Recent content by david3305

  1. david3305

    Finding x in Parallel Lines Geometry Problem

    Lol. I felt doubt about this because most of the problems i have solved had a correct answer in the choices.
  2. david3305

    Finding x in Parallel Lines Geometry Problem

    Yeah, that's what I think too. I'll leave this open just in case anybody has anything to add. Maybe another way of solving?
  3. david3305

    Finding x in Parallel Lines Geometry Problem

    Homework Statement Find x. L1 and L2 are parallel choices: a)2°40' b)2°30' c)2°45' d)2°15' e)2°20' Homework Equations Σleft∠ = Σright∠ The Attempt at a Solution 2x+1+4x-1+4x-1 = 2x-1+2x+4+3x+2 10x-1=7x+5 3x=6 x=2° Not sure how the answers include minutes ' Maybe I'm overlooking something?
  4. david3305

    How Can You Solve for x When L1 and L2 are Parallel?

    yeah tbh I was kinda lazy to think about it, just following jambaugh's way I can find the following is true: (θ) + (90) + (180 - x) + (90 - 4θ) = 180 -3θ + 360 - x = 180 180 - 3θ = x By the same reasoning we already know: θ + 2θ + 3θ + 3θ = 180 9θ = 180 θ = 20 so: 180 - 3(20) = x x = 120
  5. david3305

    How Can You Solve for x When L1 and L2 are Parallel?

    Nice. I didn't look it that way, 'follow the turns', cool. Thank you for your reply. I'm interested to see another way of solving for X that doesn't involve using the polygon like I did. And yes, the drawing is not to scale at all.
  6. david3305

    How Can You Solve for x When L1 and L2 are Parallel?

    Homework Statement Find x. L1 and L2 are parallel Choices: a)100 b)120 c)140 d)150 e)135 Homework Equations From the image, the angles of the polygon in blue should satisfy: 6θ + 90 + 4θ + 2θ + 90 + x = 540 12θ + x = 360 x = 360 - 12θ The Attempt at a Solution I couldn't figure out how to...
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