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Lapatulllka [165]
3 years ago
10

What is the LCM of the numbers 3, 6, and 9? OA. 9 OB. 18 O C. 36 OD 54

Mathematics
2 answers:
Bond [772]3 years ago
8 0

Answer:

b...................................

vaieri [72.5K]3 years ago
5 0
B. 18
:)
explanation: The LCM of 3,6,9 3 , 6 , 9 is the result of multiplying all prime factors the greatest number of times they occur in either number. The LCM of 3,6,9 3 , 6 , 9 is 2⋅3⋅3=18 2 ⋅ 3 ⋅ 3 = 18 .
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Work Shown:

\frac{5}{\sqrt{11}-\sqrt{3}}\\\\\\\frac{5(\sqrt{11}+\sqrt{3})}{(\sqrt{11}-\sqrt{3})(\sqrt{11}+\sqrt{3})}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{(\sqrt{11})^2-(\sqrt{3})^2}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{11-3}\\\\\\\frac{5\sqrt{11}+5\sqrt{3}}{8}\\\\\\

This shows why choice B is the answer.

---------------

Explanation:

  • In the second step, I multiplied top and bottom by (\sqrt{11}+\sqrt{3})
  • This is so we can apply the difference of squares rule in step 3. The difference of squares rule is (x-y)(x+y) = x^2-y^2.
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So in general, if the denominator is \sqrt{a}-\sqrt{b} and you want to rationalize the denominator, then you should multiply top and bottom by \sqrt{a}+\sqrt{b}. The same applies in reverse as well.

This leads to the denominator becoming (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) = (\sqrt{a})^2-(\sqrt{b})^2 = a-b

Keep in mind that a \ge 0 and b \ge 0

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