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anonymous

  • one year ago

FAN AND MEDAL

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  1. anonymous
    • one year ago
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    Identify square root of 2 as either rational or irrational, and approximate to the tenths place. Rational: square root of 2 ≈ 1.5 Irrational: square root of 2 ≈ 1.5 Rational: square root of 2 ≈ 1.4 Irrational: square root of 2 ≈ 1.4

  2. anonymous
    • one year ago
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    @isaac4321

  3. SolomonZelman
    • one year ago
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    Ok, do you knwo what RATIONAL and IRRATIONAL numbers are?

  4. SolomonZelman
    • one year ago
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    let me briefly introduce all numbers to you, ok?

  5. anonymous
    • one year ago
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    ok

  6. anonymous
    • one year ago
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    i think the answer is D

  7. anonymous
    • one year ago
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    hey @kiamousekia

  8. anonymous
    • one year ago
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    hey

  9. anonymous
    • one year ago
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    you are right

  10. SolomonZelman
    • one year ago
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    `Natural numbers:` \(\color{black}{ \displaystyle 1,~~2,~~3,~~4,~~5,~~6~~... }\) `Whole numbers:` \(\color{black}{ \displaystyle 0,~~1,~~2,~~3,~~4,~~5,~~6~~... }\) (same as natural numbers, BUT INCLUDES ZERO) `Integers:` \(\color{black}{ \displaystyle ...~-6,~~-5,~~-4,~~-3,~~-2,~~-1,~~0~~1,~~2,~~3,~~4,~~5,~~6~~... }\) (positive and negative whole numbers) `Rational numbers:` Fractions, like \(\color{black}{ \displaystyle 1/5,~ }\) \(\color{black}{ \displaystyle -4/9,~~~~10/7 }\) and on Also, \(\color{black}{ \displaystyle \sqrt{16} }\) since it is really a 4) and all roots, if and only if the number inside is a perfect square (perfect cube or whatever) (But Rational numbers also includes all integers) `Irrational number` Roots like, \(\color{black}{ \displaystyle \sqrt{3},~~\sqrt[8]{8} }\) Also Euler (and other) special constants such \(\color{black}{ \displaystyle \pi,~e,~\psi~ }\) Also logarithms \(\color{black}{ \displaystyle \ln(2) }\), \(\color{black}{ \displaystyle \log_9{7} }\) ...

  11. anonymous
    • one year ago
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    by the way d is the answer

  12. anonymous
    • one year ago
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    thanks

  13. anonymous
    • one year ago
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    i will give the medal and fan to @SolomonZelman because he/she sorry but i don't... because he/she used a lot of time to try to help me

  14. SolomonZelman
    • one year ago
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    You don't have to, i got enough meds and fans o~o As long as you have all of the knowledge that you were looking for ....

  15. SolomonZelman
    • one year ago
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    (not that i mind though)

  16. anonymous
    • one year ago
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    okay but thanks for helping

  17. SolomonZelman
    • one year ago
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    You welcome~!

  18. anonymous
    • one year ago
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    i don't know you

  19. anonymous
    • one year ago
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    k

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