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Mathematics
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6 over quantity 5 minus square root of 3
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\[\frac{ 6 }{ 5 - \sqrt{3} }\] You mean that?
yes thats what i mean
since you have 5 minus sq root 3, multiply top & bottom by 5 + sq root 3
\[\frac{ 6 }{ 5 - \sqrt{3} } \times \frac{ 5 + \sqrt{3} }{ 5 + \sqrt{3} }\]
\[\frac{ 6(5 + \sqrt{3}) }{ 5^{2} - (\sqrt{3})^{2} }\]
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\[\frac{ 6(5 + \sqrt{3}) }{ 25 - 3 }\]
\[\frac{ 3(5 + \sqrt{3}) }{ 11 }\]
that not one of my answers
do they have the same denominator i do? 11
theres only one with 11 quantity of 15 plus 3 square root of 3 all over 11
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That's the same as my ansewr by multiplying out the 3 on top. that's correct
thanks
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