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mariomintchev Group Title

...never got help so im reposting

  • one year ago
  • one year ago

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  1. mariomintchev Group Title
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    • one year ago
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  2. UnkleRhaukus Group Title
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    |dw:1358469689716:dw|

    • one year ago
  3. UnkleRhaukus Group Title
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    You have to work out weather path ABC or ADC is shorter. path ABC = AB + BC path ADC= AD + DC you can use Pythagorus's theorem to calculate AB and (DC)

    • one year ago
  4. mariomintchev Group Title
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    yeah i know im getting sqrt(40) for A to B and 6 for B to C

    • one year ago
  5. mariomintchev Group Title
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    B is the correct answer but i can't get the exact distance from A to B and then from B to C

    • one year ago
  6. mariomintchev Group Title
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    @UnkleRhaukus you there?

    • one year ago
  7. UnkleRhaukus Group Title
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    sorry i got distracted, you've got AB right and you've got BC right too

    • one year ago
  8. mariomintchev Group Title
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    how would i put the answer in a website? i think thats what im doing wrong.

    • one year ago
  9. mariomintchev Group Title
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    @satellite73 can you assist me here too please

    • one year ago
  10. satellite73 Group Title
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    you want the total distance right? do you know the shortest rout?

    • one year ago
  11. mariomintchev Group Title
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    yes / B is the shortest

    • one year ago
  12. satellite73 Group Title
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    from A to B is \(\sqrt{2^2+6^2}=\sqrt{40}\) and from B to C is 6

    • one year ago
  13. mariomintchev Group Title
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    ok. how would i put that in on a hw website?

    • one year ago
  14. satellite73 Group Title
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    so that distance total is\[6+\sqrt{40}\]

    • one year ago
  15. satellite73 Group Title
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    i have no idea maybe it wants a decimal

    • one year ago
  16. satellite73 Group Title
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    https://www.wolframalpha.com/input/?i=6%2Bsqrt%2840%29

    • one year ago
  17. UnkleRhaukus Group Title
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    \[\sqrt{40}=\sqrt{4\times10}=\sqrt{2^2\times2\times5}=2\sqrt{2}\sqrt{5}\]

    • one year ago
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