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HatSimulator

  • 2 years ago

A rectangular pasture has a fence around the perimeter. The length of the fence is 16x7 and the width is 48x4. What is the area of the pasture? (1 point) a) 3x^3 b)128x^11 c)768x^11 d)768x^28 I KNOW the answer is c, but i need it explained. I already found it here once but can ANYONE explain how to do this?

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  1. hartnn
    • 2 years ago
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    Area of rectangular (here pasture) = Length * Width

  2. HatSimulator
    • 2 years ago
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    Allright, so how do i do it? 16 * ^7 gives nothing and im so confused How do i do 16x^7 * 48x^4? I dont get this problem

  3. HatSimulator
    • 2 years ago
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    Same thing with this > Simplify. 5^–1(3^–2) If i type this into a calculator i get a stupid number

  4. hartnn
    • 2 years ago
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    ok, treat constants and variables separately. \(16x^7 \times 48x^4 = (16\times 48) \times(x^7 \times x^4)\) got this ?

  5. HatSimulator
    • 2 years ago
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    Okay... yes i think i get that... but where did the two Z come from

  6. hartnn
    • 2 years ago
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    Z ?? there are no Z's there....

  7. HatSimulator
    • 2 years ago
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    Whoops, sorry, okay, go on. I understand now

  8. HatSimulator
    • 2 years ago
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    I think i get it 16 * 48 = 768 And 7 + 4 = 11?

  9. hartnn
    • 2 years ago
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    good!

  10. HatSimulator
    • 2 years ago
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    Okay, 1 more?

  11. hartnn
    • 2 years ago
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    actually, its \(x^7 \times x^4 = x^{7+4}=x^{11}\) and sure :)

  12. HatSimulator
    • 2 years ago
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    Okay something like this Simplify (4xy2)3(xy)5 and (sorry) this Evaluate a–4b2 for a = –2 and b = 4

  13. HatSimulator
    • 2 years ago
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    Simplify (4xy^2)^3(xy)^5 Sorry

  14. hartnn
    • 2 years ago
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    i believe the first one looks like this : \((4xy^2)^3(xy)^5\)

  15. hartnn
    • 2 years ago
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    ok, so you need to remember/understand few thing before u start: \(\huge (ab)^n = a^nb^n\) and \(\huge (a^m)^n=a^{mn}\)

  16. HatSimulator
    • 2 years ago
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    okay

  17. HatSimulator
    • 2 years ago
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    The other problem i can't figure out how to write an equation

  18. hartnn
    • 2 years ago
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    \(\large Evaluate \: \: a^{–4}b^2 \:\: for\:\: a = –2 \:\:and\:\: b = 4\) right ?

  19. HatSimulator
    • 2 years ago
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    Here you go

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  20. hartnn
    • 2 years ago
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    ok, for the previous one : using, \(\huge (ab)^n = a^nb^n\) \(\large (4xy^2)^3= 4^3 x^3 (y^2)^3\) got this ?

  21. HatSimulator
    • 2 years ago
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    Oh man. So confused

  22. HatSimulator
    • 2 years ago
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    Are you doing Simplify (4xy^2)^3(xy)^5 ?

  23. hartnn
    • 2 years ago
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    yes.

  24. HatSimulator
    • 2 years ago
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    And you changed that to (4xy^2)^3=4^3x^3(y^2)^3....

  25. HatSimulator
    • 2 years ago
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    How did you do that... that hurts my brain

  26. hartnn
    • 2 years ago
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    |dw:1360891677558:dw|

  27. hartnn
    • 2 years ago
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    open up your brain and try to accept new things :)

  28. hartnn
    • 2 years ago
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    basically the exponent outside the bracket becomes the exponent of each of the terms inside the bracket.

  29. HatSimulator
    • 2 years ago
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    Where did averything after that equal sign come from? Where'd the 5 go?

  30. HatSimulator
    • 2 years ago
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    Howd 4 and the 4 more exponents come from

  31. hartnn
    • 2 years ago
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    (4xy^2)^3(xy)^5 <------you are asking about this 5 ? i am starting with (4xy^2)^3 part of (4xy^2)^3(xy)^5

  32. hartnn
    • 2 years ago
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    if you are asking from where does all the exponents come from take a look at this again : \((ab)^n=a^nb^n\) or perhaps : \((abc)^n=a^nb^nc^n\)

  33. HatSimulator
    • 2 years ago
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    OHHHHHHHHHHHHHHHHHH

  34. HatSimulator
    • 2 years ago
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    Yes, yes, go on (:

  35. hartnn
    • 2 years ago
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    so i assume you are clear with this diagram.|dw:1360892207194:dw| using same rule, can you tell me what u get for \((xy)^5\) ??

  36. HatSimulator
    • 2 years ago
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    |dw:1360892393038:dw|

  37. hartnn
    • 2 years ago
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    correct :) see, it isn't difficult at all. now you have \(\large 4^3x^3(y^2)^3x^5y^5\) right ? lets start simplifying with constants (numbers) 4^3 =... ?

  38. HatSimulator
    • 2 years ago
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    64

  39. hartnn
    • 2 years ago
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    yes. now comes the 2nd rule i posted. \(\huge (a^m)^n=a^{mn}\) so, what about \((y^2)^3\) ??

  40. HatSimulator
    • 2 years ago
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    |dw:1360892627948:dw|

  41. hartnn
    • 2 years ago
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    12? how ?

  42. HatSimulator
    • 2 years ago
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    Wait

  43. HatSimulator
    • 2 years ago
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    |dw:1360892663950:dw|

  44. HatSimulator
    • 2 years ago
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    y^1*2

  45. hartnn
    • 2 years ago
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    umm..no, let me give you an example \(\large (z^5)^3 = z^{5 \times 3 }=z^{15}\) do similar thing for \((y^2)^3 \)

  46. HatSimulator
    • 2 years ago
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    (y^2)

  47. hartnn
    • 2 years ago
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    :O the exponents are getting multiplied. what are the 2 exponents in \((y^2)^3\)

  48. HatSimulator
    • 2 years ago
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    Its a 3 isnt it... Sorry (y^6)

  49. hartnn
    • 2 years ago
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    yes, y^6 is correct. so, we have now \(64x^3y^6x^5y^5\) lets bring 'x' terms and y terms together.

  50. hartnn
    • 2 years ago
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    \(64 \: \: (x^3x^5) \:\: (y^6y^5)\) ok, any doubts ?

  51. HatSimulator
    • 2 years ago
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    No, i got it so far.

  52. hartnn
    • 2 years ago
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    now one of the very important rule : \(\large a^m a^n = a^{m+n}\) here, if we multiply the variables, their exponents gets ADDED . so, what about \(x^3x^5=... ?\)

  53. HatSimulator
    • 2 years ago
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    It would end up being 64(x^8)(y^11) ?

  54. HatSimulator
    • 2 years ago
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    Thats the answer

  55. hartnn
    • 2 years ago
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    wow! thats absolutely correct! :)

  56. hartnn
    • 2 years ago
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    did you get how ?

  57. HatSimulator
    • 2 years ago
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    Yep!

  58. HatSimulator
    • 2 years ago
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    I get it now

  59. HatSimulator
    • 2 years ago
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    Thank you very much!

  60. hartnn
    • 2 years ago
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    welcome ^_^

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