 km
hence if we want area we have to square the scale into area scale factor such as (3/2)^2 = x/1.5
so 1.5 * (3/2)^2 which is 3.375

 

[misscute406]

the scale is 2 cm : 3 km
hence if we want area we have to square the scale into area scale factor such as (3/2)^2 = x/1.5
so 1.5 * (3/2)^2 which is 3.375

why do you divide 3/2 ??

 

[misscute406]

how to enlargement SF=-2 of a shape

 

[masterex567]

why do you divide 3/2 ??

We're creating a ratio between the scale on the ground to distance on the map.
(3 on ground : 2 on map)^2 = (x on ground : 1.5 on map)

similarly you can do (2/3)^2 = 1.5/x and get same result

 

[misscute406]

We're creating a ratio between the scale on the ground to distance on the map.
(3 on ground : 2 on map)^2 = (x on ground : 1.5 on map)

similarly you can do (2/3)^2 = 1.5/x and get same result

oooh thanx man!!

 

[masterex567]

how to enlargement SF=-2 of a shape

first identify centre of enlargement, from where enlargment is taking place from.
Then the image co-ordinates should be twice the distance from centre than of the object.
For example if the object is at (1,1) and centre is (0,0) the image should be (2,2) which is twice as far as the (1,1) object.

Now, the important part is that the scale factor is in negative. Hence the image should be drawn on opposite side of the centre. (i.e on left, if the object is on right)

 

[misscute406]

first identify centre of enlargement, from where enlargment is taking place from.
Then the image co-ordinates should be twice the distance from centre than of the object.
For example if the object is at (1,1) and centre is (0,0) the image should be (2,2) which is twice as far as the (1,1) object.

Now, the important part is that the scale factor is in negative. Hence the image should be drawn on opposite side of the centre. (i.e on left, if the object is on right)

oooh kay thank you man!!

 

[NOneed2speedd]

Need help in these questions

1) may june 13 v.42 Q.4 (d), Q.9 (b) ii , Q.11 (d) iii

 

[misscute406]

need help in this question!! 0580/42/m/j/14, Question 11

The total area of each of the following shapes is X.
The area of the shaded part of each shape is kX.
For each shape, fi nd the value of k and write your answer below each diagram.

 

[masterex567]

need help in this question!! 0580/42/m/j/14, Question 11

The total area of each of the following shapes is X.
The area of the shaded part of each shape is kX.
For each shape, fi nd the value of k and write your answer below each diagram.

Question 11:
Okay, if the total area of the shapes is X. And the area of the shaded part is k. It means k is a number that is multiplied with the area to give shaded area.
For example in the first one, if the total area of a triangle is X, and the shaded area is 1/3 of the total area (of X). Then to calculate this shaded area it'll be 1/3 * X. In this case k is 1/3. It's a number multiplied by X (total area) to give the shaded area.
Now the second one, We can calculate the area of the sector by 72/360 * pi * radius squared. If the total area (X) of the circle is pi * radius squared, what are you multiplying with it to get shaded area? That's 72/360. So 72/360 is k.
For the third one, EF = FG, hence if we assume EF is 1, then FG will also be 1. Therefore EG will be 2. Now since they are similar triangles, (EF/EG)squared will give area. Therefore 1squared/2squared is answer. which is equal to 1/4.
For the fourth one, a hexagon is made of 6 equilateral triangles. Hence each angle will be 60 degrees. So to find total area (X) it'll be 6 * length * length * sin60. now the area that is shaded has the same area as any triangle from the center to the two sides. So they want to find area of only 1 triangle out of the 6 triangles. Hence k will be 1/6.
Last one, shaded area is area of sector - area of triangle. Which is 90/360*pi*radius squared - 1/2 * radius squared. Work this out and factorise, it'll give 1/4*radius squared (pi - 2). Now this is the shaded area. They want k which is the value you multiply with the whole sector to give this shaded area. so divide the shaded area by area of whole sector. the answer will be pi - 2/pi.

 

[misscute406]

hey masterex567 how in the fourth one 6 equilateral triangles are there?? only 3 are there

 

[masterex567]

hey masterex567 how in the fourth one 6 equilateral triangles are there?? only 3 are there

Oh yes forgot to explain that. That was a trick. A regular hexagon has 6 equilateral triangles, if you notice the shaded area in the diagram is actually equal to one of these 6 regular triangles.

 

[AnonymousX9]

What do you guys think? Will Paper 42 be very hard or moderate? I'm nervous although my paper 2 was really good.