Halp.

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can you help? :)

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AHHHH I'm too tiredddddd -_- what's going on now? are the ones you filled in already, correct? you think so?
Yeah I think they're correct, but not for sure. aww i'm sorry you're tired D:
What is c..?
c^2=a^2+b^2 c^2=36+64 oh..so 10
o, ok that looks better
\[\Large\rm P=\left(10,\frac{32}{3}\right),\qquad F_1=(-6,0),\qquad F_2=(6,0)\]So use your distance formula, ya?
we can just use the p point like that?
how do we use the distance formula here? :o
\[\Large\rm d(P,F_1)=\sqrt{(10--6)^2+\left(\frac{32}{3}-0\right)^2}\]ya? :d
Shouldn't the focus points be related to the c values? Hmm sec, reading up on some of this stuff >.<
\[\sqrt{256+\frac{ 1024 }{ 9 }}\]
?
err, dangit. Yeah I think. So it would be f1= (-10,0) f2=(10,0)
so that changes this next part we just did. oopsies.
oh true :3 heh
soo I think d(p, f1) = 68
\[\Large\rm P=\left(10,\frac{32}{3}\right),\qquad F_1=(-10,0),\qquad F_2=(10,0)\]Really? Hmm I'm getting something more like this,\[\Large\rm d(P,F_1)=\sqrt{(10--10)^2+\left(\frac{32}{3}-0\right)^2}\approx 22.7\]
\[=\sqrt{(10+10)^2+(\frac{ 32 }{ 3 }-0)^2}\] \[\sqrt{400+\frac{ 1024 }{ 9 }}\] Put 400/9 then add them up and apply sq root = 68
I messed up somewhere really bad, yeah? o.o
\[\Large\rm =\sqrt{400+113.8}=\sqrt{513.8}\approx 22.7\]Still not sure how you're getting 68 :) lol
Oh oh oh i see what You did, no no no. not 400/9 you have to multiply the top by 9 as well if u want to do it that way. 400 = (400*9)/9 = 3600/9
yeah yeah I thought i did that :o
that mistake would have given you a number smaller than 22.7 though lol so i'm not sure what's going on ;0
lol me neither. Ah well. moving on then xD
erm so it says put exact interger. but the answer is 22.7 should i round it to 23 do you think?
Oh it says exact? Ok then uhhh
oh sorry, sorry. it says "exact interger, fraction, or decimal"
so that would be.. 3 numbers after the decimal point? o.0
\[\Large\rm d=\sqrt{\frac{3600}{9}+\frac{1024}{9}}=\frac{68}{3}\]Is that where the 68 was coming from? 0_o
oh! there it is xD
I completely forgot the 3 lol
how bout the other distance? :)
would that answer be 22.666? or just two 6's?
the other distance = 10.66?
68/3 is a better form of the solution. if you want a decimal with three places though, it would be 22.667
yes, the other distance is 32/3 :)
or 22.67 for two decimals
ee the feeling of accomplishment haha yay! I'm going to try the fraction form first.
last box!
\[\Large\rm \frac{68}{3}-\frac{32}{3}=?\]Fractions seem better. Makes this last step easier.
36/3!
:P
36 over 3 factorial? :o no that can't be right. jk :x hehehe
noooo =.= haha it's 12 then xD
does 12 fit the form of 2(a)? where a=6?
naaaah
jk, yes
12 = 2(6) yayyy you did it \c:/ good job!
wait so in that last box I put 12 again?
no..it would be 6 in that box
ya 6 :O
lol I was like WAIT that doesn't make sense xD yaya *we* did it!! :) ty ty
ook let's see if it's all correct.
yep yep it's all correct :*
yay \c:/
want to help with another one? though it is quitee extensive

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