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Mathematics 18 Online
OpenStudy (anonymous):

...never got help so im reposting

OpenStudy (anonymous):

OpenStudy (unklerhaukus):

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OpenStudy (unklerhaukus):

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)

OpenStudy (anonymous):

yeah i know im getting sqrt(40) for A to B and 6 for B to C

OpenStudy (anonymous):

B is the correct answer but i can't get the exact distance from A to B and then from B to C

OpenStudy (anonymous):

@UnkleRhaukus you there?

OpenStudy (unklerhaukus):

sorry i got distracted, you've got AB right and you've got BC right too

OpenStudy (anonymous):

how would i put the answer in a website? i think thats what im doing wrong.

OpenStudy (anonymous):

@satellite73 can you assist me here too please

OpenStudy (anonymous):

you want the total distance right? do you know the shortest rout?

OpenStudy (anonymous):

yes / B is the shortest

OpenStudy (anonymous):

from A to B is \(\sqrt{2^2+6^2}=\sqrt{40}\) and from B to C is 6

OpenStudy (anonymous):

ok. how would i put that in on a hw website?

OpenStudy (anonymous):

so that distance total is\[6+\sqrt{40}\]

OpenStudy (anonymous):

i have no idea maybe it wants a decimal

OpenStudy (unklerhaukus):

\[\sqrt{40}=\sqrt{4\times10}=\sqrt{2^2\times2\times5}=2\sqrt{2}\sqrt{5}\]

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