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Mathematics: Post your doubts here!

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i just solved this but only got it on the second trial because at my first try i used 2 sec difference in the equation for SQ but then after checking the marking scheme it was supposed to be used in the equation for SP . (and i don't completely understand why until now) :S

so here is what i did (the correct one):

1- i got the equation for Sp "1/2 x a x (t+2)(squared)"

2- started to get rid of the brackets "1/2 x 1.75 x (t(squared) + 4t +4)
= 0.875(t(squared) + 4t + 4)
= 0.875t(squared) + 3.5t + 3.5

3- got an expression for Sq "1/2 at(squared)"

4- subtracted them from each other to get the distance in-between to be equal to 4.9

0.875t(squared) +3.5t +3.5 - 1/2 x 1.75 x t(squared)
= the "0.875t(squared) is crossed with the "1/2 x 1.75 x t(squared) leaving the equation to be "4.9=3.5t + 3.5"
3.5t=4.9-3.5
t=1.4/3.5 = 0.4

hope you understood what you made wrong and why :)
 
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Hahah i sort ov gt it but why ate we adding 2 to sp and why nt sq as sq was released after p has passed through A and plus why is the initial speed fr p 0 and nt 3.5 ?? :p
 
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Hahah i sort ov gt it but why ate we adding 2 to sp and why nt sq as sq was released after p has passed through A and plus why is the initial speed fr p 0 and nt 3.5 ?? :p
well that part about the 2 at Sp and not in Sq i have no idea exactly, but the initial speed has to be 0 as if you take 3.5 instead you will be getting a different distance than 4.9 which you will then have to deduct the extra distance ... its just to not complicate things, its already complicated as hell why do you want to make it more complicated :D

if you ever find out why the 2 is used in Sp not Sq please tell me....please :D
 
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what are the most challenging exams for paper 4 that i can solve (note i said "that I CAN solve" :p and that had the differentiation and integration rules in them (can't remember name of that part)
 
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This question was done by sum1 else b4, bt i dint get it....
That shit is scary indeed.. I will try to solve it in a couple hours when my mind is out of the dizziness :D
ok guys heres the deal
6 ii 5 students sit anywhere by the window... 5! =120
3 business men sit anywhere= 3!= 6
now for the married couples... there are 2 couples =2 two seats each =2 and 3 seats remaining for each couple =3p2
3p2 *2*2= 24
5!=120
3!=6

24*6*120 = 17280
 
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Btw do you know how to upload pictures rather than giving the url?
Yeah if you want to upload a picture from your computer, there is button on the right besides *Post Reply* for that. Other wise you could also insert and image by pasting it's URL in the box that appears when you click this tree sort of image above that reads Insert/edit image.
 
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Yeah if you want to upload a picture from your computer, there is button on the right besides *Post Reply* for that. Other wise you could also insert and image by pasting it's URL in the box that appears when you click this tree sort of image above that reads Insert/edit image.

Ohh thank you :D
 
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Any one can solve this for me?
 

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