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UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1358469689716:dw

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0You 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)

mariomintchev
 2 years ago
Best ResponseYou've already chosen the best response.0yeah i know im getting sqrt(40) for A to B and 6 for B to C

mariomintchev
 2 years ago
Best ResponseYou've already chosen the best response.0B is the correct answer but i can't get the exact distance from A to B and then from B to C

mariomintchev
 2 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus you there?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0sorry i got distracted, you've got AB right and you've got BC right too

mariomintchev
 2 years ago
Best ResponseYou've already chosen the best response.0how would i put the answer in a website? i think thats what im doing wrong.

mariomintchev
 2 years ago
Best ResponseYou've already chosen the best response.0@satellite73 can you assist me here too please

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1you want the total distance right? do you know the shortest rout?

mariomintchev
 2 years ago
Best ResponseYou've already chosen the best response.0yes / B is the shortest

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1from A to B is \(\sqrt{2^2+6^2}=\sqrt{40}\) and from B to C is 6

mariomintchev
 2 years ago
Best ResponseYou've already chosen the best response.0ok. how would i put that in on a hw website?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1so that distance total is\[6+\sqrt{40}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i have no idea maybe it wants a decimal

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{40}=\sqrt{4\times10}=\sqrt{2^2\times2\times5}=2\sqrt{2}\sqrt{5}\]
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