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erica.d

if jane walks 15m straight and then she turns left and walks for 5 m after that she walks another 3 m towards right then how far is she from the point of start? i am finding it tough to picturize this ..

  • one year ago
  • one year ago

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  1. erica.d
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    @across

    • one year ago
  2. erica.d
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    @cwrw238

    • one year ago
  3. across
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    You have to use the Pythagorean theorem.

    • one year ago
  4. erica.d
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    i know that but how to plot this or draw this

    • one year ago
  5. tyteen4a03
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    The diagram you're looking for is this:|dw:1348068681247:dw|

    • one year ago
  6. tyteen4a03
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    Excuse my drawing, but you get the idea.

    • one year ago
  7. erica.d
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    @tyteen4a03 thanks...now what?

    • one year ago
  8. tyteen4a03
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    Putting this in the coordinate plane, assuming that the starting point is the origin, the final point would be (-5, 18). Now plug these numbers into the distance formula \(\sqrt{(x2-x1)^2 + (y2-y1)^2}\) and you get the answer.

    • one year ago
  9. erica.d
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    @across if i'll join the two end points ..still i am unable to find the distance ...

    • one year ago
  10. across
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    If you join the two ends, you will end up with a right triangle that has a base of length 5 and a height of length 15+3.

    • one year ago
  11. erica.d
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    |dw:1348069046132:dw| ?? @across

    • one year ago
  12. across
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    |dw:1348069089040:dw|

    • one year ago
  13. erica.d
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    that's amazing....@across you are genius :) thanks a lot

    • one year ago
  14. erica.d
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    |dw:1348069219200:dw| i was doing it this way...and i guess there is no need to plot it on graph and point out coordinates ..... @across thanks alot again

    • one year ago
  15. erica.d
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    i wish i could give you all the medals lol

    • one year ago
  16. tyteen4a03
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    @erica.d One thing about mathematics is that there are many solution to the same problem. The Distance Formula, for example, is useful when you only know the starting and the ending point (imagine your question, but you're only given 2 points). In this case, using Pythagoras's theorem is like reinventing the wheel.

    • one year ago
  17. erica.d
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    i know that but what @across has done was not imagined by me ...by the way thanks

    • one year ago
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