Algebraically show that each of the given combinations are equivalent to the given functions.
1 answer:
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Add 1 to both sides:
In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that
Under this assumption, we square both sides and we have
The solutions to this equation are
But since we can only accept solutions greater than -1, we discard and accept
.
In fact, we have
and
which is the only solution.
x 12
the 12 in the numerator (top) and the 12 in the denominator (bottom) cross out so you are left with 50.
Answer: 50