anonymous
  • anonymous
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 ..
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@across
anonymous
  • anonymous
@cwrw238
across
  • across
You have to use the Pythagorean theorem.

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anonymous
  • anonymous
i know that but how to plot this or draw this
tyteen4a03
  • tyteen4a03
The diagram you're looking for is this:|dw:1348068681247:dw|
tyteen4a03
  • tyteen4a03
Excuse my drawing, but you get the idea.
anonymous
  • anonymous
@tyteen4a03 thanks...now what?
tyteen4a03
  • tyteen4a03
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.
anonymous
  • anonymous
@across if i'll join the two end points ..still i am unable to find the distance ...
across
  • across
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.
anonymous
  • anonymous
|dw:1348069046132:dw| ?? @across
across
  • across
|dw:1348069089040:dw|
anonymous
  • anonymous
that's amazing....@across you are genius :) thanks a lot
anonymous
  • anonymous
|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
anonymous
  • anonymous
i wish i could give you all the medals lol
tyteen4a03
  • tyteen4a03
@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.
anonymous
  • anonymous
i know that but what @across has done was not imagined by me ...by the way thanks

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