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Mathematics: Post your doubts here!

Maz

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i did try but i didn't get the right answer and that is a frequent question almost in all exams :s
You can solve it using log, but isn't really in our syllabus and it's a bit complex too.
I did like :-
log x/log 3 = 2 [since h^-1(x)=2]
x = 3^2
x = 9

I am sure there is another easier method, since it's just for 1 mark
 
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You can solve it using log, but isn't really in our syllabus and it's a bit complex too.
I did like :-
log x/log 3 = 2 [since h^-1(x)=2]
x = 3^2
x = 9

I am sure there is another easier method, since it's just for 1 mark
i didn't use the log methode ever so we need another way to solve it :confused:
 

NIM

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3 a part 2! the 3 mark question after the line of symmetry one! R^2:r^2...
Ok:

Let the radius of bigger circle be R

Area of bigger circle = pie.R^2 = 10(area of small circles)

Why 10? There are five circles. PLUS there are 5 areas same as the shaded area and THIS area is equal to the circle's! (as mentioned in the qp)

5 circles + 5 shaded areas
5 Circles + 5 Circles
10 Circles

10(Pie.r^2) : Pie.R^2
Cancel Pie
10r^2 : R^2

They asked us the ration of R^2:r^2
so:
1 : 10
 
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Ok:

Let the radius of bigger circle be R

Area of bigger circle = pie.R^2 = 10(area of small circles)

Why 10? There are five circles. PLUS there are 5 areas same as the shaded area and THIS area is equal to the circle's! (as mentioned in the qp)

5 circles + 5 shaded areas
5 Circles + 5 Circles
10 Circles

10(Pie.r^2) : Pie.R^2
Cancel Pie
10r^2 : R^2

They asked us the ration of R^2:r^2
so:
1 : 10
Thanks a lot! I didn't get the whole 10 circle connection initially :D
 
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