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hybrik

  • one year ago

Find x+y+z if x(square root of y)=108 y(square root of z)=80(square root of 3) z(square root of x)=225(square root of 3)

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  1. hybrik
    • one year ago
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    in other terms: \[x \sqrt{y}=108, y \sqrt{z}=80\sqrt{3}, and, z \sqrt{x}=225\sqrt{3}\]

  2. hybrik
    • one year ago
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    @ganeshie8

  3. hybrik
    • one year ago
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    @Hero

  4. hybrik
    • one year ago
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    @welshfella

  5. hybrik
    • one year ago
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    any ideas?

  6. hybrik
    • one year ago
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    @undeadknight26

  7. undeadknight26
    • one year ago
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    This is mind boggling o.o

  8. hybrik
    • one year ago
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    I got 6 minutes then i have lunch.

  9. undeadknight26
    • one year ago
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    Getting confused from just looking at it.

  10. hybrik
    • one year ago
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    its a algebra 2 question i believe

  11. undeadknight26
    • one year ago
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    You pretty much have to do them all at once...And the x(sqrt y) thing confuses me.

  12. hybrik
    • one year ago
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    i got lunch, i was thinking to square the whole equations

  13. hybrik
    • one year ago
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    any ideas?

  14. freckles
    • one year ago
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    \[x \sqrt{y}=108 \\ y \sqrt{z}=80 \sqrt{3} \\ z \sqrt{x}=225 \sqrt{3}\] this is right?

  15. hybrik
    • one year ago
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    ye

  16. freckles
    • one year ago
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    and is x and y and z integers are real?

  17. hybrik
    • one year ago
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    I don't know :|

  18. freckles
    • one year ago
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    and is x and y and z integers or real? *

  19. hybrik
    • one year ago
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    I just want to learn how to do these questions, because I like to learn.

  20. freckles
    • one year ago
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    I understand that but I was asking if you knew anything about the variables

  21. freckles
    • one year ago
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    I guess we will assume they are real

  22. hybrik
    • one year ago
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    ok

  23. freckles
    • one year ago
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    \[x^2 y=108^2 \\ y^2 z=80^2(3) \\ z^2 x=225^2 (3) \\ \text{ here I just squared both sides of equation equation } \] \[y=\frac{108^2}{x^2} \text{ solved first equation for } y \\ \text{ replacing } y \text{ \in second equation with this } \\ (\frac{108^2}{x^2})^2z=80^2(3) \\ \text{ by law of exponents you have } \frac{108^4}{x^4} z=80^2 (3) \\ \text{ multiplying both sides by } \frac{x^4}{108^4} \\ z=\frac{80^2(3)x^4}{108^4} \\ \text{ so replacing the } z \text{ \in the last equation with this } \\ (\frac{80^2(3)x^4}{108^4})^2x=225^2(3) \text{ you can solve this one for } x \text{ now }\] this seems like a pretty long way there might be a shorter way but you can solve the last equation for x then go back and find z and then go back and find y then add them up to find x+y+z

  24. hybrik
    • one year ago
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    This is taking a bit of time, need 2 minutes.

  25. freckles
    • one year ago
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    @ganeshie8 do you see any kind of short cut here

  26. hybrik
    • one year ago
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    x=27

  27. hybrik
    • one year ago
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    y=16

  28. freckles
    • one year ago
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    cool I got x=27 too

  29. hybrik
    • one year ago
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    Can't find Z though..

  30. hybrik
    • one year ago
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    5 radical 3

  31. freckles
    • one year ago
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    \[z=\frac{80^2 (3)x^4}{108^4}\]

  32. hybrik
    • one year ago
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    x+y+z = 42+\[5\sqrt{3}\]

  33. freckles
    • one year ago
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    \[y=\frac{108^2}{x^2} \\ y=\frac{108^2}{(27)^2}=16 \\ z=\frac{80^2(3)x^4}{108^4} \\ z=\frac{80^2(3)(27)^4}{108^4}\]

  34. hybrik
    • one year ago
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    o

  35. hybrik
    • one year ago
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    75

  36. freckles
    • one year ago
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    cool stuff

  37. freckles
    • one year ago
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    x+y+z=27+16+75

  38. hybrik
    • one year ago
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    42+75=127

  39. freckles
    • one year ago
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    so i guess they were integers after all

  40. hybrik
    • one year ago
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    What if they weren't what would you do?

  41. freckles
    • one year ago
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    isn't 27+16 equal to 43?

  42. freckles
    • one year ago
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    and isn't 43+75=118

  43. hybrik
    • one year ago
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    bad math confirmed.

  44. freckles
    • one year ago
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    well we just assumed they were real numbers in the beginning

  45. hybrik
    • one year ago
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    So thats why I got a 145/150 on this Mu Alpha contest

  46. freckles
    • one year ago
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    that includes the integers and non-integers

  47. freckles
    • one year ago
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    I was thinking that there should be a short cut for some reason but I do not see one

  48. hybrik
    • one year ago
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    Two awards, [Fan medal], https://www.youtube.com/watch?v=jH-miNEnIgE

  49. hybrik
    • one year ago
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    One more question maybe?

  50. freckles
    • one year ago
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    ok I will try to answer

  51. hybrik
    • one year ago
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    Given that a^2 sub 1 + a^2 sub 2 + a^2 sub 3= 7, find the maximum possible value of 3a sub 1 + 2a sub 2 + a sub 3

  52. freckles
    • one year ago
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    \[a^2_1+a^2_2+a^2_3=7 \\ \text{ find max of } 3a_1+2a_2+a_3\]

  53. hybrik
    • one year ago
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    yeah

  54. freckles
    • one year ago
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    I'm thinking that we can try to use that one thing it starts with an L lebniz?

  55. hybrik
    • one year ago
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    Leibniz, I am in 6th grade, I dont know this stuff well, explian more of it.

  56. freckles
    • one year ago
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    oh I was thinking of a calculus thing

  57. hybrik
    • one year ago
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    I do pre calculus

  58. freckles
    • one year ago
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    i'm so dumb it was another l word

  59. freckles
    • one year ago
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    lagrange multiplers

  60. hybrik
    • one year ago
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    What is that exactly?

  61. freckles
    • one year ago
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    http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx this is what it is

  62. hybrik
    • one year ago
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    K then?

  63. freckles
    • one year ago
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    \[g(z,y,z)=x^2+y^2+z^2=7 \\ f(x,y,z)=3x+2y+z \\ f_x=3 \\ f_y=2 \\ f_z=1 \\ g_x=2x \\ g_y=2y \\ g_z=2z \\ \text{ so we have the equations } \\ 3= \lambda 2x \\ 2= \lambda 2y \\ 1 =\lambda 2z\]

  64. freckles
    • one year ago
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    we need to solve that bottom system for (x,y,z)

  65. freckles
    • one year ago
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    and we also have that first equation

  66. freckles
    • one year ago
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    \[3=\lambda 2x \\ 2=\lambda 2y \\ 1=\lambda 2z \\ x^2+y^2+z^2=7\]

  67. hybrik
    • one year ago
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    3x+2y+z = what number?

  68. freckles
    • one year ago
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    well if you can solve that system of equations above for (x,y,z) you can find out you might need to find lambda first though

  69. hybrik
    • one year ago
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    I honestly don't know what that is :(, I read it but it doesnt make my brain into order

  70. freckles
    • one year ago
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    so maybe we need another way to solve besides lagrange multiplers then

  71. freckles
    • one year ago
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    any ideas out there

  72. freckles
    • one year ago
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    that is the only one I have unfortunately

  73. freckles
    • one year ago
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    that lagrange multipler way is pretty cute I found lambda and it wasn't too bad it was just a quadratic equation and then as result I was able to find x,y, and z :(

  74. hybrik
    • one year ago
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    It is hard to learn pre cal in 6th grade :(

  75. freckles
    • one year ago
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    yeah I think lagrange multipler is a cal 3 method

  76. freckles
    • one year ago
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    or cal 2

  77. hybrik
    • one year ago
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    :_:

  78. hybrik
    • one year ago
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    I have, one more >_<

  79. hybrik
    • one year ago
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    Let x and y be real numbers satisfying 2/x = y/3 = x/y. Determine x^3

  80. hybrik
    • one year ago
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    Im assuming 0, but that just seems wrong

  81. freckles
    • one year ago
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    \[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\] maybe we can play with these equations

  82. imqwerty
    • one year ago
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    \[x^3+y^3+z^3=(x+y+z)^2-3(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z)-9xyz\]

  83. hybrik
    • one year ago
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    @imqwerty the question i just asked is that the simplified part of it or the one before ">_<"

  84. imqwerty
    • one year ago
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    is there any way to find\[x^3+y^3+z^3\]?

  85. hybrik
    • one year ago
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    @freckles Where did you get the values to equal from (6)

  86. freckles
    • one year ago
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    \[6=xy \\ 2y=x^2 \ \text{ note multiply both sides by } x \text{ and then use first equation }\]

  87. imqwerty
    • one year ago
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    :O ok :)

  88. freckles
    • one year ago
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    \[\frac{2}{x}=\frac{y}{3} \\ \text{ multiply both sides by } 3 \\ \text{ multiply both sides by } x \\ \text{ this is how I got } 6=xy \] Is this the equation with 6 you are talking about @hybrik

  89. hybrik
    • one year ago
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    Or did you cross multiply 2 * 3 and x*y

  90. freckles
    • one year ago
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    that is what some people like to call it

  91. freckles
    • one year ago
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    that is what I did for each pair of fractions that were equal

  92. freckles
    • one year ago
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    I only needed the first two equations though

  93. freckles
    • one year ago
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    \[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\ \] Do you see multiplying the second one on both sides by x gives 2xy=x^3 and you see that xy is given as 6 in the first equation?

  94. hybrik
    • one year ago
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    Im just going with my method, proceed your explanation

  95. freckles
    • one year ago
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    my method is done :p

  96. freckles
    • one year ago
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    you just replace xy with 6

  97. freckles
    • one year ago
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    2(6)=x^3

  98. hybrik
    • one year ago
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    Huh? I dont remember 2xy=x^3?

  99. freckles
    • one year ago
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    did you read anything I said?

  100. freckles
    • one year ago
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    that isn't meant as a rude question by the way

  101. freckles
    • one year ago
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    \[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\] So you see the second equation? 2y=x^2 ?

  102. freckles
    • one year ago
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    I multiplied both sides by x 2y(x)=x^2(x) 2xy=x^3

  103. freckles
    • one year ago
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    the first equation says xy=6

  104. freckles
    • one year ago
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    replace xy with 6 so you have 2(6)=x^3

  105. hybrik
    • one year ago
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    12 = x^3, x = 2 radical 3

  106. freckles
    • one year ago
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    you don't need to find x

  107. hybrik
    • one year ago
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    ok

  108. freckles
    • one year ago
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    \[x^3=12 \text{ also this doesn't give a solution of } x=2 \sqrt{3}\]

  109. hybrik
    • one year ago
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    ik that was x^2 :/

  110. freckles
    • one year ago
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    you do have to show work on these right?

  111. hybrik
    • one year ago
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    i guess

  112. freckles
    • one year ago
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    lol what does that mean?

  113. hybrik
    • one year ago
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    yes 3:

  114. freckles
    • one year ago
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    since you have to show work, I will show you the answer for the second one under one condition you come back and show me your method that you were suppose to use. ok how about this then... I show you the lagrange multipler way on that second one and show you my result... then you come up with the right method for your level and compare our answers

  115. freckles
    • one year ago
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    oops I didn't realize I was repetitive in that paragraph I was doing some cutting and pasting and rearranging and forgot I said some stuff when I did :p

  116. hybrik
    • one year ago
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    3;

  117. freckles
    • one year ago
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    what kind of face is that? an agreeing face? or a " I don't know" face?

  118. hybrik
    • one year ago
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    its a 3 and a type of colon 3;

  119. freckles
    • one year ago
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    well I might go to sleep now then goodnight

  120. hybrik
    • one year ago
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    me too 1:24 am is hurting me head

  121. phi
    • one year ago
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    Find x+y+z if \[ x\sqrt{y}=108 \\ y\sqrt{z}=80\sqrt{3} \\ z\sqrt{x}=225\sqrt{3}\] I would solve it this way: A \(y\sqrt{z}=2^4 5^1\sqrt{3} \) z has exactly one factor of 3 B \( z\sqrt{x}=3^2 5^2 \sqrt{3}\) z has exactly one factor of 3 , so \( \sqrt{x}\) must contain \( 3\sqrt{3}\) and x has the factor 9*3=27. Also, 2 is not a factor of x C \( x\sqrt{y}=2^2 3^3\) we know from B, 2 is not a factor of x, but 27 is. That leaves \( \sqrt{y}=4 \) and y=16 D \( z\sqrt{x}=3^2 5^2 \sqrt{3} \) with x=27 , we find z= 75 x+y+z= 27+16+75= 118

  122. hybrik
    • one year ago
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    I can see what you did, very easy. @phi

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