Check my answer? Almost done! c:

http://i59.tinypic.com/8x6lhf_th.png

Question

Check my answer? Almost done! c:

http://i59.tinypic.com/8x6lhf_th.png

anonymous · Answer

@hartnn

anonymous · Answer

what school and year
i have the answers for like full lessons

hartnn · Answer

again wrong answer choices

anonymous · Answer

It's FLVS, 5.10, but I'd just like to know if I did it correctly and if not figure out what i did wrong :x

anonymous · Answer

its right

anonymous · Answer

you are correct

hartnn · Answer

it'd be just $\tan \theta$
and not $\pm \tan \theta $

hartnn · Answer

and cos is not correct -.-

zepdrix · Answer

why no plus/minus?

anonymous · Answer

^

hartnn · Answer

$\sqrt{x^2} = |x|$
not +/- x

hartnn · Answer

answer to a square root of real number can never be negative

anonymous · Answer

so it's technically Tan still? out of these choices?

zepdrix · Answer

|x| = +/-x
it's positive some places, negative in others.

example:
-tan(3pi/4) = sqrt(tan^2(3pi/4))

I think that's the point of the negative though.
See how this negative is changing the tan(3pi/4) to a positive?
making sure the root is positive.

Maybe I don't have my head on straight today :P
I dunno, I'll think about it some more.

hartnn · Answer

last option is correct, yes tan.

anonymous · Answer

ok im posting one more and im done =3

hartnn · Answer

example $\sqrt {(x)^2}$
when x is negative, say -6

hartnn · Answer

answer won't be negative

hartnn · Answer

also |-6| =6

zepdrix · Answer

$$\Large\rm \sqrt{x^2}=\pm x$$So in your example, when x=-6$$\Large\rm \sqrt{(-6)^2}=\pm (-6)=-(-6)$$See how we have to use the negative root in order to get the correct answer?

|-6|=+/-(-6) = -(-6)=6