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Mathematics
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Simplify the radical expression.
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\[\sqrt{\frac{ 20x^13y^5 }{ 5xy^7}} = \sqrt{\frac{ 4x^{12}}{ y^2}}\]
5 goes in 20, x in x^13, y^7 in y^5
Corrected:\[\sqrt{\frac{ 20x^{13}y^5 }{ 5xy^7}} = \sqrt{\frac{ 4x^{12}}{ y^2}}\]
Thank you c: Can you help with like 3 more?
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\[\frac{ \sqrt{4x^{12}} }{ \sqrt{y^2}}\]
Finally\[ \frac{ 2x^6}{ y }\]
Let's see :)
a:\[(3+\sqrt{6})(3-\sqrt{6}) \rightarrow 9 - (\sqrt6)^2 \rightarrow 9-6 =3\]
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b:\[5\sqrt7+2\sqrt{175} \rightarrow 5\sqrt7+2\sqrt{7\times25} = 5\sqrt7 + 2*5\sqrt7=15\sqrt7\]
c: You'll have to square both sides to remove the radicals... which does leave you with\[5x -1= 4x+9\]collecting like terms:\[x=10\]
I hope you understood the steps.
You're amazing ^.^ Thank you
Ah, you too! Thanks a bunch :)
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