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√8y^5 x √40y^2 can you help me simplify the product algebra 2
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√8y^5 x √40y^2 um...
the "x" is a multiplication sign
8y=5x+8 i think if not...
you know the √ is a radical right?
Do you mean: \[\sqrt{8y^5}\times \sqrt{40y^2}\]?
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Yes, thank you
Use this rule for radicals: \[\sqrt{a}\times \sqrt{b}=\sqrt{ab}\]
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