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anonymous

  • one year ago

The table shows the proof of the relationship between the slopes of two perpendicular lines. What is the missing statement in step 2?

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  1. anonymous
    • one year ago
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    1. is similar to . given 2. property of similar triangles 3. = property of proportion 4. slope of = slope of = definition of slope 5. slope of × slope of = multiplying the slopes 6. (This statement is intentionally left blank.) Substitution Property of Equality 7. slope of × slope of = -1 simplifying the right side

  2. anonymous
    • one year ago
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    couldnt get all of it but tried my best

  3. jim_thompson5910
    • one year ago
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    a screenshot is best because we can see the full problem (even the missing symbols)

  4. anonymous
    • one year ago
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    1. AC←→⊥CD←→ (view diagram) ΔABC is similar to ΔCED. given 2. (This statement is intentionally left blank.) property of similar triangles 3. ABBC = ECDE property of proportion 4. slope of AC←→ = -ABBC slope of DC←→ = DEAC definition of slope 5. slope of AC←→ × slope of DC←→ = -ABBC×DEEC multiplying the slopes 6. Substitution Property of Equality 7. slope of AC←→ × slope of DC←→ = -1 simplifying the right side

  5. anonymous
    • one year ago
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    sorry here it is

  6. jim_thompson5910
    • one year ago
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    well step 3 has AB*BC = EC*DE, or so it seems the step before that would be some proportion and cross multiplying would lead to step 3

  7. jim_thompson5910
    • one year ago
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    does that make sense?

  8. anonymous
    • one year ago
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    not really but u can explain it in detail. mabye ill get it

  9. jim_thompson5910
    • one year ago
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    let's say you had a/b = c/d

  10. jim_thompson5910
    • one year ago
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    if you cross multiplied, you'd get a*d = b*c agreed?

  11. anonymous
    • one year ago
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    yea

  12. jim_thompson5910
    • one year ago
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    so what proportion could be set up to lead to AB*BC = EC*DE ?

  13. anonymous
    • one year ago
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    they seem to look like this but in fraction AB/BC;. something like that.

  14. anonymous
    • one year ago
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    AB/DE, and EC/DE

  15. jim_thompson5910
    • one year ago
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    more like AB/EC = DE/BC. There are other possibilities |dw:1436929680423:dw|

  16. jim_thompson5910
    • one year ago
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    |dw:1436929728958:dw|

  17. anonymous
    • one year ago
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    oops sorry. thats not on my answer choices tho

  18. anonymous
    • one year ago
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    the closest choice has a fraction DE/EC. would that work?

  19. jim_thompson5910
    • one year ago
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    well as long as they cross multiply to AB*BC = EC*DE, then it will work

  20. jim_thompson5910
    • one year ago
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    there are more possibilities than what I drew up there (maybe 4 possibilities?)

  21. anonymous
    • one year ago
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    ok hold on

  22. anonymous
    • one year ago
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    AB/EC=BC/DE AB/BC=DE/EC AB=DE BC=EC AB/BC= -EC/DE

  23. anonymous
    • one year ago
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  24. jim_thompson5910
    • one year ago
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    choice C looks funky...there are no fractions there?

  25. jim_thompson5910
    • one year ago
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    try posting a screenshot if you can

  26. anonymous
    • one year ago
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    there are no fractions in c, and sorry i cant post a screenshot.

  27. jim_thompson5910
    • one year ago
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    ok let me think

  28. jim_thompson5910
    • one year ago
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    in line 3, where it says `3. ABBC = ECDE property of proportion` there are no fractions here either?

  29. anonymous
    • one year ago
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    actually yea those are fractions

  30. anonymous
    • one year ago
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    its really hard to paste those kinds of things on here.

  31. jim_thompson5910
    • one year ago
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    so line 3 actually says this right? AB/BC = EC/DE

  32. anonymous
    • one year ago
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    yes yes

  33. jim_thompson5910
    • one year ago
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    |dw:1436930874242:dw|

  34. jim_thompson5910
    • one year ago
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    you can multiply both sides by BC and divide both sides by EC that will make BC and EC swap |dw:1436930925056:dw|

  35. jim_thompson5910
    • one year ago
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    giving |dw:1436930949685:dw|

  36. anonymous
    • one year ago
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    omg thanks so much!

  37. anonymous
    • one year ago
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    can you help me with one more? its a mastery, and i didnt really understand this lesson

  38. jim_thompson5910
    • one year ago
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    sure

  39. anonymous
    • one year ago
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    AB and BC form a right angle at point B. If A = (-3, -1) and B = (4, 4), what is the equation of BC

  40. jim_thompson5910
    • one year ago
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    first find the equation of the line through A and B

  41. anonymous
    • one year ago
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    slope?

  42. anonymous
    • one year ago
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    5/7 is the slope

  43. jim_thompson5910
    • one year ago
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    use that slope to find the equation of the line through those two points

  44. jim_thompson5910
    • one year ago
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    you can use y = mx+b

  45. anonymous
    • one year ago
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    i think its y=5/7x+8/7

  46. jim_thompson5910
    • one year ago
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    yep

  47. jim_thompson5910
    • one year ago
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    |dw:1436931657035:dw|

  48. jim_thompson5910
    • one year ago
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    AB and BC form a right angle, so |dw:1436931712083:dw|

  49. jim_thompson5910
    • one year ago
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    what is the slope of BC ?

  50. anonymous
    • one year ago
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    x + 3y = 16 2x + y = 12 -7x − 5y = -48 7x − 5y = 48

  51. anonymous
    • one year ago
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    these are the answer choices btw. and the slope for BC is -7/5 right?

  52. jim_thompson5910
    • one year ago
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    what is the equation of the line going through B and has slope of -7/5

  53. jim_thompson5910
    • one year ago
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    use y = mx+b

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