Round 205,000 to the nearest ten thousand
2 answers:
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Suppose first that is a normal subgroup. Then by definition we must have for all
,
for every
. Let
and choose
(
). By hypothesis we have
, i.e.
for some
, thus
. So we have
. You can prove
in the same way.
Suppose for all
. Let
, we have to prove
for every
. So, let
. We have that
for some
(by the hypothesis). hence we have
. Because
was chosen arbitrarily we have the desired
.
Answer:
<h2>If we placed the number 10 in the box, we obtain a system of equations with infinitely many solutions.</h2>Step-by-step explanation:
The given system is
<em>It's important to know that a system with infinitely many solutions, it's a system that has the same equation</em>, that is, both equation represent the same line, or as some textbooks say, one line is on the other one, so they have inifinitely common solutions.
Having said that, the first thing we should do here is reorder the system
This way, you can compare better both equations. If you look closer, observe that the second equation is double, that is, it can be obtained by multiplying a factor of 2 to the first one, that is
So, by multiplying such factor, we obtaine the second equation. Observe that must be equal to 10, that way the system would have infinitely solutions.
Therefore, the answer is 10.