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Is that a minus sign before the 5?Solve
x-2[3x-2(x+1)_5] = 16
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Is that a minus sign before the 5?Solve
x-2[3x-2(x+1)_5] = 16
x-2(3x-2(x+1)+5)=16+ sign
The only chapter that taunts me.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
You asked the same question last week.Solve the following equation by using quadratic formula
y2/2 - y/6 = 1/12
Sorry. Don't know Crammer's law.Solve the following equation by Crammers law
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.
Ok so the y-component is (24-12)=12 so distance = 12*4 = 48km. Yeah you guys are right.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)
How did you calculate the time?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
Great. Thanks a lot for your consideration.i got exactly same answer when i solved it. probably we both cant be wrong.
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
thnx m not good in these angles chapits 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 transformationsits 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
I am ok with that, not really goodbtw are u good at transformations
mmmm can u tell me how to calculate stetch factor and shear factor???I am ok with that, not really good
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