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blast234 Group Title

....

  • one year ago
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  1. Laura<3 Group Title
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    Uh... Sorry. I don't know. What math are you taking?

    • one year ago
  2. phi Group Title
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    the first step might be to use this rule \[ \frac{\sqrt{a}}{\sqrt{b} } = \sqrt{\frac{a}{b}} \]

    • one year ago
  3. phi Group Title
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    that is a good way to get rid of the sqrt in the bottom. you can then use \[ \sqrt{a}\cdot \sqrt{b}= \sqrt{a \cdot b} \] to merge the 2 sorts then find the prime factors of the number, and "pull out" pairs of the same number from the sort for the x's pull out pairs of x's (remember x^16 is x*x*x...*x (x times itself 16 times, so lots of pairs)

    • one year ago
  4. ximki Group Title
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    125x^15

    • one year ago
  5. phi Group Title
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    Let's do it this way \[ \frac{\sqrt{250 x^{16}}}{\sqrt{2x}}= \sqrt{\frac{250 x^{16}}{2x}}\] simplify to get rid of the fraction. divide 2 and x into the top. what do you get?

    • one year ago
  6. phi Group Title
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    write out the whole answer, with the sort

    • one year ago
  7. phi Group Title
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    what happened to the x in the bottom ? \[\sqrt{\frac{250 x^{16}}{2x}} \]

    • one year ago
  8. phi Group Title
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    remember x*x*x/x is x*x one x from the top cancels out the x in the bottom

    • one year ago
  9. phi Group Title
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    we start with \[ \sqrt{\frac{250 x^{16}}{2x}} \] now divide 2 into 250. you get 125 divide x into x^16 . what do you get?

    • one year ago
  10. phi Group Title
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    x^16 is short hand for x*x*x*x*x*x*x*x*x*x*x*x*x*x*x*x if you divide that by x what do you get?

    • one year ago
  11. phi Group Title
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    yes. Let's do a simple example \[ \frac{x}{x}\] that is 1 \[ \frac{x\cdot x}{x} = \frac{\cancel{x}\cdot x}{\cancel{x}} =x\]

    • one year ago
  12. phi Group Title
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    In your problem we cancel out 1 x from the top and bottom that leaves 15 x's multiplied together. we would use short hand to write it x^15

    • one year ago
  13. phi Group Title
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    closer but when you divide, you don't have the denominator anymore. in other words, we had 250/2 we get 125 NOT 125/2 and x*x/x is x NOT x/x can you fix your answer?

    • one year ago
  14. phi Group Title
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    yes now let's do the easy part: the x^15 Here is an example with x^2: \[ \sqrt{x\cdot x} = x \] we "pull out" a pair of x's and make it just x with no square root Here is an example with x^3 \[ \sqrt{x\cdot x\cdot x} = x\sqrt{x} \] pull out a pair of x's , but leave the last one inside the square root what do we do with x^15 ?

    • one year ago
  15. phi Group Title
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    yes, but we can think about it like this. 15 is an odd number, but 14 is even 14 is 2*7 so we see there are 7 pairs of x's and 1 left over to make 15 \[ \sqrt{x^{15}} = x^7 \sqrt{x} \]

    • one year ago
  16. phi Group Title
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    now the 125. we factor it into prime numbers, and look for pairs. 5 divides into 125. can you find the factors?

    • one year ago
  17. phi Group Title
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    125 is 5*5*5 NO 7 involved. Can you simplify \[ \sqrt{5 \cdot 5 \cdot 5} \]?

    • one year ago
  18. phi Group Title
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    you can "pull out" one pair of 5's

    • one year ago
  19. phi Group Title
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    except when the pair comes out, you drop one of the 5's

    • one year ago
  20. blast234 Group Title
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    is that right

    • one year ago
  21. phi Group Title
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    yes, now put it all together.

    • one year ago
  22. phi Group Title
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    \[ \sqrt{125 x^{15} }\] is ?

    • one year ago
  23. blast234 Group Title
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    How can i put them together

    • one year ago
  24. phi Group Title
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    \[ \sqrt{125 x^{15} } = \sqrt{125} \cdot \sqrt{x^{16}} \] and you know the answer to each of the square roots (we just figured them out)

    • one year ago
  25. phi Group Title
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    that is the simplified sort(125), but you need to multiply it by sqrt(x^15) (simplified)

    • one year ago
  26. phi Group Title
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    sqrt(125) * sqrt(x^15) = 5 * sqrt(5) * x^7 * sqrt(x) we can combine the sqrt(5)*sqrt(x) into sqrt(5x) so the final answer is 5 x^7 sqrt(5x)

    • one year ago
  27. phi Group Title
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    Try some more, because it takes practice.

    • one year ago
  28. blast234 Group Title
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    Ok, i will

    • one year ago
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