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Mathematics: Post your doubts here!

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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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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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