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OpenStudy (anonymous):
i got 2a-1/1-2a=H(x) do i need to do anything else
OpenStudy (anonymous):
its (2a-1)/(1-2a) =-1 , so -1 is the answer
OpenStudy (unklerhaukus):
here are the steps akash_809 has used
\[h(x)=\frac{x-1}{1-x}\]\[h(2a)=\frac{2a-1}{1-2a}\]\[\qquad\quad=\frac{-(1-2a)}{1-2a}\]\[\qquad\quad=\frac{-\cancel{(1-2a)}}{\cancel{1-2a}}\]\[\qquad\quad =-1\]
OpenStudy (anonymous):
what if I used A+1 as my x value?
OpenStudy (anonymous):
platinum ring again :)
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OpenStudy (unklerhaukus):
ring ring; hello?
OpenStudy (anonymous):
@jethro210 it would be same
OpenStudy (anonymous):
i.e =-1
OpenStudy (unklerhaukus):
the function gives the same value for all inputs
(except x=1, which has indeterminate value )
OpenStudy (anonymous):
thanks for the medal un(c)kle
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OpenStudy (unklerhaukus):
anytime
OpenStudy (anonymous):
@jethro210 u wl always get quick replies due to the pic :) you have used
OpenStudy (anonymous):
cool.
OpenStudy (anonymous):
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