Hi! I'm a canadian student in america.
Please help me on division of radicals =)
1.) 6√18 ÷ 12√40
2.) 8√19 ÷ 4√38
3.) 20√3 ÷ 5√ 3
4.) 42√ 6 ÷ 3√6
5.) -4√20 ÷ √2
6.) 1 / 2+√5
7.) 1 / 3-√11
8.) 3 / √3-1
9.) 5 / 2-√3
10.) 2 / 4-√14
thanks!
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Hi! I'm a canadian student in america.
Please help me on division of radicals =)
1.) 6√18 ÷ 12√40
2.) 8√19 ÷ 4√38
3.) 20√3 ÷ 5√ 3
4.) 42√ 6 ÷ 3√6
5.) -4√20 ÷ √2
6.) 1 / 2+√5
7.) 1 / 3-√11
8.) 3 / √3-1
9.) 5 / 2-√3
10.) 2 / 4-√14
thanks!
Hello,
$\displaystyle \frac{6\sqrt{18}}{12\sqrt{40}}=\frac{6 \cdot 3 \cdot \sqrt{2}}{12 \cdot 2 \cdot \sqrt{5}}=\frac{3\sqrt{2}}{4\sqrt{5}}=\frac{3\sqrt {10}}{20}$
$\displaystyle \frac{-4\sqrt{20}}{\sqrt{2}}=\frac{-4 \cdot 2 \sqrt{5}}{\sqrt{2}}=\frac{-8 \cdot \sqrt{10}}{2}=-4\sqrt{10}$
From #6 on I presume, that the denominators are sums(?):
$\displaystyle \frac{1}{2+\sqrt{5}} \cdot \frac{2-\sqrt{5}}{2-\sqrt{5}}=\frac{2-\sqrt{5}}{-1}=-2+\sqrt{5}$
$\displaystyle \frac{1}{3-\sqrt{11}} \cdot \frac{3+\sqrt{11}}{3+\sqrt{11}}=\frac{3+\sqrt{11}} {-2}$
$\displaystyle \frac{3}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1}=\frac{3 \sqrt{3} + 3}{2}$
$\displaystyle \frac{5}{2-\sqrt{3}} \cdot \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{10+5\sqrt{3}}{ 1}=10+5\sqrt{3}$
$\displaystyle \frac{2}{4-\sqrt{14}} \cdot \frac{4+\sqrt{14}}{4+\sqrt{14}}=\frac{8+2\sqrt{14} }{2}=4+\sqrt{14}$
EB