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anonymous

  • one year ago

Find the perimeter of the following shape: Shape ABCD is shown. Point A is at 7, 5. Point B is at 6, 3. Point C is at 3, 2. Point D is at 4, 4. 10.8 11 11.4 11.6

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  1. anonymous
    • one year ago
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    http://learn.flvs.net/webdav/assessment_images/educator_geometry/v15/module06/0600_g12_q4.jpg

  2. anonymous
    • one year ago
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    @Here_to_Help15

  3. Here_to_Help15
    • one year ago
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    I um failed but @johnweldon1993 it the best user in OS hes the recommendation i was talking about lol

  4. anonymous
    • one year ago
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    okay haha thank you

  5. johnweldon1993
    • one year ago
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    Lol I appreciate that @Here_to_Help15 So sadly the link wont load for me however let me draw a rough sketch |dw:1433883395422:dw|

  6. anonymous
    • one year ago
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    oh crap dude im sorry i coul have screen shotted it for you. sorry u had to draw it out

  7. johnweldon1993
    • one year ago
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    Let me label it with the ABCD lol |dw:1433883546238:dw|

  8. johnweldon1993
    • one year ago
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    Lol no worries...at least there weren't like decimals or crap like that lol

  9. anonymous
    • one year ago
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    haha right!!

  10. johnweldon1993
    • one year ago
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    Okay...so we need the distance between all these points Lets work with just points A and B for now Point A is at (7,5) Point B is at (6,3) How far down and how far over from point A do you have to go to get to point B? You have to go down 2 units...and over to the left 1 unit right?

  11. anonymous
    • one year ago
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    yes

  12. johnweldon1993
    • one year ago
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    Okay...so now |dw:1433883752245:dw| Look familiar?

  13. anonymous
    • one year ago
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    umm kinda yea

  14. johnweldon1993
    • one year ago
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    It's a right triangle Which means we can use the Pythagorean Theorem to solve for that missing length \[\large a^2 + b^2 = c^2\] we need 'c' so \[\large c = \sqrt{a^2 + b^2}\] which here would be \[\large c = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\] That is the distance between A and B...did that process make sense?

  15. anonymous
    • one year ago
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    yes that makes sence

  16. johnweldon1993
    • one year ago
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    Okay good...so we just do that for each of these points... Distance from A to B 2 down, 1 over ...then just \(\large c = \sqrt{2^2 + 1^2} = \sqrt{5}\) Distance from B to C 1 down, 3 over so \(\large c = \sqrt{1^2 + 3^3} = \sqrt{10}\) Distance from C to D 2 up, 1 over so \(\large c = \sqrt{2^2 + 1^2} = \sqrt{5}\) and Distance from D to A 1 up, 3 over so \(\large c = \sqrt{1^2 + 3^2} = \sqrt{10}\) SO! Now that we have all those distances.. |dw:1433884295940:dw| We just add them all together to get the perimeter

  17. johnweldon1993
    • one year ago
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    So \[\large Perimeter = \sqrt{5} + \sqrt{5} + \sqrt{10} + \sqrt{10}\] \[\large Perimeter = 2\sqrt{5} + 2\sqrt{10} = ?\]

  18. anonymous
    • one year ago
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    i got 10.79 so it would be 10.8? SO A??

  19. johnweldon1993
    • one year ago
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    Correct :)

  20. anonymous
    • one year ago
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    yess thank you so much!!!!

  21. johnweldon1993
    • one year ago
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    No problem :)

  22. anonymous
    • one year ago
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    Can u help me with another?

  23. johnweldon1993
    • one year ago
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    Maybe...not very good in math :P lol

  24. anonymous
    • one year ago
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    haha umm what? r u crazy? u r really good at math

  25. johnweldon1993
    • one year ago
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    Lol I try :P

  26. anonymous
    • one year ago
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  27. johnweldon1993
    • one year ago
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    Okay so Arc length can be found using the formula \[\large s = r\theta\]

  28. anonymous
    • one year ago
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    okay

  29. Here_to_Help15
    • one year ago
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    Told ya he was good lol

  30. anonymous
    • one year ago
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    haha i know @Here_to_Help15 thank you alos very very much

  31. johnweldon1993
    • one year ago
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    BUT! 1 thing that may trip you up is we need the angle in radiens to convert the angle to radiens we need to do \(\large \theta \times \frac{\pi}{180}\) so here \(\large 30 \times \frac{\pi}{180} = \frac{30\pi}{180} = \frac{\pi}{6}\) So all we need to do is \[\large s = r\theta = 6ft \times \frac{\pi}{6} = ?\]

  32. Here_to_Help15
    • one year ago
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    Dont thank me thank him ^ lol

  33. Here_to_Help15
    • one year ago
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    and pfft not good at math thats a good one lol

  34. anonymous
    • one year ago
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    hold on i think i did this wrong... im trying to get this into my calculator but its being dumb

  35. anonymous
    • one year ago
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    i got 1.13....

  36. johnweldon1993
    • one year ago
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    Not quite :) Want an easier hint? \[\large \cancel{6}\times \frac{\pi}{\cancel{6}} \]

  37. anonymous
    • one year ago
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    ohhh haha wow i am dumb i got it now thank you.....

  38. johnweldon1993
    • one year ago
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    No you're not...you absolutely can go through the calculator...but why go through all that when there are shortcuts :P

  39. anonymous
    • one year ago
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    haha yes u just set it up in a way that i understand the second time!! thank you soooooo freaking much u have no idea how much of a help u are!!!!

  40. Here_to_Help15
    • one year ago
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    \(\Huge\tt\color{#f9bec7}{He's~Right~ lol }\)

  41. johnweldon1993
    • one year ago
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    Well that's what I'm here for :) if you need anything else feel free to message me or tag me in a question :)

  42. anonymous
    • one year ago
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  43. anonymous
    • one year ago
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    this is the last one i need help with!!!

  44. Here_to_Help15
    • one year ago
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    \(\Huge\tt\color{#3c0808}{Can~i ~Borrow~you~now~Lol !}\)

  45. anonymous
    • one year ago
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    @johnweldon1993

  46. anonymous
    • one year ago
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    u can help @Here_to_Help15 fist though

  47. Here_to_Help15
    • one year ago
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    \(\Huge\tt\color{#f9bec7}{Ladies~ First :) }\)

  48. johnweldon1993
    • one year ago
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    Oh geez another one!? ;D haha just kidding Okay... |dw:1433885518952:dw|

  49. johnweldon1993
    • one year ago
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    With a right triangle...what do we know about sin and cos?

  50. anonymous
    • one year ago
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    haha im sorry last one i promis :p

  51. Here_to_Help15
    • one year ago
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    Can i join this post i am also learning this :)

  52. johnweldon1993
    • one year ago
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    Lol you already are part of the post :P but just work along and then we can work with yours :D

  53. Here_to_Help15
    • one year ago
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    Sine Function: sin(θ) = Opposite / Hypotenuse Cosine Function: cos(θ) = Adjacent / Hypotenuse Tangent Function: tan(θ) = Opposite / Adjacent

  54. anonymous
    • one year ago
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    sin = opposite over hypotenuse cos= adjacent over hypotenuse

  55. Here_to_Help15
    • one year ago
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    I might as well throw tangent in the mix as well lol

  56. johnweldon1993
    • one year ago
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    Perfect @CayleeS23 Okay...so in your question...it asks for sin(x) So we look at your triangle...go to the 'x' degree....and look what is opposite that angle...and look at what the hypotenuse is so |dw:1433885706708:dw| So that would mean \[\large sin(x) = \frac{opposite}{hypotenuse} = \frac{3}{5}\]

  57. johnweldon1993
    • one year ago
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    clearly I cannot spell -_- lol

  58. Here_to_Help15
    • one year ago
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    haha my teacher just literally did a livelesson on this at 12 pm

  59. Here_to_Help15
    • one year ago
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    So then that means you need a calculator

  60. johnweldon1993
    • one year ago
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    No not at all (here anyways) this question only wants the ratios....which that is :)

  61. anonymous
    • one year ago
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    wait wouldnt it be A?

  62. Here_to_Help15
    • one year ago
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    Oh lol i thought it was going more further

  63. johnweldon1993
    • one year ago
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    So that is the process to find the sin(x) @CayleeS23 now you show me how to find cos(y) *or just answer it correctly..pfft okay >.< XD Yes A is correct :)

  64. Here_to_Help15
    • one year ago
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    lol

  65. anonymous
    • one year ago
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    haha sorry!! i got it once u posted how to find sin i just did the same with cos using its rule then looked back at the answers :p i would draw it out like u but i have no idea how

  66. johnweldon1993
    • one year ago
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    Yeah it takes some pretty advanced drawing skills ;) XD But either way awesome job! See you didnt even need my help :P

  67. Here_to_Help15
    • one year ago
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    \(\Huge\tt\color{#82ffd6}{Oh~I~See!}\)

  68. Here_to_Help15
    • one year ago
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    sorry guys i love doing latex lol

  69. johnweldon1993
    • one year ago
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    \[\Huge\tt\color{#82ffd6}{OIC!}\]

  70. anonymous
    • one year ago
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    haha yes i did need your help :p your drawing skills uff ya know they pef hahahaha

  71. Here_to_Help15
    • one year ago
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    \(\large\tt\color{#99d3f0}{LIKE~IK~RIGHT! }\)

  72. johnweldon1993
    • one year ago
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    XD lol mmhmm we're gonna be good friends @CayleeS23 :P

  73. Here_to_Help15
    • one year ago
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    \(\Huge\tt\color{#f9bec7}{Hahaha!}\)

  74. anonymous
    • one year ago
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    haha i thin soo ;)

  75. anonymous
    • one year ago
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    think* damn i cant type either

  76. johnweldon1993
    • one year ago
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    As long as you dont.. <.< >.> Stalk me :O XD no jk that's invited too XDDD

  77. anonymous
    • one year ago
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    hahahha oh my!! imma dm you

  78. Here_to_Help15
    • one year ago
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    \(\Huge\tt\color{#f9bec7}{Im~so~lonely~got~nobody~*cries* }\)

  79. anonymous
    • one year ago
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    whattt nooo u got me @Here_to_Help15 your chill man!

  80. Here_to_Help15
    • one year ago
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    \(\Huge\tt\color{#f9bec7}{*Heart~Breaks </3* !}\)

  81. anonymous
    • one year ago
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    brb i have like 2 questions then i can turn this in and then i can chat with yall!!

  82. johnweldon1993
    • one year ago
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    \[\bbox [10pt, black, border:5pt solid blue]{\Huge\cal\color{red}{\bigstar}\color{orange}{I}\color{yellow}{T}\space \color{green}{W}\color{blue}{I}\color{indigo}{L}\color{violet}{L}\space\color {red}{B}\color{orange}{E}\space \color{yellow}{O}\color{green}{K}\color{blue}{A}\color{indigo}{Y}\space \color{violet}{H}\color{red}{A}\color{orange}{H}\color{yellow}{A}\color{red}{\bigstar}}\]

  83. Here_to_Help15
    • one year ago
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    OMG LOOK WHAT SOMEONE SENT ME please help me our else i will go to the open study board and say your name that you didn't help.So please help me in english please right now.

  84. Here_to_Help15
    • one year ago
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    How do you that one lol @johnweldon1993

  85. johnweldon1993
    • one year ago
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    Secrets :P

  86. Here_to_Help15
    • one year ago
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    Argh i tend to discover them lol

  87. Here_to_Help15
    • one year ago
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    Out of the blue question but john do you play clash of clans ?

  88. anonymous
    • one year ago
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    omg poeple on here r crazy af.. report that person

  89. Here_to_Help15
    • one year ago
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    lol i told him "is that a virtual threat because im loving it :)"

  90. anonymous
    • one year ago
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    hahha yess thats great

  91. johnweldon1993
    • one year ago
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    Lol and no I dont play clash of clans @Here_to_Help15

  92. Here_to_Help15
    • one year ago
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    John you happen to speak french?

  93. johnweldon1993
    • one year ago
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    I took 3 years of French in High school but other than introducing myself I would die quickly in France XD

  94. Here_to_Help15
    • one year ago
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    LOL

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