SOLUTION: Picture yourself in a ’special’ world where {{{c^2+s^2=1}}} Under this assumption, select the true statement/s: A) Every {{{1 + (s^2)/(c^2)}}} can be ’exchanged’ fo

Algebra ->  Linear-equations -> SOLUTION: Picture yourself in a ’special’ world where {{{c^2+s^2=1}}} Under this assumption, select the true statement/s: A) Every {{{1 + (s^2)/(c^2)}}} can be ’exchanged’ fo      Log On


   



Question 1151425: Picture yourself in a ’special’ world where
c%5E2%2Bs%5E2=1
Under this assumption, select the true statement/s:
A) Every
1+%2B+%28s%5E2%29%2F%28c%5E2%29
can be ’exchanged’ for
1%5E%22%22%2Fc%5E2
B) Every
%28s%5E2%29%2F%28c%5E2%29
can be ’exchanged’ for
1%5E%22%22%2Fc%5E2+-+1
C) None of these

Found 2 solutions by Edwin McCravy, MathTherapy:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Picture yourself in a ’special’ world where 
c%5E2%2Bs%5E2=1
Under this assumption, select the true statement/s:
We try out each possibility:

A) Every
1+%2B+%28s%5E2%29%2F%28c%5E2%29
can be ’exchanged’ for
1%5E%22%22%2Fc%5E2
We see if this equation
1+%2B+%28s%5E2%29%2F%28c%5E2%29=1%5E%22%22%2Fc%5E2
is equivalent to the original.
We multiply through by the common denominator c²


c%5E2%2Bs%5E2=1
That is the correct choice.  But let's see why the other one is not correct

B) Every
%28s%5E2%29%2F%28c%5E2%29
can be ’exchanged’ for
1%5E%22%22%2Fc%5E2+-+1
We see if this equation
%28s%5E2%29%2F%28c%5E2%29=1%5E%22%22%2Fc%5E2+-+1
is equivalent to the original.
We multiply through by the common denominator c²
c%5E2%2Aexpr%28%28s%5E2%29%2F%28c%5E2%29%29=c%5E2%2Aexpr%281%5E%22%22%2Fc%5E2%29+-+c%5E2%2A1


c%5E2=s%5E2-1
And even if I subtract s² from both sides, I get
c%5E2-s%5E2=-1
that is not the same as the original.
So A) is the only correct answer.

Edwin

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Picture yourself in a ’special’ world where
c%5E2%2Bs%5E2=1
Under this assumption, select the true statement/s:
A) Every
1+%2B+%28s%5E2%29%2F%28c%5E2%29
can be ’exchanged’ for
1%5E%22%22%2Fc%5E2
B) Every
%28s%5E2%29%2F%28c%5E2%29
can be ’exchanged’ for
1%5E%22%22%2Fc%5E2+-+1
C) None of these

Edwin made a mistake in B)
The way I did it was:
We need to determine IF: matrix%281%2C3%2C+s%5E2%2Fc%5E2%2C+%22=%22%2C+highlight%281%5E%22%22%2Fc%5E2+-+1%29%29
------ Substituting matrix%281%2C3%2C+1+-+c%5E2%2C+for%2C+s%5E2%29

<====== As you can clearly see, this is TRUE!