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OpenStudy (anonymous):
what 1.75?
myininaya (myininaya):
do you see the \[\emptyset\]
what does it mean
what information isn't giving us?
OpenStudy (anonymous):
for 1st one let the intersection of tangent and line joining be point A. find the distance from A to any of the centre and then u will find angle x
OpenStudy (anonymous):
its diameter
myininaya (myininaya):
i mean it is*
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myininaya (myininaya):
oh lol i never seen that symbol for diameter
OpenStudy (anonymous):
so u can solve it.
OpenStudy (anonymous):
hello
myininaya (myininaya):
i;m here
im just thinking about it
OpenStudy (anonymous):
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OpenStudy (anonymous):
look we have sinx =r1/(5-d)=r2/d
do u agree?
OpenStudy (anonymous):
d is the distance from one centre of circle to point A
OpenStudy (anonymous):
the radius through point of tangency and tangent are always perpendicular. do u know it?
myininaya (myininaya):
yes
OpenStudy (anonymous):
so sinx =r1/(5-d)=r2/d
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myininaya (myininaya):
what are you calling point A again?
OpenStudy (anonymous):
right he is
myininaya (myininaya):
okay yoda
OpenStudy (anonymous):
u know r1 and r2 so u find d
myininaya (myininaya):
i dont care about point A
i get it dip
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OpenStudy (anonymous):
its lengthy
OpenStudy (anonymous):
once u get d then u can find X
OpenStudy (anonymous):
got it myininaya??
myininaya (myininaya):
yes i totally got it dip lol
myininaya (myininaya):
d=35/12
so
x=17.46 approximately
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OpenStudy (anonymous):
are u trying the second??
myininaya (myininaya):
you were right not hard
you could email that guy and let him know you did his problem lol
myininaya (myininaya):
i want to see if there is anything harder
OpenStudy (anonymous):
u can email..
myininaya (myininaya):
i dont care about emaling him
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OpenStudy (anonymous):
i really don't have any interest to let him know that we have done.
OpenStudy (anonymous):
yeh hui na baat
myininaya (myininaya):
so no calculus dip?
myininaya (myininaya):
how is you integration skills?
OpenStudy (anonymous):
they must be good
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OpenStudy (anonymous):
i know a lil bit
myininaya (myininaya):
see if you want to attempt
\[\int\limits_{}^{}e^\sqrt{3x+9} dx\]
OpenStudy (anonymous):
put 3x+9=y^2
myininaya (myininaya):
omg so smart
OpenStudy (anonymous):
2ye^ydy /3
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OpenStudy (anonymous):
\[\frac23 \int\limits ye^ydy\]use by parts
OpenStudy (anonymous):
yeah cmon test us
OpenStudy (anonymous):
\[\frac23 (y-1)e^y+c\]
OpenStudy (anonymous):
got it?
myininaya (myininaya):
lol you suck jk gj
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OpenStudy (anonymous):
lol
OpenStudy (anonymous):
what jk gj?
myininaya (myininaya):
just kidding
good job
OpenStudy (anonymous):
oh..
myininaya (myininaya):
ok lets try another i have problem please hold
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OpenStudy (anonymous):
dipankar wt branch r u in?
OpenStudy (anonymous):
yeah post it
OpenStudy (anonymous):
any new problem @ him?
OpenStudy (anonymous):
not right now
OpenStudy (anonymous):
ok
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OpenStudy (anonymous):
u in mechanical?
myininaya (myininaya):
OpenStudy (anonymous):
wts the qstn?
myininaya (myininaya):
find the height of the lamppost
myininaya (myininaya):
the thing that looks like a lamppost
there may be more than one way of solving this
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OpenStudy (anonymous):
2
myininaya (myininaya):
how did you do that so fast
OpenStudy (anonymous):
practice
myininaya (myininaya):
2 is right
what did you do
OpenStudy (anonymous):
assumed the eqn of tangent to ellipse in terms of slope m and then plugged in -5,0 to find m to get the eqn
then plugged in 3, y to find y
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myininaya (myininaya):
this is the way i did it
OpenStudy (anonymous):
whats that??
myininaya (myininaya):
its in the file
OpenStudy (anonymous):
ur right as well
OpenStudy (anonymous):
the equation of tangent is y=mx+-sqrt(m^2a^2+b^2)
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OpenStudy (anonymous):
is that @ him..
OpenStudy (anonymous):
i think so.
OpenStudy (anonymous):
yes thats he eqn of tangent
OpenStudy (anonymous):
u have to take only the +ve sign and then put -5,0 as 2him said.
OpenStudy (anonymous):
ur in uni arent u myininaya?
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myininaya (myininaya):
uni?
OpenStudy (anonymous):
university?
myininaya (myininaya):
yes
myininaya (myininaya):
say i give you two circles
the center of these circles has y=x line running through there centers
the big circle has center at the origin and has radius 3
the small circle has a center that we want to find and has radius 1
give me one possible center for the center of the small circle
this one i don't think is hard but who knows
myininaya (myininaya):
actually this one is pretty easy i think
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OpenStudy (anonymous):
1,1
myininaya (myininaya):
oh wait let me say one more thing....
OpenStudy (anonymous):
any think (a,a) like points may be
myininaya (myininaya):
lol
nevermind :(
OpenStudy (anonymous):
there is no constrain to judge the exact position
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myininaya (myininaya):
circle is tangent to the other circle
myininaya (myininaya):
they only touch at one point
but nevermind this one is too easy