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hal_stirrup

in a graph where a liner equation intercept another liner equation. why is the intersection make the two equation equal.

  • one year ago
  • one year ago

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  1. hal_stirrup
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    |dw:1346965908373:dw|

    • one year ago
  2. ifrah34
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    because the two angles reflect eachother

    • one year ago
  3. asnaseer
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    a linear equation can be expressed in the form:\[y=mx+c\]which you have correctly drawn as a straight line graph

    • one year ago
  4. hal_stirrup
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    explain more please . thanks

    • one year ago
  5. asnaseer
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    now think of two linear equations - these would represent, say:\[y=m_1x+c_1\]and:\[y=m_2x+c_2\]

    • one year ago
  6. hal_stirrup
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    ok

    • one year ago
  7. asnaseer
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    now, if these two lines intersect, then, at the point of intersection, the y-values of both must be equal - agreed?

    • one year ago
  8. hal_stirrup
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    agreed .

    • one year ago
  9. asnaseer
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    therefore, we can say:\[m_1x+c_1=m_2x+c_2\]at the point of intersection

    • one year ago
  10. asnaseer
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    which is the same as saying:\[y=y\]

    • one year ago
  11. hal_stirrup
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    because y=y thats grate . amazing . tell if we have quadratic equations |dw:1346967366273:dw|

    • one year ago
  12. asnaseer
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    same applies for quads - except this time you will get two solutions for x instead of one that you got for the linear case

    • one year ago
  13. hal_stirrup
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    |dw:1346967476493:dw|

    • one year ago
  14. asnaseer
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    correct

    • one year ago
  15. hal_stirrup
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    lets say

    • one year ago
  16. asnaseer
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    and each of these, in theory could give you two different y values

    • one year ago
  17. hal_stirrup
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    y1=x^2 +mx + c and y2=x^2 +mx + c

    • one year ago
  18. asnaseer
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    quads are usually written as:\[y=a_1x^2+b_1x+c_1\]and:\[y=a_2x^2+b_2x+c_2\]

    • one year ago
  19. hal_stirrup
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    ok

    • one year ago
  20. asnaseer
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    so at the points of intersection we will have:\[a_1x^2+b_1x+c_1=a_2x^2+b_2x+c_2\]

    • one year ago
  21. asnaseer
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    which leads to a quadratic equation in x - solve to get the two values for x where these curves intersect

    • one year ago
  22. asnaseer
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    in general, we could have two curves defined by:\[y=f(x)\]and:\[y=g(x)\] |dw:1346967756823:dw|

    • one year ago
  23. asnaseer
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    |dw:1346967787868:dw|

    • one year ago
  24. hal_stirrup
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    ya

    • one year ago
  25. asnaseer
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    and, at each point of intersection, the y and x values on each curve are equal

    • one year ago
  26. asnaseer
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    I hope I have explained it well enough for you - let me know if you require any more explanation

    • one year ago
  27. hal_stirrup
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    have outdone yourself

    • one year ago
  28. hal_stirrup
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    many thanks.

    • one year ago
  29. asnaseer
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    thx - and you are welcome my friend! :)

    • one year ago
  30. hal_stirrup
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    what is the highest think your done in maths??

    • one year ago
  31. asnaseer
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    think == thing?

    • one year ago
  32. hal_stirrup
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    ya sorry

    • one year ago
  33. asnaseer
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    :) I have studied for a PhD in Aeronautical Engineering. but that was many years ago. I now teach maths as a hobby because I still enjoy it.

    • one year ago
  34. hal_stirrup
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    WOW, thats great. can send you some of my question from time to time?

    • one year ago
  35. asnaseer
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    I'm not sure what you mean by "send", but if you mean message me now and then if no one else is helping you - then yes, by all means do - I'll be glad to help out when I can.

    • one year ago
  36. hal_stirrup
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    by that i meant , the message on the top left corner. thanks

    • one year ago
  37. asnaseer
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    ok - thats fine.

    • one year ago
  38. hal_stirrup
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    am going to do a lot of calculus. nice one

    • one year ago
  39. asnaseer
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    but bear in mind that I may not always respond immediately because I also moderate this site so please don't think I'm ignoring you in those cases. :)

    • one year ago
  40. hal_stirrup
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    thanks. I will be patient . :)

    • one year ago
  41. asnaseer
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    gr8! speak to you later then...

    • one year ago
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