3 years ago

This is past due I need help will give brainlist

Mathematics

2 answers:

Licemer1 [7]3 years ago

6 0

Answer:

Step-by-step explanation:

Flauer [41]3 years ago

5 0

Your answer is A hope this helps! Have a great day!

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EastWind [94]

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Step-by-step explanation:

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3 years ago

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3 years ago

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Delvig [45]

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3 years ago

Rationalize the denominator and simplify. StartFraction 5 minus StartRoot 3 EndRoot Over 4 plus 2 StartRoot 3 EndRoot EndFractio

nasty-shy [4]

Answer:

$\frac{13-7\sqrt{3}}{2}$

Step-by-step explanation:

We need to rationalize the denominator of $= \frac{5-\sqrt{3}}{4+2\sqrt{3}}$ . For rationalizing we multiply the equation by $\frac{4-2\sqrt{3}}{4-2\sqrt{3}}$

So, solving

$= \frac{5-\sqrt{3}}{4+2\sqrt{3}}*\frac{4-2\sqrt{3}}{4-2\sqrt{3}} \\=\frac{(5-\sqrt{3})(4-2\sqrt{3})}{4+2\sqrt{3}*4-2\sqrt{3}}\\=\frac{(5-\sqrt{3})(4-2\sqrt{3})}{(4)^2-(2\sqrt{3})^2}\\= \frac{5(4-2\sqrt{3})-\sqrt{3}(4-2\sqrt{3})}{16-(4*3)}\\=\frac{20-10\sqrt{3}-4\sqrt{3}+2*3}{16-12}\\=\frac{20+6-14\sqrt{3}}{4}\\=\frac{26-14\sqrt{3}}{4}\\= \frac{2(13-7\sqrt{3})}{4}\\=\frac{13-7\sqrt{3}}{2}$