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Micheellexox

  • 2 years ago

Algebra help pleasee

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  1. Micheellexox
    • 2 years ago
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  2. dumbcow
    • 2 years ago
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    the square root is same as exponent of (1/2) ... so divide the exponent in half to nearest integer , then leave the remainder inside square root \[\sqrt{x^{7}} = x^{7/2} = x^{3} x^{1/2} = x^{3} \sqrt{x}\]

  3. Jhannybean
    • 2 years ago
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    \[\frac{ 7q^4 }{{\sqrt{q ^{15}}} }\] so now we're focusing on just q.\[\frac{ q^4 }{ q ^{\frac{ 15 }{ 2 }} }\]So now we have \[q ^{(15/2)-4}= q ^{(15/2)-(8/2)}= q ^{7/2}\] wecan also rewrite q under a radical. \[\sqrt{q ^{7}}\] putting it altogether, we would have \[\frac{ 7 }{ \sqrt{q ^{7}}}\] or you can write this as \[\frac{ 7 }{ q ^{7/2} }\]

  4. Jhannybean
    • 2 years ago
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    Hope that's not too complicated! :(

  5. dumbcow
    • 2 years ago
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    never leave a square root in denominator either... \[\frac{1}{\sqrt{x}}*\frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}}{x}\]

  6. Micheellexox
    • 2 years ago
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    Im confused... Lol. This whole thing is confusing. I am just trying to finish up this last question. /:

  7. Jhannybean
    • 2 years ago
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    What part is confusing?

  8. Jhannybean
    • 2 years ago
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    Oh, in correlation to what @dumbcow was referring to, \[\frac{ 7 }{ \sqrt{q ^{7}} }*\frac{ \sqrt{q ^{7}} }{ \sqrt{q ^{7}} }=\frac{ 7\sqrt{q ^{7}} }{ q ^{7} }\] As he said,you can't have a radical in the denominator.

  9. mahdi2020
    • 2 years ago
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    \[\frac{ 7q ^{4} }{ \sqrt{q ^{8}}*\sqrt{q ^{7}} }\] = \[\frac{ 7 }{ \sqrt{q ^{7}} }= \frac{ 7 }{ q ^{3}*\sqrt{q} }\] \[ans*\frac{ \sqrt{q} }{ \sqrt{q} }\] =\[\frac{ 7\sqrt{q} }{ q ^{3}*q }=\frac{ 7\sqrt{q} }{ q ^{4} }\]

  10. mahdi2020
    • 2 years ago
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    q^4 = sqrt(q*8)

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