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Kathatesmath94

  • 2 years ago

can someone check my work??

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  1. Kathatesmath94
    • 2 years ago
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    sqrt3(sqrt6+sqrt3)+(sqrt8-5)

  2. Kathatesmath94
    • 2 years ago
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    |dw:1354245450082:dw|

  3. Lime
    • 2 years ago
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    I believe it is much easier to use a calculator for Square root, but let's break it down. \[3 + 3 \sqrt2 +(\sqrt 8 - 5)\]

  4. Kathatesmath94
    • 2 years ago
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    well from there i got 3+3sqrt2= 6sqrt2 + sqrt8-5

  5. Kathatesmath94
    • 2 years ago
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    and the sqrt of 8 is 4 and sqrt of 4 is 2 so it will be 6sqrt2+sqrt2-5

  6. Kathatesmath94
    • 2 years ago
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    whoah where did that come from

  7. scarydoor
    • 2 years ago
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    |dw:1354246002329:dw|

  8. Kathatesmath94
    • 2 years ago
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    ohh i see i took the sqaure root of 8 wrong

  9. Kathatesmath94
    • 2 years ago
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    can i ask ya'll another question??

  10. scarydoor
    • 2 years ago
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    yep.

  11. Kathatesmath94
    • 2 years ago
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    Part 1: Find the perimeter of a rectangular object which has a length of sqrt 128 feet and a width of sqrt 200 feet. Part 2: Explain, in complete sentences, how you arrived at the answer and give the final solution in simplified radical form. Part 3: What type of object in your home or school might this be? i am not going to lie I don't even know where to start....

  12. scarydoor
    • 2 years ago
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    |dw:1354246494891:dw| Just add up the sides, for part one. Simplify the square roots a little.

  13. Kathatesmath94
    • 2 years ago
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    sqrt of 128= 8sqrt 2?

  14. scarydoor
    • 2 years ago
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    That's right.

  15. Kathatesmath94
    • 2 years ago
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    and the sqrt of 200 = 10sqrt 2???

  16. scarydoor
    • 2 years ago
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    yep.

  17. Kathatesmath94
    • 2 years ago
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    what do I do from there?

  18. Kathatesmath94
    • 2 years ago
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    do I add them together?

  19. scarydoor
    • 2 years ago
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    perimeter of a rectangle.

  20. Kathatesmath94
    • 2 years ago
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    how do i do that?

  21. scarydoor
    • 2 years ago
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    You want to know how long the edge is... |dw:1354247036064:dw|

  22. Kathatesmath94
    • 2 years ago
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    so multiply 8sqrt2*8sqrt2 and 10sqrt2*10sqrt2? and then add them together?

  23. Kathatesmath94
    • 2 years ago
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    are you still there?

  24. Kathatesmath94
    • 2 years ago
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    am I still doing this right? hello?

  25. scarydoor
    • 2 years ago
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    Yeah to find the perimeter of a rectangle, you add the lengths of all sides together.

  26. Kathatesmath94
    • 2 years ago
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    after you multiply?

  27. scarydoor
    • 2 years ago
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    Just take the lengths of the sides and add them together.

  28. Kathatesmath94
    • 2 years ago
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    so 8sqrt2*8sqrt2=64sqrt2 10sqrt2*10sqrt2= 100sqrt 2 64sqrt2+100sqrt2= 164sqrt2?

  29. Kathatesmath94
    • 2 years ago
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    am i right so far?

  30. Kathatesmath94
    • 2 years ago
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    was I just suppose to add them all together or multiply? like i did above?

  31. jim_thompson5910
    • 2 years ago
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    |dw:1354248044017:dw|

  32. jim_thompson5910
    • 2 years ago
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    |dw:1354248083431:dw|

  33. jim_thompson5910
    • 2 years ago
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    |dw:1354248126981:dw| Perimeter = Sum of all sides P = s1 + s2 + s3 + s4 P = 8*sqrt(2) + 10*sqrt(2) + 8*sqrt(2) + 10*sqrt(2) P = 36*sqrt(2)

  34. jim_thompson5910
    • 2 years ago
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    So the perimeter is \[\Large 36\sqrt{2}\]

  35. Kathatesmath94
    • 2 years ago
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    oh so I was suppose to add them not multiply

  36. jim_thompson5910
    • 2 years ago
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    yes, or you can add the length and width and multiply by 2 this is because P = L + W + L + W P = 2L + 2W P = 2*(L + W)

  37. Kathatesmath94
    • 2 years ago
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    oh ok that what i was thinking of

  38. jim_thompson5910
    • 2 years ago
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    Explain, in complete sentences, how you arrived at the answer and give the final solution in simplified radical form you essentially simplify both sqrt(128) and sqrt(200) to get 8*sqrt(2) and 10*sqrt(2) you can either add up the four sides or you can add up the length and width and multiply by 2 either way, you get the answer of \[\Large 36\sqrt{2}\]

  39. Kathatesmath94
    • 2 years ago
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    ok thats what i wrote so far what is the next step?

  40. jim_thompson5910
    • 2 years ago
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    What type of object in your home or school might this be? \[\Large 36\sqrt{2} \approx 36(1.41421356)\] \[\Large 36\sqrt{2} \approx 50.91168816\] so the perimeter is roughly 50.91168816 feet

  41. jim_thompson5910
    • 2 years ago
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    \[\Large 8\sqrt{2} \approx 8(1.41421356)\] \[\Large 8\sqrt{2} \approx 11.31370848\] the width is roughly 11.31370848 feet

  42. Kathatesmath94
    • 2 years ago
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    the funny equal sign is a approximate sign right?

  43. jim_thompson5910
    • 2 years ago
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    \[\Large 10\sqrt{2} \approx 10(1.41421356)\] \[\Large 10\sqrt{2} \approx 14.1421356\] the length is roughly 14.1421356 feet

  44. jim_thompson5910
    • 2 years ago
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    so you just have to name an object that's roughly 14.1421356 ft by 11.31370848 ft

  45. Kathatesmath94
    • 2 years ago
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    so it could be a pool right? for the object?

  46. jim_thompson5910
    • 2 years ago
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    yeah or a large table/counter of some sort

  47. Kathatesmath94
    • 2 years ago
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    ohhh wow that wasnt too bad there is another one very similiar and if I post all my work will you tell me if I am right or not?

  48. jim_thompson5910
    • 2 years ago
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    sure go for it

  49. Kathatesmath94
    • 2 years ago
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    awesome that way I know I really understand it

  50. Kathatesmath94
    • 2 years ago
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    Part 1: Find the area of a rectangular object which has a length of 4sqrt2 inches and a width of 6sqrt6 inches. Part 2: Explain, in complete sentences, how you arrived at the simplified answer and give the final solution in simplified radical form. Part 3: What type of object in your home or school might this be? |dw:1354248883689:dw|

  51. jim_thompson5910
    • 2 years ago
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    ok it's a bit different because they want the area this time

  52. jim_thompson5910
    • 2 years ago
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    and not the perimeter

  53. Kathatesmath94
    • 2 years ago
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    isn't area multiplying the sides together?

  54. jim_thompson5910
    • 2 years ago
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    exactly

  55. jim_thompson5910
    • 2 years ago
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    Area = Length times Width

  56. Kathatesmath94
    • 2 years ago
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    ok so first i have to simplify the sqrts 6sqrt 6 = 6+3sqrt2?

  57. Kathatesmath94
    • 2 years ago
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    9sqrt 2?

  58. jim_thompson5910
    • 2 years ago
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    how are you getting 6sqrt 6 = 6+3sqrt2

  59. Kathatesmath94
    • 2 years ago
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    uhhh i might of factored wrong because i took the factors 2 and 3 and pulled the 3 out to add to the 6 and left the 2 inside the radical

  60. jim_thompson5910
    • 2 years ago
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    you can't do that

  61. jim_thompson5910
    • 2 years ago
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    6*sqrt(6) is as simple as it gets

  62. Kathatesmath94
    • 2 years ago
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    i was trying to make it so both of my radicals have the same square root thing of sqrt2

  63. jim_thompson5910
    • 2 years ago
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    notice that 6 has no factors that are perfect squares (ignore 1)

  64. jim_thompson5910
    • 2 years ago
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    so sqrt(6) can't be simplified

  65. Kathatesmath94
    • 2 years ago
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    oh..... i was wondering how that worked cause nothing times itself equals 6

  66. jim_thompson5910
    • 2 years ago
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    Multiply the coefficients: 6*4 = 24 Multiply the radicands: 6*2 = 12

  67. Ammarah
    • 2 years ago
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    what is ur question, i mean what r u trying to find exactly?

  68. Ammarah
    • 2 years ago
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    so u multiply and then plugg it back in to the equation.

  69. jim_thompson5910
    • 2 years ago
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    so 6*sqrt(6) times 4*sqrt(2) = 24*sqrt(12) simplify 24*sqrt(12)

  70. Ammarah
    • 2 years ago
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    yes dont forget!! Simplify!

  71. Kathatesmath94
    • 2 years ago
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    so I can go straight to part 2 right? and find the area?

  72. Ammarah
    • 2 years ago
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    yes.

  73. Kathatesmath94
    • 2 years ago
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    sorry idk why that post just showed up i have a major delay

  74. Ammarah
    • 2 years ago
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    ?

  75. jim_thompson5910
    • 2 years ago
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    yes since you can't simplify the length/width just multiply them out

  76. Kathatesmath94
    • 2 years ago
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    now is 24sqrt12 the radical area that needs to be converted into the approximate length?

  77. jim_thompson5910
    • 2 years ago
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    simplify that 24*sqrt(12) 24*sqrt(4*3) 24*sqrt(4)*sqrt(3) 24*2*sqrt(3) 48*sqrt(3)

  78. Kathatesmath94
    • 2 years ago
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    and do you get the approximate length by putting the number in the calculator? i meant to ask you how you did that last time

  79. jim_thompson5910
    • 2 years ago
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    so the area is 48*sqrt(3) square inches

  80. jim_thompson5910
    • 2 years ago
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    you just type 6*sqrt(6) and you get 14.6969 roughly

  81. Kathatesmath94
    • 2 years ago
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    and 4sqqrt of 2 is approximately 5.65685424949238

  82. jim_thompson5910
    • 2 years ago
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    correct

  83. Kathatesmath94
    • 2 years ago
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    so the object could be like a table? o ra whiite board maybe?

  84. jim_thompson5910
    • 2 years ago
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    yeah that's a reasonably sized whiteboard

  85. Kathatesmath94
    • 2 years ago
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    you have helped me so much and thank you for helping me with this someone was helping me but they left and I didnt know where to go from there you are definitley a life SAVER!

  86. jim_thompson5910
    • 2 years ago
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    i'm glad i did and hopefully it's all making sense now

  87. Kathatesmath94
    • 2 years ago
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    yeah they were a little different but there steps were very similar and I took notes while we talked so hopefully that will help me if I have a problem like this later on in the course

  88. danielcvalencia
    • 2 years ago
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    youre WELCOME

  89. Ammarah
    • 2 years ago
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    no thanks

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