Find the greatest common factor of 8a^ 3 b ^2 and 12ab^ 4

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

6 0

<span>8a^ 3 b ^2 = 4ab^2 (2a^2)
and
12ab^ 4 = 4ab^2(3b^2)

answer: GCF = </span>4ab^2

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Conjugate/Rational Number?

hram777 [196]

Answer:

1) $\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

Step-by-step explanation:

The rationalization of the denominator of the surds are found as follows;

1) $\dfrac{2}{\sqrt{5} }$

$\dfrac{2}{\sqrt{5} } \times \dfrac{\sqrt{5} }{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

$\dfrac{2}{\sqrt{5} } = \dfrac{2 \cdot \sqrt{5} }{5}$

2) $-\dfrac{5}{\sqrt{3} }$

$-\dfrac{5}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

$-\dfrac{5}{\sqrt{3} } = -\dfrac{5 \cdot \sqrt{3} }{3}$

3) $\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} }$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } \times \dfrac{ \sqrt{10} }{\sqrt{10} } = \dfrac{\sqrt{20} + \sqrt{50} }{10 } = \dfrac{2\cdot \sqrt{5} + 5 \cdot \sqrt{2} }{10} = \dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

$\dfrac{\sqrt{2} + \sqrt{5} }{\sqrt{10} } =\dfrac{\sqrt{5} }{5} + \dfrac{ \sqrt{2} }{2}$

4) $\dfrac{3 + \sqrt{2} }{\sqrt{3} }$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \dfrac{3 \cdot \sqrt{3}+\sqrt{6} }{3 } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

$\dfrac{3 + \sqrt{2} }{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} } = \sqrt{3} + \dfrac{\sqrt{6} }{3}$

5) $\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} } = \dfrac{\sqrt{5} - \sqrt{2} }{\sqrt{5} - \sqrt{2} } = \dfrac{\sqrt{15} -\sqrt{6} }{5 - 2} = \dfrac{\sqrt{15} - \sqrt{6} }{3}$

$\dfrac{\sqrt{3} }{\sqrt{5} + \sqrt{2} }= \dfrac{\sqrt{15} - \sqrt{6} }{3}$

6) $\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} }$

$\dfrac{\sqrt{7} }{\sqrt{3} - \sqrt{5} } \times \dfrac{\sqrt{3} + \sqrt{5}}{\sqrt{3} + \sqrt{5}} = \dfrac{\sqrt{21} + \sqrt{35}}{{3} + {5}} = \dfrac{\sqrt{21} + \sqrt{35}}{8}$