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3\sqrt{11}a^{\frac{3}{2}}
Kimi Pārōnaki e ai ki a
\frac{9\sqrt{11a}}{2}
Pātaitai
Algebra
5 raruraru e ōrite ana ki:
\sqrt{99a^3}
Ngā Raru Ōrite mai i te Rapu Tukutuku
How do you simplify \displaystyle\sqrt{{{49}{a}^{{8}}}} ?
https://socratic.org/questions/how-do-you-simplify-sqrt-49-a-8
Alan P. Apr 7, 2015 \displaystyle\sqrt{{{49}{a}^{{8}}}}=\sqrt{{{7}^{{2}}\cdot{\left({a}^{{4}}\right)}^{{2}}}} \displaystyle={7}{a}^{{4}}
How do you simplify \displaystyle\sqrt{{{9}{a}{b}^{{4}}}} ?
https://socratic.org/questions/how-do-you-simplify-sqrt-9ab-4
\displaystyle={3}\cdot{b}^{{2}}\sqrt{{a}} Explanation: \displaystyle\sqrt{{{9}{a}{b}^{{4}}}} \displaystyle=\sqrt{{{3}\cdot{3}\cdot{a}\cdot{b}\cdot{b}\cdot{b}\cdot{b}}} \displaystyle=\sqrt{{{3}^{{2}}\cdot{a}\cdot{b}^{{2}}\cdot{b}^{{2}}}} ...
How do you simplify \displaystyle{4}{\sqrt[{3}]{{{a}^{{3}}}}} ?
https://socratic.org/questions/how-do-you-simplify-4root3-a-3
\displaystyle{4}{\sqrt[{{3}}]{{{a}^{{3}}}}}={4}{a} Explanation: \displaystyle{4}{\sqrt[{{3}}]{{{a}^{{3}}}}} = \displaystyle{4}{\sqrt[{{3}}]{{{a}\times{a}\times{a}}}} = \displaystyle{4}{\sqrt[{{3}}]{\underline{{{\left({a}\times{a}\times{a}\right)}}}}} ...
What's the easiest way to make \frac{\sqrt{5a} \sqrt{10a^5}}{\sqrt{2 a}} into 5a^2 \sqrt{a}?
https://www.quora.com/Whats-the-easiest-way-to-make-frac-sqrt-5a-sqrt-10a-5-sqrt-2-a-into-5a-2-sqrt-a
Thank you for showing your work, so that we know we are not doing your homework for you. Let’s look at your fraction: \frac{\sqrt{5a}\cdot\sqrt{10a^5}}{\sqrt{2a}} Do you remember that \sqrt{AB} ...
How do you simplify \displaystyle\sqrt{{{49}{a}^{{{15}}}}} ?
https://socratic.org/questions/how-do-you-simplify-sqrt-49a-15
\displaystyle{7}{a}^{{7.5}} Explanation: As \displaystyle{49}={7}\cdot{7}={7}^{{2}} , \displaystyle\sqrt{{49}}={7} And \displaystyle\sqrt{{{a}^{{15}}}}=\sqrt{{{a}^{{14}}}}\cdot\sqrt{{a}}={a}^{{7}}\sqrt{{a}}={a}^{{7.5}} ...
sqrt(18a^3)
https://www.tiger-algebra.com/drill/sqrt(18a~3)/
sqrt(18a3) Simplified Root : 3 a • sqrt(2a) Simplify : sqrt(18a3) Step 1 :Simplify the Integer part of the SQRT Factor 18 into its prime factors 18 = 2 • 32 To simplify a ...
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Ngā Raru Ōrite
\sqrt{40}
\sqrt{99a^3}
\sqrt{\frac{16}{25}}
\sqrt{3} \times \sqrt{3a^4}
\sqrt{\sqrt{256a^8}}
\sqrt{196}
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