Approximate, to nearest 10', the solutions of the equation in the interval [0 (degrees), 360 (Degrees)) 12 tan^2t-5tant -2 = 0
@optiquest @zepdrix Sorry you guuuys
ooo crapper... i dont remember how to do `degree minutes seconds`. Are you allowed to use a calculator on this? I would assume so..
Yeah, so i just need the solutions and then i'm good to do the minute thing
oh ok cool
so factoring tant(12tant-5)-2 = 0
no no no silly billy.
aw -.-
you have a quadratic equation. how do we solve quadratics? it either factors nicely, or it doesn't (in which case we use our quadratic function to solve).
so how do I solve it?
your problem really looks like this:\[\Large\rm 12x^2-5x-2=0\]Where \(\Large\rm x=\tan t\)
Hmm won't factor nicely ;p quadratic function time.
plug in all the bits and pieces, tell me what numbers you get
dangit...quad formula was a long time ago lol
do it -_- nao!
I doing eeet I doing eet
\[(-5\pm \sqrt{5^{2}-4(12)(2)})/2(12)(2)\]
right?
too many 2's in the bottom and hold up, your c value is -2, not 2.
and it's -5 too right? :|
you gotta learn to use the equation tool sometime XD lol
oh yes, -5 as well
soo ( -5 (+/-) 11)/48
no, still too many 2's in the bottom
bottom is 2(a), yah?
i think you accidentally put 2(a)(c)
isn't it 2(a)(c)?
nooo :O
ooh...
so -5(+/-)11/24
yay good \c:/
Simplify it further to 2 values
hmm \[\huge\rm 12 x^2 - 5 x-2 =0\] tan t = x find two number if you multiply them you should get product of ac and if you add or subtract them you should get middle term which is -5 heehe how about headphone method ??;3
Nooo, they're fractions >.< that sounds like a lot of work. *grumble grumble*
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wait what? the answers are 6/24, and -16/24 right?
oh take note that this time they want the answers within degrees
oh oh we made a mistake back here: -5(+/-)11/24 should have been 5(+/-)11/24
oopsies
the first part is -(b) plugging in your b gives you -(-5)
So i guess you get 16/24 and -6/24 ya? simplify furtherrrrrrr -_-
0.66 and -0.25
So we have:\[\Large\rm x=\frac{2}{3},\qquad -\frac{1}{4}\]So,\[\Large\rm \tan t=\frac{2}{3},\qquad -\frac{1}{4}\]
To solve for t, we'll have to use our inverse function.
\[\Large\rm \tan t=\frac{2}{3}\]So,\[\Large\rm \tan^{-1}\left(\frac{2}{3}\right)=t\]
make sure you're in degree mode
33.690
good good good. tangent is period in 180 degrees, so we get the same 2/3 if we add 180 degrees to our angle, ya?
So our first two solutions for our angle t are, 33.690 and 213.690
How bout the others?\[\Large\rm \tan^{-1}\left(-\frac{1}{4}\right)=t\]
hrm my calculator doesn't like that it says syntax error -.-
it doesn't like the negative in front of the 1/4 but is it - 14.036?
are you using the subtraction sign instead of the (-) button by mistake?
....meybee
yesss -14 degrees. that doesn't lie in our interval though, right? so we'll add 180 degrees to find the next one.
165.963
and 345.963
mmm ok good
i THINK we did these correctly. i'm trying to see if i can check our work on a graph or something >.< sec
kks, ah, you always go the extra mile. :D
ehhh im not really sure on this one >.< we'll just go with this, it's probably right. you understand how to convert your angles to degrees, minutes, seconds from that point?
oh shooo i forgot to do that haha THANK YOU
yay team \c:/
yayaa
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