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ປັດໄຈທົ່ວໄປທີ່ຍິ່ງໃຫຍ່ທີ່ສຸດ
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2\sqrt{10}\approx 6,32455532
ເບິ່ງຂັ້ນຕອນການແກ້ໄຂ
ຂັ້ນຕອນການແກ້
\sqrt{40}
ຕົວປະກອບ 40=2^{2}\times 10. ຂຽນຮາກຂັ້ນສອງຂອງຜົນຄູນ \sqrt{2^{2}\times 10} ເປັນຜົນຄູນຂອງຮາກຂັ້ນສອງ \sqrt{2^{2}}\sqrt{10}. ເອົາຮາກຂັ້ນສອງຂອງ 2^{2}.
2\sqrt{10}
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Arithmetic
\sqrt{40}
ບັນຫາທີ່ຄ້າຍຄືກັນຈາກWeb Search
How do you simplify \displaystyle-\sqrt{{40}} ?
https://socratic.org/questions/how-do-you-simplify-sqrt40
\displaystyle-{2}\sqrt{{10}} Explanation: \displaystyle-\sqrt{{40}}\rightarrow Look for any perfect squares \displaystyle-\sqrt{{{4}\cdot{10}}}\rightarrow \displaystyle{4} is a ...
How do you simplify \displaystyle{5}\sqrt{{40}} ?
https://socratic.org/questions/how-do-you-simplify-5-sqrt40
\displaystyle{10}\sqrt{{{10}}} Explanation: Starting with \displaystyle\sqrt{{{40}}} \displaystyle\sqrt{{{4}\cdot{10}}} \displaystyle\sqrt{{{2}^{{2}}\cdot{10}}} \displaystyle{2}\sqrt{{{10}}} ...
sqrt400
http://www.tiger-algebra.com/drill/sqrt400/
sqrt(400) Simplified Root : 20 Simplify : sqrt(400) Factor 400 into its prime factors 400 = 24 • 52 To simplify a square root, we extract factors which are squares, i.e., ...
If an ice rink is \displaystyle\sqrt{{{404}}} across then what is its perimeter?
https://socratic.org/questions/59e82233b72cff05e66021ea
It could be \displaystyle{44} , but I'm not particularly convinced. Explanation: Since there is not enough information in the question, let us work with the following assumptions: The ice rink is ...
How do you write \displaystyle\sqrt{{{405}}} in simplified radical form?
https://socratic.org/questions/how-do-you-write-sqrt-405-in-simplified-radical-form
\displaystyle{9}\sqrt{{{5}}} Explanation: Factoring: \displaystyle{405} \displaystyle{\left(\text{XXX}\right)}={5}\times{81} \displaystyle{\left(\text{XXX}\right)}={5}\times{9}^{{2}} ...
sqrt(408)
https://www.tiger-algebra.com/drill/sqrt(408)/
sqrt(408) Simplified Root : 2 • sqrt(102) Simplify : sqrt(408) Factor 408 into its prime factors 408 = 23 • 3 • 17 To simplify a square root, we extract factors which are ...
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2\sqrt{10}
ຕົວປະກອບ 40=2^{2}\times 10. ຂຽນຮາກຂັ້ນສອງຂອງຜົນຄູນ \sqrt{2^{2}\times 10} ເປັນຜົນຄູນຂອງຮາກຂັ້ນສອງ \sqrt{2^{2}}\sqrt{10}. ເອົາຮາກຂັ້ນສອງຂອງ 2^{2}.
ບັນຫາທີ່ຄ້າຍຄືກັນ
\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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