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anonymous
 4 years ago
Let me draw it.
Jim has put a fence along the side AC of the triangular patch of land shown below.
anonymous
 4 years ago
Let me draw it. Jim has put a fence along the side AC of the triangular patch of land shown below.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342705835156:dw To solve this do I do tan(x) = 8 ÷ 15? and if not what should I do?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what are we trying to find here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The length of the fence (AC)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0use pythogorous theorem no?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh just 8^2 + 15^2 = 64 + 225 = 289 or √17

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sont you mean 64 + 225 = 289 or 17^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh yeah sory I meant 17^2 Is that no bueno?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's not the length of AC though

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[AC = \sqrt{x^2 + y^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know what x and y are, though? Unless 8 and 15 are x and y?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry that was the distance formula

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha no worries. I'm just glad you're helping :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[AC = \sqrt{AB^2 + BC^2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah! Okay! So, 225+64 = √289? so the length of AC is 17?
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