Maths, Addmaths and Statistics: Post your doubts here!

[Saad Mughal]

Solve
x-2[3x-2(x+1)_5] = 16

Is that a minus sign before the 5?

 

[sweetiepie]

+ sign

 

[Saad Mughal]

+ sign

x-2(3x-2(x+1)+5)=16
x-2(3x-2x-2+5)=16
x-2(x+3)=16
x-2x-6=16
-x=22
x=-22

 

[sweetiepie]

Solve the following equation by using quadratic formula

y2/2 - y/6 = 1/12

 

[sweetiepie]

Solve the following equation by Crammers law

2x-6y = -12
3x-2y = -4

 

[Haris Bin Zahid]

Two ships, A and B leave their ports simultaneously at 12 00 hours. The ports are 104 km apart with one port due west of the other. The speeds and directions of the two ships are shown in the diagram, where tan a = 0.75 and tan b = 2.4. Find
(a) the speed and direction of the velocity of A relative to B,
(b) the time at which A is due south of B and the distance between A and B at this instant,
(c) the distance between the two ships at 17 00 hours.
My answers are
(a) 28.6 km/h and 114.8 degrees
(b) 16 00 and 48 km
(c) 65.4 km
But I've got no means to check them. I would appreciate if someone can attempt it so that I can cross-check my answers.
Suchal Riaz Saad Mughal

 

[Saad Mughal]

Two ships, A and B leave their ports simultaneously at 12 00 hours. The ports are 104 km apart with one port due west of the other. The speeds and directions of the two ships are shown in the diagram, where tan a = 0.75 and tan b = 2.4. Find
(a) the speed and direction of the velocity of A relative to B,
(b) the time at which A is due south of B and the distance between A and B at this instant,
(c) the distance between the two ships at 17 00 hours.View attachment 23654
My answers are
(a) 28.6 km/h and 114.8 degrees
(b) 16 00 and 48 km
(c) 65.4 km
But I've got no means to check them. I would appreciate if someone can attempt it so that I can cross-check my answers.
Suchal Riaz Saad Mughal

The only chapter that taunts me.
I'm afraid I can't help you because I am not very good at relative velocity. Sorry.
However, I did manage to do part (a) and I'm getting the same answer, and for (b) shouldn't the distance be 24 km?
Distance AB = Distance Traveled by B - Distance Traveled by A = 26*4 - 20*4 = 24km.

 

[Saad Mughal]

Solve the following equation by using quadratic formula

y2/2 - y/6 = 1/12

You asked the same question last week.
y^2/2 - y/6 = 1/12
3y^2 - y = 1/2 (by lcm)
6y^2 - 2y - 1 = 0
Using quadratic eq.,

 

[Saad Mughal]

Solve the following equation by Crammers law

Sorry. Don't know Crammer's law.

 

[Suchal Riaz]

Two ships, A and B leave their ports simultaneously at 12 00 hours. The ports are 104 km apart with one port due west of the other. The speeds and directions of the two ships are shown in the diagram, where tan a = 0.75 and tan b = 2.4. Find
(a) the speed and direction of the velocity of A relative to B,
(b) the time at which A is due south of B and the distance between A and B at this instant,
(c) the distance between the two ships at 17 00 hours.View attachment 23654
My answers are
(a) 28.6 km/h and 114.8 degrees
(b) 16 00 and 48 km
(c) 65.4 km
But I've got no means to check them. I would appreciate if someone can attempt it so that I can cross-check my answers.
Suchal Riaz Saad Mughal

i got exactly same answer when i solved it. probably we both cant be wrong.

 

[Suchal Riaz]

The only chapter that taunts me.
I'm afraid I can't help you because I am not very good at relative velocity. Sorry.
However, I did manage to do part (a) and I'm getting the same answer, and for (b) shouldn't the distance be 24 km?
Distance AB = Distance Traveled by B - Distance Traveled by A = 26*4 - 20*4 = 24km.

no ur wrong, u cant find the distance like that, u r subtracting their distance from different locations.
as they are vertical, we need to take their y-component find their difference then multiply by 4(t)

 

[Saad Mughal]

no ur wrong, u cant find the distance like that, u r subtracting their distance from different locations.
as they are vertical, we need to take their y-component find their difference then multiply by 4(t)

Ok so the y-component is (24-12)=12 so distance = 12*4 = 48km. Yeah you guys are right.
I was bound to be wrong in this question, I hate this chapter

 

[Saad Mughal]

Ok so the y-component is (24-12)=12 so distance = 12*4 = 48km. Yeah you guys are right.
I was bound to be wrong in this question, I hate this chapter

How did you calculate the time?

 

[Haris Bin Zahid]

i got exactly same answer when i solved it. probably we both cant be wrong.

Great. Thanks a lot for your consideration.

 

[asma tareen]

can anyone explain me (b) 1 and 2 Q2
http://papers.xtremepapers.com/CIE/...Calculator Version) (4024)/4024_s10_qp_22.pdf

 

[usama321]

can anyone explain me (b) 1 and 2 Q2
http://papers.xtremepapers.com/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_s10_qp_22.pdf

its kinda easy. Take the pentagon ABCDE. Its total internal angle = (5-2)180 = 540. Now suppose AED = x, and EDC= 2x

so,
x+2x+120+70+110 = 540

x= 80

AED = 80

Now second part.


10+80+EDA = 180

EDA = 90

 

[asma tareen]

its kinda easy. Take the pentagon ABCDE. Its total internal angle = (5-2)180 = 540. Now suppose AED = x, and EDC= 2x

so,
x+2x+120+70+110 = 540

x= 80

AED = 80

Now second part.


10+80+EDA = 180

EDA = 90

thnx m not good in these angles chap

 

[asma tareen]

its kinda easy. Take the pentagon ABCDE. Its total internal angle = (5-2)180 = 540. Now suppose AED = x, and EDC= 2x

so,
x+2x+120+70+110 = 540

x= 80

AED = 80

Now second part.


10+80+EDA = 180

EDA = 90

btw are u good at transformations

 

[usama321]

btw are u good at transformations

I am ok with that, not really good

 

[asma tareen]

I am ok with that, not really good

mmmm can u tell me how to calculate stetch factor and shear factor???