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mariomintchev

  • one year ago

...never got help so im reposting

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

  3. UnkleRhaukus
    • one year ago
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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)

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

  5. mariomintchev
    • one year ago
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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

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

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

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

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

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

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

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

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

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

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

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

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

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