Mathematics
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OpenStudy (anonymous):
Given the triangle below, what is sec ∡B?
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OpenStudy (anonymous):
OpenStudy (anonymous):
i'llk type it in the calculater
OpenStudy (lgbasallote):
so you need to solve for adjacent side first
\[AB = \sqrt{7^2 - 4^2}\]
OpenStudy (lgbasallote):
ugh i typed it wrong again =_=
sec = hypotenuse/adjacent
OpenStudy (anonymous):
wait so what do i put in the calculater?
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OpenStudy (lgbasallote):
\[\sqrt{7^2 - 4^2}\]
OpenStudy (lgbasallote):
soleve that first then tell me the answer
OpenStudy (lgbasallote):
note: retain the sqrt symbol
OpenStudy (anonymous):
i got 5.74456264653803
OpenStudy (lgbasallote):
i said retain the sqrt symbol so just solve 7^2 - 4^2
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
33
OpenStudy (lgbasallote):
so the adjacent side is \(\sqrt{33}\) now do hypotenuse/adjacent
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
isn't it the same 7^-4^ ?
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OpenStudy (lgbasallote):
no
OpenStudy (lgbasallote):
hypotenuse over adjacent is \[\frac{7}{\sqrt{33}}\]
OpenStudy (lgbasallote):
7 is hypotenuse
\(\sqrt{33}\) is adjacent
OpenStudy (anonymous):
do i put everything over 33?
OpenStudy (lgbasallote):
no you rationalize
\[\frac{7}{\sqrt{33}} \times \frac{\sqrt{33}}{\sqrt{33}}\]
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OpenStudy (anonymous):
ok so its going to be|dw:1345362932506:dw|