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Can anyone help me please? In a right triangle: The altitude=y The base=y The hypotenuse=3 Find the value of y.
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Use Pythagorean theorem. y^2+y^2=3^2
So what is the answer? @allank \[\sqrt{y^4}= \sqrt{9}\]
sqrt(2y^2)=3
y^2+y^2=3^2 that becomes 2(y^2)=9 then y^2=(9/2) thus y=sqrt(9/2)
@timo86m I don't see that answer on my hw... a. y=2sqrt3/4 b.y=3sqrt2/4 c.y=sqrt6/2 d.y=3sqrt2/2
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it is not completed lol just a hint
y = (3/2)*sqrt(2)
d
@timo86m SO my answer should be d?
yeah but is it over 2 or 4?@timo86m
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y = (3/2)*sqrt(2)
\[y=3\sqrt{2}/2\]
it d :)
|dw:1342815698822:dw|
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