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#26 Help Me ! » conics » 2015-03-25 18:27:15
- niharika_kumar
- Replies: 22
The tangent and normal at the point P(at²,2at) to the parabola y²=4ax meet the x-axis in T and G respectively, then find the angle at which the tangent at P to the parabola is inclined to tangent at p to the circle through P,T,G.
#27 Re: Help Me ! » permutation and combination » 2015-03-14 04:23:12
thanks I got it too. 
#28 Help Me ! » permutation and combination » 2015-03-12 16:35:57
- niharika_kumar
- Replies: 25
Consider the set of eight vectors V={
}.Three non-coplanarvectors can be chosen from V in ways.Then p is?
#29 Re: Help Me ! » circle » 2015-03-05 15:15:11
can you give me a diagram of the double ordinate of a parabola.
I am not able to visualize it.
#30 Re: Help Me ! » circle » 2015-03-05 15:13:55
thanks.
and
HAPPY HOLI to all of you.
#31 Help Me ! » circle » 2015-03-04 18:56:29
- niharika_kumar
- Replies: 12
If two distinct chords of a circle
drawn from a point (p,q) is bisected by x-axis ,then which of the following is correct.a)
b)
c)
d)
#32 Re: Help Me ! » find minimum percentage » 2015-01-02 03:15:16
please post it.
#33 Re: Help Me ! » find minimum percentage » 2014-12-31 02:15:59
well there was a solution too which he gave.
But that solution is not being digested by me.
The answer is 10%.
And the solution said that convert the percentage to the opposite of what it said. for example: If 70% have lost their left hand then there are 30% who have not lost it.And do the same with others.Then he added all these data claiming that maximum percentage who didn't loose any part is 90% ,hence minimum percentage who lost all parts is 100-90=10%.
When I asked him if the data would have been given like 50% each,he told me there won't be such question.
help.
#34 Help Me ! » find minimum percentage » 2014-12-30 03:12:30
- niharika_kumar
- Replies: 6
hi friends.
I am back after few months. Feeling really good to be here again.
How are you all?
So here is my question.
In second world war 70% soldiers lost their left arm, 75% lost their right arm, 80% lost their left eye and 85% lost their right eye.
Find the minimum percent who lost all four parts?
pls help.
Niharika Kumar
#35 Re: Help Me ! » Find the value » 2014-11-17 03:20:01
thank you so much bob
#36 Re: Help Me ! » Find the value » 2014-11-15 15:48:06
anyhow, manually.
Did you appear in KVPY @agnishom.
#37 Help Me ! » Find the value » 2014-11-15 15:40:11
- niharika_kumar
- Replies: 38
Find the value of
#38 Re: Science HQ » collision » 2014-10-21 15:11:47
thank you so much bob.
#39 Re: Science HQ » collision » 2014-10-18 22:42:27
Now for the horizontal motion. At impact with the block, the velocity is still V and the block is moving at -V/4. We aren't told the mass of the block so no momentum equation is possible. But we can use the elasticity equation. We'll have to assume e=1 again as we aren't told anything else. So
You'll need to be careful with negatives here, but you should be able to get the velocity of the ball after the impact (assume block is still -v/4).
Bob
I am not sure if I could understand the way of using that equation properly.
Will you pls show how you placed all the terms in that.
#40 Science HQ » collision » 2014-10-18 03:10:40
- niharika_kumar
- Replies: 5
The block A begin to move with a velocity v/4.At the same time a ball is thrown from the wall of height h with a velocity v which traces the path as shown before coming back to the original position.Find x.
#41 Re: Introductions » Hi Everyone » 2014-10-18 02:49:52
Welcome to the forum!
#42 Re: Help Me ! » coordinate » 2014-10-03 16:23:26
here is one more:-
Show that all the chords of the curve
which subtend a right angle at the origin pass through a fixed point.Also find that point.#43 Re: Help Me ! » coordinate » 2014-10-02 22:34:55
my thanks to him too.
Even I was unable to do that and hence my 2nd equation was not in latex form.
#44 Re: Help Me ! » coordinate » 2014-10-02 16:16:34
thanks bob.
I didn't consider fact that y-mx is the factor of both the equations, but yes you used it, so
if we have that (y-mx) is the factor of both the given equations,
why don't we put y=mx in both of them, then eliminate the term x^2 and solve by cross multiplication.
that gives m^2 =(hA-Ha)/(bH-hB)={(aB-Ab)/2(bH-Bh)}^2
That has given a clear solution without m,n and p and the steps are less complicated.
#45 Re: Help Me ! » coordinate » 2014-10-01 19:11:03
Here is another question:-
Obtain the condition that one of the straight lines given by the equation
may coincide with one of those given by the equation
a'x^2+2h'xy+b'y^2=0
#46 Re: Science HQ » find the max. height. » 2014-09-28 14:40:16
thank you bob.
#47 Re: Introductions » Good Evening » 2014-09-27 16:13:49
Welcome to the forum!
#48 Re: Science HQ » find the max. height. » 2014-09-27 16:12:04
h is the height to which the ball is raised up from its hanging position.
#50 Science HQ » find the max. height. » 2014-09-26 20:15:34
- niharika_kumar
- Replies: 4
In the given figure the ball of mass m is held above as shown.It is then allowed to swing freely.Find the maximum height attained by the block.
