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LaynaMae

  • 2 years ago

Helpp ?

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  1. LaynaMae
    • 2 years ago
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  2. djaydeej88
    • 2 years ago
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    #1. Peter did not simplify his answer completely because he did not reach the conclusion of \[x ^{12}\] second part of the problem is that \[x^3 x^3 x^3 x^3 \neq x^3 + x^3 + x^3 + x^3\] so Peter's work would NOT be the same.

  3. djaydeej88
    • 2 years ago
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    \[x^3 + x^3 + x^3 + x^3 = 4x^3\]

  4. djaydeej88
    • 2 years ago
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    #2. use property of negative exponents in addition to fractional exponents. #3. use quotient property since the base is the same.

  5. djaydeej88
    • 2 years ago
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    #4.\[\sqrt[3]{x^3} = x ^{3*(1/3)}\] involves powers of a product \[x ^{1/3}x ^{1/3}x ^{1/3} = x ^{(1/3) + (1/3) + (1/3)} = x ^{3/3} = x\] involves product of powers \[1/x ^{-1} = x ^{1} = x\] involves negative exponents

  6. LaynaMae
    • 2 years ago
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    Could you help me on showing my work for #2 and #3 ? I don't even know where to begin ..

  7. surjithayer
    • 2 years ago
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    \[\sqrt[n]{x}=\left( x \right)^{\frac{ 1 }{n }}\]

  8. surjithayer
    • 2 years ago
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    \[\frac{ 1 }{\sqrt[3]{x ^{-6}} }=\frac{ 1 }{\left( x ^{-6} \right)^{\frac{ 1 }{3 }} }=\frac{ 1 }{\left( x \right)^{-6*\frac{ 1 }{ 3 }} } =\frac{ 1 }{ x ^{-2 } } =x ^{2}\] i think i have cleared everything,if any doubt ask me.

  9. LaynaMae
    • 2 years ago
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    I'm starting to understand

  10. LaynaMae
    • 2 years ago
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    Would number 3 be |dw:1378942627571:dw| ??

  11. surjithayer
    • 2 years ago
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    \[\frac{ x ^{\frac{ 2 }{ 3 }} }{ x ^{\frac{ 4 }{9 }} }=x ^{\frac{ 2 }{3 }}*x ^{\frac{ -4 }{9 }}\] \[=x ^{\left( \frac{ 2 }{ 3 }-\frac{ 4 }{9 } \right)}\] take L.C.M and solve.

  12. LaynaMae
    • 2 years ago
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    L C M ?

  13. surjithayer
    • 2 years ago
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    when bases are same powers are added during multiplication.

  14. LaynaMae
    • 2 years ago
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    Ohh okay

  15. LaynaMae
    • 2 years ago
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    Okay no I don't get it ..... None of this makes sense :( What's being added and what's being multiplied here ?

  16. surjithayer
    • 2 years ago
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    \[x ^{a}*x ^{b}=x ^{a+b}\]

  17. djaydeej88
    • 2 years ago
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    for #3, when you have the same bases, and it is a division problem, the powers are subtracted. the power from the denominator is subtracted from the power in the numerator like so \[x ^{2/3}/x ^{4/9} = x ^{(2/3)-(4/9)}\]

  18. surjithayer
    • 2 years ago
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    \[\frac{ 2 }{3 }and \frac{ -4 }{ 9 } are added\]

  19. LaynaMae
    • 2 years ago
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    I would be able to figure that out if I could find my freaking scientific calculator :'c

  20. surjithayer
    • 2 years ago
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    \[\frac{ 2 }{3 }and \frac{ -4 }{ 9 } are added\]

  21. LaynaMae
    • 2 years ago
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    2/9 ?

  22. surjithayer
    • 2 years ago
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    \[yes x ^{\frac{ 2 }{ 9 }}\]

  23. djaydeej88
    • 2 years ago
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    use this property when you see a problem like #3 \[x ^{m}/x ^{n} = x ^{m-n}\]

  24. LaynaMae
    • 2 years ago
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    Okay

  25. LaynaMae
    • 2 years ago
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    So is 2/9 the answer then ? :o

  26. djaydeej88
    • 2 years ago
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    yep

  27. LaynaMae
    • 2 years ago
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    Okay thank you so much

  28. surjithayer
    • 2 years ago
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    note that answer is not 2/9 it is as i have given above x ^2/9

  29. LaynaMae
    • 2 years ago
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    Ohh okay well that makes more sense now . Thank YOU

  30. surjithayer
    • 2 years ago
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    yw

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