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anonymous

  • 5 years ago

rationalize denominator : please help 1.) 4/ sqrt. 7p 2.) srqt.98/x 3.) 9/3 sqrt.y 4.) sqrt.7 / sqrt.5 +7

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  1. anonymous
    • 5 years ago
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    1.) \[4/ \sqrt{7p}\] 2.) \[\sqrt{98}/x\] 3.) \[9/ 3\sqrt{y}\] 4.) \[\sqrt{7}/ \sqrt{5} + 7\]

  2. anonymous
    • 5 years ago
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    please hlep , i dont understand them

  3. dumbcow
    • 5 years ago
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    ok the idea is to get the sqrt out of the denominator For 1) multiply by sqrt(7p)/sqrt(7p)

  4. dumbcow
    • 5 years ago
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    sqrt(x) * sqrt(x) = x

  5. anonymous
    • 5 years ago
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    so it would be p ?

  6. dumbcow
    • 5 years ago
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    7p everything within the radical sorry imagine if i said x = 7p ->4*sqrt(7p)/7p that is the rationalised form for the first one

  7. dumbcow
    • 5 years ago
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    For number 4 you have to use a conjugate which is basically the denominator except you flip the sign ->sqrt(5)-7 multiply this on top and bottom by distributing (using FOIL) and the sqrts in denominator will go away

  8. dumbcow
    • 5 years ago
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    number 2 is already rationalised but you can simplify sqrt(98) by factoring out a perfect square number 3 can be done same way as num 1, dont forget to reduce hope this helps

  9. anonymous
    • 5 years ago
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    so how would it be done ?

  10. anonymous
    • 5 years ago
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    and thank you for helping , im just really bad in math

  11. dumbcow
    • 5 years ago
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    which part are you stuck on

  12. dumbcow
    • 5 years ago
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    figure it out??

  13. anonymous
    • 5 years ago
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    all of it , especially foil

  14. dumbcow
    • 5 years ago
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    ok lets look at num 2 since there is no sqrt in denominator, we don't have to do anything there but we can simplify sqrt(98) Factors of 98 = 2*49 =2*7*7 rewrite sqrt(98) as sqrt(2)*sqrt(7)*sqrt(7) remember sqrt(7)*sqrt(7) = 7 leaving 7sqrt(2)

  15. anonymous
    • 5 years ago
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    7sqrt. 2x

  16. anonymous
    • 5 years ago
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    ?

  17. anonymous
    • 5 years ago
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    for number one i got : 4sqrt.7p/7p ? correct ?

  18. dumbcow
    • 5 years ago
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    7sqrt(2)/x sorry no the x doesnt change it stays on the bottom

  19. dumbcow
    • 5 years ago
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    correct

  20. dumbcow
    • 5 years ago
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    for num 3 we want to get rid of sqrt(y) by multiplying by sqrt(y) on top and bottom ->9/3sqrt(y) * sqrt(y)/sqrt(y) = 9sqrt(y)/3y Now reduce what does 9/3 reduce to?

  21. anonymous
    • 5 years ago
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    or would 7p cancel itself ?

  22. anonymous
    • 5 years ago
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    to 3

  23. dumbcow
    • 5 years ago
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    correct ohh no the 7p wouldnt cancel because the one on top is inside the sqrt, that is important

  24. dumbcow
    • 5 years ago
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    answer for 3) is 3sqrt(y)/y

  25. anonymous
    • 5 years ago
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    oh no your answer for 3 is wrong , it doesnt come out in my exam as an option

  26. dumbcow
    • 5 years ago
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    hmm maybe the sqrt is in the numerator try 3sqrt(y)

  27. anonymous
    • 5 years ago
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    there is no 3 , only 9 sqrt

  28. dumbcow
    • 5 years ago
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    hmm maybe the sqrt is in the numerator try 3sqrt(y)

  29. dumbcow
    • 5 years ago
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    interesting, what are the possible answers they give

  30. anonymous
    • 5 years ago
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    number 3 : A) 9^3 sqrt.y^2 /y b.) 9y c.) 9^3 sqrt. y^2 d.) 9^3 sqrt. y /y

  31. dumbcow
    • 5 years ago
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    and your sure the question was 9/(3sqrt(y))

  32. anonymous
    • 5 years ago
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    yes

  33. anonymous
    • 5 years ago
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    it better view higher above ^^^^ with the equation button

  34. anonymous
    • 5 years ago
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    all the way up^^^^^^^^^^^^^^^^^^^^^^

  35. dumbcow
    • 5 years ago
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    well in that case the answer is 3sqrt(y)/y i'd bet on d) , weird you might have to talk to teacher on this one

  36. anonymous
    • 5 years ago
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    and for the rest of the problems ?

  37. dumbcow
    • 5 years ago
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    well what about the first 2 are those answers available :)

  38. anonymous
    • 5 years ago
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    do you know how to solve this though ?

  39. dumbcow
    • 5 years ago
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    rationalise fractions with sq roots? yes i know how to solve those but maybe they're looking for something else look at prev examples and see if it lines up with what i've shown

  40. anonymous
    • 5 years ago
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    correction : \[9 / ^{3} \sqrt{y}\]

  41. anonymous
    • 5 years ago
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    so it changes everything ?

  42. dumbcow
    • 5 years ago
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    little bit the answer is A think of it now as fractional exponents cube root of y = y^(1/3) to cancel that we multiply top and bottom by y^2/3 ->y^1/3 * y^2/3 = y (add the exponents) leaving (9*y^2/3)/y

  43. anonymous
    • 5 years ago
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    you mean : 9^3 sqrt.y^2 / y ?

  44. dumbcow
    • 5 years ago
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    oh y^2/3 can be written as cube root of y^2

  45. anonymous
    • 5 years ago
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    its the same than ?

  46. dumbcow
    • 5 years ago
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    yes

  47. anonymous
    • 5 years ago
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    thank youuu ! its so complicated to me. and the rest ?

  48. dumbcow
    • 5 years ago
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    oh yes FOIL is just way to remember how to distribute. what is result of (x+2)*(x-3)? _>distribute the x to both the x and -3 ->x*x + x*-3 = x^2 - 3x ->then distribute the 2 to the x and -3 ->2*x + 2*-3 = 2x -6 put it together ->x^2 -3x +2x -6 = x^2 -x -6 make sense?

  49. anonymous
    • 5 years ago
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    is there an easier way ?

  50. anonymous
    • 5 years ago
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    i have to use (x+2)*(x-3) in the equation to cancel out sqrt ?

  51. dumbcow
    • 5 years ago
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    no that was just an example and not really you still have to do the multiplications For this problem we would be multiplying (sqrt(5)+7) * (sqrt(5) - 7)

  52. dumbcow
    • 5 years ago
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    sorry i gotta go the answer is (7sqrt(7) - sqrt(35))/44 try to figure it out on your own by multiplying what i put above for denominator then multiply sqrt(7)*(sqrt(5)-7) for numerator Then at the end you have to distribute a negative by flipping the signs good luck

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