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Mathematics 17 Online
myininaya (myininaya):

ok dipankarstudy.. i have no clue if this is really difficult i haven't attempted it yet

myininaya (myininaya):

http://www.stirlingsouth.com/richard/trig1m.jpg

myininaya (myininaya):

http://www.stirlingsouth.com/richard/math.htm

myininaya (myininaya):

i wonder what that symbol by the 1.250 means

OpenStudy (anonymous):

ist one is not so difficult.

myininaya (myininaya):

and the 1.75

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*

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):

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

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

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

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

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

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

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

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

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

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

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

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)

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?

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

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

myininaya (myininaya):

circle is tangent to the other circle

myininaya (myininaya):

they only touch at one point but nevermind this one is too easy

OpenStudy (anonymous):

(2sqrt(2),2sqrt(2))

myininaya (myininaya):

yes thats right

OpenStudy (anonymous):

(sqrt(2),sqrt(2)) is also possible

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