Determine the centre (h,k) and the radius r of the circle: 4x^2 – x + 4y^2 – 7y – 9 = 0
write in (x-h)^2+(y-k)^2=r^2 form
that's it?
not going to divide by 4, complete the square and all that? c'mon myininaya!
you want me to do their problem?
well of course! the form is easy, it is getting there that is hard
how do i play this song on the piano? hit the right keys...
i figure they wanted to try first
yeah well this one is really ugly. maybe there is a typo?
but if you insist that i do it for them \[(4x^2-x)+(4y^2-7y)=9\] \[(4x^2-\frac{4}{4}x)+(4y^2-\frac{4}{4}*7y)=9\] \[4(x^2-\frac{1}{4}x)+4(y^2+\frac{1}{4}*7y)=9\] \[4(x^2-\frac{1}{4}x+(\frac{1}{2*4})^2)+4(y^2+\frac{7}{4}y+(\frac{7}{2*4})^2)=9+4*(\frac{1}{2*4})^2+4*(\frac{7}{2*4})^2\] \[4(x-\frac{1}{8})^2+4(y+\frac{7}{8})^2=9+4*\frac{1}{64}+4*\frac{49}{64}\] \[4(x-\frac{1}{8})^2+4(y+\frac{7}{8})^2=9+\frac{1}{16}+\frac{49}{16}\] \[4(x-\frac{1}{8})^2+4(y+\frac{7}{8})^2=9*\frac{16}{16}+\frac{50}{16}\] \[4(x-\frac{1}{8})^2+4(y+\frac{7}{8})^2=\frac{194}{16}\] \[\frac{1}{4}[4(x-\frac{1}{8})^2+4(y+\frac{7}{8})^2]=\frac{1}{4}*\frac{196}{16}\] \[(x-\frac{1}{8})^2+(y+\frac{7}{8})^2=\frac{49}{16}\]
Center is \[(\frac{1}{8},-\frac{7}{8})\] \[r^2=\frac{49}{16}=> r=\frac{7}{4}\]
Thanks for you help :)
that must have kept you busy
not really it was easy
interesing wikepedia and my answer have a different radius
oops i see my center is diffent too i changed something that i wasnt suppose to
yeah i cheated too at first. that is why i said it was really ugly. but maybe i entered it wrong
a minus to a plus
kept you occupied for a while though, didn't it?
see second to third i changed something
no it didn't
but i got the wrong answer lol
like always
i would have divided by 4 first
http://openstudy.com/users/satellite73#/users/satellite73/updates/4e4056bd0b8bfb83f3cbf936
we could have done that first
i'm not typing that over again just to show john my mistake
he can read this and he can make the correction himself
no don't bother
btw did you notice that the wiz gave ictrees a medal?
lol
why is the radius different though?
no lol
my radius and wikepedia's radius differ by like .03125
(x-\frac{1}{8})^2+(y-\frac{7}{8})^2=\frac{49}{16} i think this is suppose to be it
it got \[\sqrt{\frac{97}{32}}\]
darn
\[(x-\frac{1}{8})^2+(y-\frac{7}{8})^2=\frac{49}{16}\]
really?
hmmm
yes
let me look at this again
complete the square is a pain that is why
and that is why we all love \[-\frac{b}{2a}\]
ok wikepedia is right lol
its hard to beat a computer
you must be tired. especially if you are calling wolfram wiki
oh lol
yes they are both w words
i am too. off to bed.
i got them confused
i like my wrong radius better thought since it had a pretty answer
though*
it was nicer :)
why are teachers so mean algebra students can't do this
i don't know
my students would hate me for those fractions
yep...that's for sure
i am sure your students think you are cute
lol i hope they do because thats what im there for
to look cute
i remember when i was a student correcting my teachers when they made a mistake it made me feel so smart
it gave me a high
i hate when i make mistakes
zarkon i see a fun problem
I don't make mistakes...I make it look like I'm making a mistake to see if my students are paying attention ;) (at least that is what I tell them )
where is that
lol
Join our real-time social learning platform and learn together with your friends!