V=2.28943
I am not sure though
Let i = sqrt(-1) which is the conventional notation to set up an imaginary number
The idea is to break up the radicand, aka stuff under the square root, to simplify
sqrt(-8) = sqrt(-1*4*2)
sqrt(-8) = sqrt(-1)*sqrt(4)*sqrt(2)
sqrt(-8) = i*2*sqrt(2)
sqrt(-8) = 2i*sqrt(2)
<h3>Answer is choice A</h3>
Answer:
true
Step-by-step explanation:
The first case occurs in
for
and
. Extending the domain to account for all real
, we have this happening for
and
, where
.
The second case occurs in
when
, and extending to all reals we have
for
, i.e. any even multiple of
.
