Janu16
  • Janu16
Which of the following are equivalent? Justify your reasoning Wait till i draw all the options
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Janu16
  • Janu16
|dw:1438040615062:dw|
Janu16
  • Janu16
@DanJS
jim_thompson5910
  • jim_thompson5910
What does choice B simplify to?

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More answers

Janu16
  • Janu16
x1?
jim_thompson5910
  • jim_thompson5910
x^1 you mean? if so, then yes x^1 = x |dw:1438041275973:dw|
jim_thompson5910
  • jim_thompson5910
how about choice C?
Janu16
  • Janu16
ya that
Janu16
  • Janu16
how do you simplify tht? add?
jim_thompson5910
  • jim_thompson5910
|dw:1438041368560:dw|
jim_thompson5910
  • jim_thompson5910
Due to the rule \[\LARGE x^a*x^b = x^{a+b}\]
Janu16
  • Janu16
|dw:1438041074473:dw|
jim_thompson5910
  • jim_thompson5910
good
Janu16
  • Janu16
wait 3x
jim_thompson5910
  • jim_thompson5910
so C is not equal to A, and not equal to B
jim_thompson5910
  • jim_thompson5910
no there's no 3x
Janu16
  • Janu16
oh ok
jim_thompson5910
  • jim_thompson5910
|dw:1438041475304:dw|
Janu16
  • Janu16
ok i see
jim_thompson5910
  • jim_thompson5910
now simplify D
Janu16
  • Janu16
so you just add the top right so 3/3?
Janu16
  • Janu16
is there 3x for this
jim_thompson5910
  • jim_thompson5910
yeah 1/3+1/3+1/3 = (1+1+1)/3 = 3/3
jim_thompson5910
  • jim_thompson5910
no 3x
jim_thompson5910
  • jim_thompson5910
x*x = x^2 NOT x*x = 2x
Janu16
  • Janu16
so x3/3 only?
jim_thompson5910
  • jim_thompson5910
\[\Large x^{1/3}*x^{1/3}*x^{1/3} = x^{1/3+1/3+1/3}\] \[\Large x^{1/3}*x^{1/3}*x^{1/3} = x^{(1+1+1)/3}\] \[\Large x^{1/3}*x^{1/3}*x^{1/3} = x^{3/3}\] \[\Large x^{1/3}*x^{1/3}*x^{1/3} = x^{1}\] \[\Large x^{1/3}*x^{1/3}*x^{1/3} = x\]
jim_thompson5910
  • jim_thompson5910
|dw:1438041705953:dw|
Janu16
  • Janu16
so b and d are equivalent?
jim_thompson5910
  • jim_thompson5910
correct
Janu16
  • Janu16
so how do i justify the reasoing
Janu16
  • Janu16
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
You're simplifying each expression using the rule \[\LARGE x^a*x^b = x^{a+b}\]
jim_thompson5910
  • jim_thompson5910
for part B, you also use the fact that \[\Large \frac{1}{x^{-k}} = x^k\]
Janu16
  • Janu16
ok thanks!
jim_thompson5910
  • jim_thompson5910
no problem
Janu16
  • Janu16
i will just out it in words

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