i didnt think you would for that one :) lets me guide you to an easier way
OpenStudy (anonymous):
ok i think that (-10) goes first
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OpenStudy (amistre64):
the inner most grouping is:
(10 div(-10))
what can you tell me about this value?
OpenStudy (amistre64):
have you done fractions yet?
\[(10\div (-10))=\frac{10}{-10}\]
OpenStudy (anonymous):
ummm....[(10div(-10) the -10 turns into positive
OpenStudy (amistre64):
the -10 does not turn positive; that is not in any rule of math :)
OpenStudy (anonymous):
ohh sorry im confused
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OpenStudy (amistre64):
10 times ___ = -10 can you fill in the blank?
OpenStudy (anonymous):
_10
OpenStudy (anonymous):
imean -1
OpenStudy (amistre64):
-1 is correct; which tells us that: \(10\div(-10)= -1\)
OpenStudy (anonymous):
do i write that down in my paper
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OpenStudy (amistre64):
that is part of the process yes, there are 2 more groupings to go
OpenStudy (amistre64):
lets go to the next in line, that ^7
OpenStudy (anonymous):
ok
OpenStudy (amistre64):
is this correct so far?
\[(10\div(-10))^7=(-1)^7\]
OpenStudy (anonymous):
are the ones at the top 7 or?
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OpenStudy (anonymous):
because i wrot 7 is it ok
OpenStudy (amistre64):
we have:
\[\left((numbers)^7\right)^2\]
OpenStudy (amistre64):
we figured out that the "numbers" = -1 so lets move outwards to the \(()^7\) part right?
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
imma show you what i wrote
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OpenStudy (anonymous):
(10div(-10))^7=(-1)
OpenStudy (amistre64):
\((-1)^7\) means: -1 x -1 x -1 x -1 x -1 x -1 x -1
(-1 x -1) x (-1 x -1) x (-1 x -1) x (-1)
(1 x 1 x 1) x -1
1 x -1
-1
thats fine as long as you know that: \((-1)^7=(-1)\)
OpenStudy (anonymous):
ok do i have my work correct
OpenStudy (amistre64):
so far so good, just one last grouping to take care of
OpenStudy (anonymous):
ok
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OpenStudy (amistre64):
\((-1)^2=?\)
OpenStudy (anonymous):
is it 2
OpenStudy (amistre64):
when we work it thru the groupings it looks like this:
\[\left((10\div(-10))^7\right)^2\]
\[\left((-1)^7\right)^2\]
\[(-1)^2\]
\[1\]
OpenStudy (amistre64):
and no, -1 x -1 = 1, not 2 :)
OpenStudy (anonymous):
is that the answer
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OpenStudy (amistre64):
well, it is the problem i have been working with you on hasnt it?
OpenStudy (anonymous):
yes
OpenStudy (amistre64):
then i believe that that is the answer :)
OpenStudy (anonymous):
can you help me on #2 or are you too busy
OpenStudy (amistre64):
my time is up, i gotta get ready for tomorrow, good luck tho ;)
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