anonymous
  • anonymous
math
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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pooja195
  • pooja195
What about it?
anonymous
  • anonymous
|dw:1440720978922:dw|
jim_thompson5910
  • jim_thompson5910
are you familiar with the rule \[\Large x^{-k} = \frac{1}{x^k}\] ??

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More answers

anonymous
  • anonymous
no
jim_thompson5910
  • jim_thompson5910
the rule is that if you have a negative exponent, you take the reciprocal of the base to make the exponent positive in this case,\[\Large 3^{-3} = \frac{1}{3^3}\]
jim_thompson5910
  • jim_thompson5910
other examples \[\Large 2^{-7} = \frac{1}{2^7}\] \[\Large 9^{-4} = \frac{1}{9^4}\]
jim_thompson5910
  • jim_thompson5910
|dw:1440721304965:dw|
jim_thompson5910
  • jim_thompson5910
now you need to simplify the following |dw:1440721342332:dw|
anonymous
  • anonymous
|dw:1440721303495:dw|
jim_thompson5910
  • jim_thompson5910
is that a p? or a q? or a 9? (see circled) |dw:1440721502919:dw|
anonymous
  • anonymous
that was a p sorry got connfused |dw:1440721443290:dw|
jim_thompson5910
  • jim_thompson5910
ok now simplify that and you'll be done
jim_thompson5910
  • jim_thompson5910
think of p as p/1
jim_thompson5910
  • jim_thompson5910
|dw:1440721767667:dw|
jim_thompson5910
  • jim_thompson5910
when dividing 2 fractions, flip the second fraction and multiply
anonymous
  • anonymous
what would i get?
anonymous
  • anonymous
like p=27?
jim_thompson5910
  • jim_thompson5910
|dw:1440721912391:dw|
anonymous
  • anonymous
2 and 27p
jim_thompson5910
  • jim_thompson5910
|dw:1440721937428:dw|
anonymous
  • anonymous
yea but i still get 1/27p
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{1}{27p}\] is the answer
anonymous
  • anonymous
oh ok thank you
anonymous
  • anonymous
if i have a diff question later ill post it
jim_thompson5910
  • jim_thompson5910
no problem
anonymous
  • anonymous
i need help
anonymous
  • anonymous
|dw:1440724201473:dw|
anonymous
  • anonymous
needs positive exponents,simplefy
jim_thompson5910
  • jim_thompson5910
use the same rule to get |dw:1440724750331:dw|
jim_thompson5910
  • jim_thompson5910
so you need to simplify \[\LARGE \frac{1}{(-8v)^2}*w^3\]
anonymous
  • anonymous
so no multiplying?
jim_thompson5910
  • jim_thompson5910
you do multiply think of the w^3 as w^3 over 1
jim_thompson5910
  • jim_thompson5910
\[\LARGE \frac{1}{(-8v)^2}*w^3 = \frac{1}{(-8v)^2}*\frac{w^3}{1}\]
jim_thompson5910
  • jim_thompson5910
then multiply straight across
jim_thompson5910
  • jim_thompson5910
also, simplify (-8v)^2
anonymous
  • anonymous
|dw:1440725573970:dw|
jim_thompson5910
  • jim_thompson5910
it should be \[\Large \frac{w^3}{64v^2}\]
jim_thompson5910
  • jim_thompson5910
you square the -8 to get 64 you square the v to get v^2
anonymous
  • anonymous
ok
anonymous
  • anonymous
what about this|dw:1440725832639:dw|
jim_thompson5910
  • jim_thompson5910
Use the rule \[\LARGE \frac{x^a}{x^b} = x^{a-b}\]
jim_thompson5910
  • jim_thompson5910
|dw:1440726015179:dw|
jim_thompson5910
  • jim_thompson5910
so for example \[\Large \frac{B^{12}}{B^{8}} = B^{12-8} = B^4\] this is just an example and not the answer
anonymous
  • anonymous
ok
anonymous
  • anonymous
i think thats all thank you
jim_thompson5910
  • jim_thompson5910
glad to be of help
anonymous
  • anonymous
can i ask you 2 more questions?
anonymous
  • anonymous
|dw:1440729886308:dw|
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
think of x as x^1 y as y^1 then use that last rule I posted above
jim_thompson5910
  • jim_thompson5910
|dw:1440730407539:dw|
jim_thompson5910
  • jim_thompson5910
do the same for y and z
jim_thompson5910
  • jim_thompson5910
and of course, 32 over -8 = -4

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