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-x is (0.2)^1/z
z=200.

For z=1, -x is 0.2
For z=2, -x is -0.45

For z =very large number say infinity, -x is (0.2)^0=1

Hence -x would be very close to 1 for z=200

x should be close to -1 & -0.9

Posted from my mobile device
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Bunuel
­If \(x = -\sqrt[200]{0.2}\), then \(x\) satisfies which of the following inequalities?
A. \(x < -1\)
B. \(-1 < x < -0.9\)
C. \(-0.9 < x < -0.5\)
D. \(-0.5 < x < -0.1\)
E. \(-0.1 < x < 0\)­
 
Numbers in the range 0 to 1, when taken under the radical sign will move towards 1 but never reach it. 
Square root of 1/4 is 1/2 = .5 (closer to 1) 
Cube root of 1/4  is .63 (closer to 1) 
Fourth root of 1/4 is 0.7 (even closer to 1)
and so on...
When you take 0.2 under the radical sign, it will move towards 1 too. With a number as large as 200, the value will be almost 1 though not quite. I would worry if it was the sixth root or something of 0.2. Since they have given you outright 200, you know that they are messing around. It has to be almost 1.
What they are saying is - if we take a big big root of a number between 0 and 1, what will be its value? It will be almost 1. Hence in teh question x will be a tiny tiny bit greater than -1. 
Answer (B)
I have discussed these relations in my Exponents and Roots module. You can access it for free tomorrow under the Super Sundays program. Check this video on my YouTube channel for details.
­
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­If \(x = -\sqrt[200]{0.2}\), then \(x\) satisfies which of the following inequalities?

\(x = -\sqrt[200]{0.2} = -1/\sqrt[200]{5} = -1/1.0001 = -.99...\)

-1 < x < -.9

IMO B
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