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safia21 Group Title

− Calculate the length of LM in the isosceles right triangle ∆ KLM

  • 2 years ago
  • 2 years ago

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  1. safia21 Group Title
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    • 2 years ago
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  2. heisenberg Group Title
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    You can use the pythagorean theorem since it has a right angle. Are you familiar with it?

    • 2 years ago
  3. safia21 Group Title
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    no im kinda confused

    • 2 years ago
  4. heisenberg Group Title
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    do you know what a hypotenuse is?

    • 2 years ago
  5. safia21 Group Title
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    yes

    • 2 years ago
  6. heisenberg Group Title
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    so the pythagorean theorem is: \[ a^2 + b^2 = c^2 \] where c is the hypotenuse. since this is an isosceles triangle, a = b, don't you think?

    • 2 years ago
  7. safia21 Group Title
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    yes

    • 2 years ago
  8. heisenberg Group Title
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    so can you take it form here? since a = b, \( a^2 + a^2 = c^2 \)

    • 2 years ago
  9. safia21 Group Title
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    what is a and what is like what do you plug in

    • 2 years ago
  10. heisenberg Group Title
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    well let me ask, what is a hypotenuse?

    • 2 years ago
  11. safia21 Group Title
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    the longset side of a right triangle

    • 2 years ago
  12. heisenberg Group Title
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    very good! which is the side *directly* across from the right angle. so the other two sides would be 'a' and 'b', but this triangle is isosceles so a = b. therefore a is an unknown that we want to solve for and 'c' is the length of the hypotenuse

    • 2 years ago
  13. heisenberg Group Title
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    so if we have \(a^2 + a^2 = c^2\) where \( c = 36 \) we only have 1 unknown so we should be able to solve this like a regular algebra problem.

    • 2 years ago
  14. safia21 Group Title
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    ok

    • 2 years ago
  15. heisenberg Group Title
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    are you still confused? all that's left is to simplify this equation and solve.

    • 2 years ago
  16. safia21 Group Title
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    okay i got 36?

    • 2 years ago
  17. heisenberg Group Title
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    show me your steps. i don't that's right.

    • 2 years ago
  18. heisenberg Group Title
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    think*

    • 2 years ago
  19. heisenberg Group Title
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    what is \( a^2 + a^2 \) ?

    • 2 years ago
  20. heisenberg Group Title
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    (@victor, please consider that the best way for a person to learn something is to have the person discover it on their own rather than just giving that person an answer.)

    • 2 years ago
  21. safia21 Group Title
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    yes i need the steps!

    • 2 years ago
  22. heisenberg Group Title
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    there's no reason you can't do it yourself, though. you have an equation. where are you getting stuck? show me your work and i can help you.

    • 2 years ago
  23. safia21 Group Title
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    can we start from the beginning im soo sorry!

    • 2 years ago
  24. heisenberg Group Title
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    the pythagorean theorem states that (for a right triangle): \[ a^2 + b^2 = c^2\]where c is the hypotenuse. we have the hypotenuse in this case, as you pointed out, and it equals 36, agree?

    • 2 years ago
  25. Victor_Hugo Group Title
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    It's an isosceles triangle so it has two equal sides and therefor two equal angles. If you add the inner angles of any triangle the result will be 180º. You already know 1 angle = 90º. 180-90=90 And since the two other angles are equal you have 90/2=45. Now that you have all the angles you can use trigonometry and achieve the value of the sides. I recommend you use "sin" or "tan" (as expressed on a calculator).

    • 2 years ago
  26. safia21 Group Title
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    yes

    • 2 years ago
  27. heisenberg Group Title
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    but this is an isosceles triangle, so the two remaining sides are equal in length, would you agree?

    • 2 years ago
  28. safia21 Group Title
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    yes

    • 2 years ago
  29. Victor_Hugo Group Title
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    Of course, you only need the value of one side.

    • 2 years ago
  30. heisenberg Group Title
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    so since the two remaining sides are \(a\) and \(b\), we know that \(a =b \) therefore we can simplify the equation to this: \( a^2 + a^2 = c^2 \)

    • 2 years ago
  31. safia21 Group Title
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    okay and c2 is 36 right so a2+a2= 36^2

    • 2 years ago
  32. heisenberg Group Title
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    exactly! so what does \( a^2 + a^2\) equal? we can reduce this to one term.

    • 2 years ago
  33. safia21 Group Title
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    a^3

    • 2 years ago
  34. heisenberg Group Title
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    not quite. how about this, what does \( x + x \) equal?

    • 2 years ago
  35. heisenberg Group Title
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    what if i wrote it this way: \( 1x+ 1x \)

    • 2 years ago
  36. safia21 Group Title
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    2x^2?

    • 2 years ago
  37. heisenberg Group Title
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    that wouldn't work. think about if x = 2, 1x + 1x = 1(2) + 1(2) = 4, but 2x^2 = 2(2)^2 = 8 so those two expressions are not equal. when you add terms of the same variable, you just add their "coefficients," the number in front of the variable.

    • 2 years ago
  38. safia21 Group Title
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    o okay

    • 2 years ago
  39. heisenberg Group Title
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    so what is x + x?

    • 2 years ago
  40. safia21 Group Title
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    4

    • 2 years ago
  41. heisenberg Group Title
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    no x is a variable, not a number. x can be *any* number

    • 2 years ago
  42. safia21 Group Title
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    okay x^2

    • 2 years ago
  43. heisenberg Group Title
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    no just add the numbers *in front* of the variable. x = 1x

    • 2 years ago
  44. safia21 Group Title
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    2x

    • 2 years ago
  45. heisenberg Group Title
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    right! so now let's look at the original equation, what is \( a^2 + a^2 \)

    • 2 years ago
  46. heisenberg Group Title
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    keep in mind that \(a^2 \) is the same thing as \(1a^2\)

    • 2 years ago
  47. safia21 Group Title
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    2x^4

    • 2 years ago
  48. heisenberg Group Title
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    there are no x's in this equation. that was just as an aside example. the question here is to simplify \(a^2 + a^2\) you know that \( x + x = 2x\), so use that same *idea* and apply it to the other equation.

    • 2 years ago
  49. safia21 Group Title
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    2a^4

    • 2 years ago
  50. heisenberg Group Title
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    you don't add the exponents, just the coefficients. \( x = 1x = x^1 = 1x^1 \) are all the same same

    • 2 years ago
  51. safia21 Group Title
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    2a^2

    • 2 years ago
  52. safia21 Group Title
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    isnt it a2=B2

    • 2 years ago
  53. heisenberg Group Title
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    very good! so let's look at our original equation: \(a^2 + a^2 = 36^2 \) = \(2a^2 = 36^2\)

    • 2 years ago
  54. heisenberg Group Title
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    yes it is, that's how we eliminated the b^2 and replaced it with another a^2

    • 2 years ago
  55. safia21 Group Title
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    ok

    • 2 years ago
  56. heisenberg Group Title
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    so now our equation is \(2a^2 = 36^2\) do you know how to solve algebra problems? that's all this is.

    • 2 years ago
  57. safia21 Group Title
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    so i multiply 36 times 36 and divide it by 2

    • 2 years ago
  58. heisenberg Group Title
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    absolutely!

    • 2 years ago
  59. heisenberg Group Title
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    that would be represented like so: \(a^2 = \frac{36^2}{2} \) then you just take the square root of both sides to get 'a' = something

    • 2 years ago
  60. heisenberg Group Title
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    a^2 = 36^2 / 2

    • 2 years ago
  61. safia21 Group Title
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    36 right

    • 2 years ago
  62. heisenberg Group Title
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    not quite, remember, it's: 36 * 36 / 2

    • 2 years ago
  63. safia21 Group Title
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    ok

    • 2 years ago
  64. heisenberg Group Title
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    first do 36 * 36, then divide that answer by 2

    • 2 years ago
  65. safia21 Group Title
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    648

    • 2 years ago
  66. heisenberg Group Title
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    very good! so we're left with: \( a^2 = 648\) just take the square root of both sides and you have your answer!

    • 2 years ago
  67. safia21 Group Title
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    i got 26

    • 2 years ago
  68. safia21 Group Title
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    but thats not one of the anwsers

    • 2 years ago
  69. heisenberg Group Title
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    what are the answers?

    • 2 years ago
  70. safia21 Group Title
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    a) 18 b) 18√2 c) 36 d) 36√2

    • 2 years ago
  71. heisenberg Group Title
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    well one of those answers roughly equals the number you got.

    • 2 years ago
  72. safia21 Group Title
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    b

    • 2 years ago
  73. heisenberg Group Title
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    that's right! congrats :)

    • 2 years ago
  74. safia21 Group Title
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    thanks soo much!

    • 2 years ago
  75. heisenberg Group Title
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    no problem :) i'm glad you wanted to learn rather than just want the answer. it'll always work out better that way. trust me ;)

    • 2 years ago
  76. safia21 Group Title
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    haha thanks! :)

    • 2 years ago
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