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Mathematics: Post your doubts here!

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i have a question, the paper its from isnt here on xtremepapers, its june 2002 paper 6, q5,
question is: the digits of number 1223678 can be rearraged to give many different 7 digit umbers. find how many 7-digit diferent numbers can be made if
i) there are no restrictions on the order of digits
ii) the digits 1,3,7 are (in any order) are next to each other
iii) these 7 digit numbers are even.

any help?
i) 7!/2!
ii) consider 1 3 and 7 to be in a box and consider that box to be one dash
_ * 7 *6*5*4*3*2*1

number if dashes 8 and inside the box 3!
so 8!*3! /2! because 2 comes twice thats y we divide

iii) that means they wud end at 2,2,6,8
6! *4 /2!
 
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I got the diameter and radius... I was wondering how would we find out the mid- point and I just understood!! We had to use the coordinate geometry formula for mid-point of the two points! Can't believe I didn't think of that before ! Thank you soo much. May Allah bless you and grant you success in this life and in the Hereafter :)
ohh dat was easy ! i totally forgot that haha
 
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question 2 ) 5^(x-1) = 5^x - 5
now take the 5^(x-1) on the right side and the -5 to the left side so that all the x's are on one side :)
now you get 5=5^x-5^(x-1)
open up 5^(x-1) you get 5^x * 5^-1 substitute this in equation
take 5^x as a common factor
5=5^x(1-5^-1)
5/(1/5^-1)=5^x
log 6.25 = x log 5
x=1.14
brillilant and simply awesome, May Allah give you immense Success Ameen
 
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question 5 part 4
you have to integrate 2sec^2x -1 +2secxtanx
after integrating this you get 2 tanx -x + 2 secx
put in the limits >>>> 2tan(pi/4)-(pi/4)+2sec(pi/4) - (2tan0-0-2sec0)
=2-pi/4+2sqrt2-2
=2sqrt2-pi/4
take 1/4 as a common factor and you get 1/4(8sqrt2-pi)
what is the integral of 2secxtanx ?
 
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For June 2012, p63, q3-
:) aaah thaaaank u so much for explaination, yeah i do understand, wil try to practise other,,, god bless abunduntly ameen... best guider ever seen!
p.s=if i got problems then i will ask fromm u, be ready heheh :sneaky:
 

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[quote="2,
thaaaaaaaaaaaank u so much once again, u guide very good...... thnkuu..... got very less time but atleast concept about these ambigous questions got clear! be blessed
:X3:
god help those who help others
 
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