Maths, Addmaths and Statistics: Post your doubts here!

[snowbrood]

Q.4 under the rotation the image of A is A' while the image of B is on the line segment A'X copy figure 1.11 and construct accurately.

 

[Waleed007]

Anyone can Solve this ?

Reflect the point A(-1,3) in the x-axis and then in the line y=4.What are the coordinates of the final image?
Iam a bit confused in it

 

[Anon]

From where is it?
Whats the answer?

EDIT: Also which topic is it?

 

[Waleed007]

From where is it?
Whats the answer?

EDIT: Also which topic is it?

Transformation Reflection.d3 random topic Q2

 

[Anon]

Oh, Thats why, I couldnt solve it.
We havent started transformation yet. Lagging in syllabus as you already know...
=(

 

[Anon]

I think,

when we reflect it at first by x axis, it becomes: (1, 3)
when we reflect again at y=4, it becomes: (1,5)

Thats what I deduce from wording of question. I havent done transformation, so better wait for someone else...
Is the Ans right?

 

[Waleed007]

I think,

when we reflect it at first by x axis, it becomes: (1, 3)
when we reflect again at y=4, it becomes: (1,5)

Thats what I deduce from wording of question. I havent done transformation, so better wait for someone else...
Is the Ans right?

Nah! the answer is (-1,11)

 

[Anon]

Just notice an wrror in my calc,

When we reflect (-1,3) on x axis, it should be, (-1, -3)
When we reflect (-1, -3) on y=4, it should be, (-1, 11)

y=4 is a straight horizontal line at y=4.
Distance between -3 and 4 is 7.
now as it would be on the other side of the line. we will add 7 to 4, which is 11.

-1 would remain same as all the reflection were of horizontal lines...

Hope you understand. I dont know proper working as I havent started.

 

[snowbrood]

hey anon can u solve my question please

 

[Anon]

post em, I will see.

 

[abcde]

Q.4 under the rotation the image of A is A' while the image of B is on the line segment A'X copy figure 1.11 and construct accurately.

The question asks for simple construction so you don't need any calculations at all. Use a compass, first at centre A and then at centre A', to construct arcs both above and below the line AA'. Draw a straight line (PQ) through the two points where the arcs intersect. The point where AA' intersects PQ is the centre of rotation. Find the angle of rotation. These two findings will allow you to locate B'.

 

[snowbrood]

i would be grateful to u if u could draw this in that figure... draw it roughly i wud get that coz i cant locate B'

 

[snowbrood]

post em, I will see.[/quote
i had posted it earlier

 

[Anon]

I have a problem in Add Maths,
Trignometry

I have successfully solved the question till a point,
but after taking out value of "alpha",
as cos(z + π/6)= - 1/2
So it should be in 2nd quadrant or 3 quadrant.

In the marking scheme, they have solved for 1 and 4th quadrant.

I havent done trig in a long time,
Can anyone explain why we will find values of quadrant 1 and 4, which are actually positive, and not 2 and 3, which are negative.

P.S: to write symbol of "pie" (π), turn on the numlock, keep alt key pressed, and then write "227" in the numpad.
Or you can simply copy paste it from my post..

 

[Anon]

Nevermind...
I got my mistake..

 

[2412957]

upper and lower bond?

 

[Gémeaux]

upper and lower bond?

for upper bound add half the value of limit of accuracy to the actual value. for lower subtract it.
eg, if it is 5 cm nearest to 1 cm, its upper bound is 5+0.5=5.5, and lower bound is 5-0.5=4.5

 

[scouserlfc]

now what if the ques asks for find the greatest area of the rectangle for the upper bound what values we use provided that the post above contains the lengths of a rectangle ! ?

 

[VelaneDeBeaute]

Area = Length x Width!
Now see, length and width, both if increased will register an increase in the area of the rectangle!
Therefore, to find the greatest possible area, you need to take the greatest values of length and width, that are, their upper bounds!

 

[Silent Hunter]

yes..take upper bounds for both..... leads to final increased area...........same case for lower bound area