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Callisto
Group Title
Consider the system:
x + 2z =1
x+2y+4z =1
2xy+3z=2
For all solutions {x,y,z}, find the range of k such that \(k(x^23) > yz \)
 one year ago
 one year ago
Callisto Group Title
Consider the system: x + 2z =1 x+2y+4z =1 2xy+3z=2 For all solutions {x,y,z}, find the range of k such that \(k(x^23) > yz \)
 one year ago
 one year ago

This Question is Closed

abb0t Group TitleBest ResponseYou've already chosen the best response.0
I don't know!! :'(
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
hmm i can find values of x , y and z but don;t know what to do after wards with them
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Same here...
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
lol sorry guess not much help i am
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Wait, can you find the numerical values for x, y, z?
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
sure , give me a moment.
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
what about first finding the solution set of the system using algebra matrix
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
Gaussian elimination method.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
dw:1363361660822:dw
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
dw:1363361796138:dw
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
That's what I can do and that's why I ask if you can find the numerical value for {x,y,z}
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
k honestly i have done sole me this kind before where we get a who row zero but hard to explain the logic behind it..that whole zero row just made hard to find definite answer.. i normally would do this with the simultaneous method now. so we can get x, y , and z in terms of each other.
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
if i solve simultaneously, i got z = y and x = 1 2z
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Yes, I got the same too, but how can we utilize it to find range of k?
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
dw:1363362152973:dw
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
so i think we can get value of x , y and z if we suppose any value of z .. according to it , we will get other values.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
\[k[(12z)^23]>z^2\]\[k>\frac{z^2}{(12z)^23}\]Hmm..
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
hmm got same thing.. after this i did the long division.. but lol not possible way out
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I'm thinking of differetiating the fraction to find the min/max/range..
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
can try tht
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Suppose wolf. is right, then we get \(\frac{z(z+1)}{(2z^2+2z+1)^2}\) Put that derivative = 0, z=0 or z=1
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Oh wait.. Why would I differentiate that!!!
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
hmm thought u were trying to get max and min.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
But it won't work?
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
yeah.. i still feel we using wrong approach.
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.3
\[x + 2z =1[1]\] \[x+2y+4z =1[2]\] \[2xy+3z=2[3]\] In Equation [1] \[x=12z\] In Equation [2] \[12z+2y+4z=1\] \[2y+2z=0[4]\] In Equation [3] \[2(12z)y+3z=2\] \[24zy+3z=2\] \[yz=0\] \[y+z=0[5]\] Therefore: \[x=12z\] or \[x=1+2y\] \[2z=1x\] \[z=\frac{1x}{2}\] \[2y=x1\] \[y=\frac{x1}{2}\] Therefore: \[yz=\frac{(1x)(x1)}{4}\] In: \[k(x^2−3)>yz\] \[k(x^23)>\frac{(1x)^2}{4}\]
 one year ago

Azteck Group TitleBest ResponseYou've already chosen the best response.3
\[4k(x^23)>(1x)^2\] \[4kx^212k+(12x+x^2)>0\] \[4kx^212k+12x+x^2>0\] \[(4k+1)x^22x(12k1)>0\] Using the discriminant: \[b^24ac>0\] \[4+4(12k1)(4k+1)>0\] Solve as quadratic and find the range of values for k.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Almost there, but still not correct.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Man at work.
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
am I late? :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I dont' think so.
 one year ago

linshan789 Group TitleBest ResponseYou've already chosen the best response.0
Can you just solve for x,y,and z (using a system of equations) then plug the variables back in?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I've tried solving the system above...
 one year ago

linshan789 Group TitleBest ResponseYou've already chosen the best response.0
I ended up with x=1, y=0, z=0. So, \[k(1 ^{2}3)>(0)(0)\] =>\[k(2)>0\] =>k<0
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
It shouldn't have a unique solution.
 one year ago

linshan789 Group TitleBest ResponseYou've already chosen the best response.0
Oh. What level of math is this?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
High school
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
k belongs to (  infinity , 1/4 ] union (1/6 , 0 ) ?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
No, almost there. Can you tell us how you get there?
 one year ago

linshan789 Group TitleBest ResponseYou've already chosen the best response.0
wait, how is k<0 unique? (inf, 0)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
k belongs to (  infinity , 1/4 ] union (1/6 , 0 ] ?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
can I post here after 30 minutes please? I have to go for dinner ?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Sure, take your time :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I am back to work.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Any guideline?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Well first of all I solved for x and then put it into the eqn and then took diff. conditions .
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
k belongs to (  infinity , 1/4 ] union [1/6 , 0 ] ?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
No.. diff = different/differentiate?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
different.
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
looks easy wait
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I am getting diff. answers. Wait!
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I have some patience, so I keep waiting, and trying :)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
That is good.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
is it [0, 1/6] ?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Very very close!!! How do you get it??
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
(x1)^2 /(4x^2 12) tried to work with this :
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
k belongs to [1/6,0] union (infinity ,1/4] .. i am still getting it
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
is it wrong^ ?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@Callisto ?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
It is wrong.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Did you get what I did ?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
Coz I am not very confident myself! :P
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I didn't get what you did :(
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Did you start from this:\[k(x^23)>\frac{(1x)^2}{4}\]?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
I am calling that f(x) k> f(x) whatever be the range of f(x), k's range will be opposite of that, i.e. maxima of f(x) is min of k and vice versa. so, is the ans (0,1/6) ?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
The answer is correct, but I don't understand your explanation.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
But aren't 0 and 1/6 satisfying the equation ?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
The inequality itself is > , not >=
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
k(x^2  3)>yz Put, x = 12z and y = z Then k{1+4z^2 4z  3} > z^2 => k(4z^2  4z  2) > z^2 => (4k+1)z^2  4kz  2k>0 Now put, k = 0 you get z^2 > 0 Which is true ...... So k = 0 satisfies your inequality ....
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@Callisto
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
lol x=1 btw I was having dinner z=0 and y =1/2
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
do this, eliminate 1 from lowest row in matrix you get two equations of x and z,solve them then u wil get answers of x,y,z
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
And MY z = 1/6 have a look: (4k+1)z^2  4kz  2k>0 Put k =1/6 We get z^2 + 2z + 1 >0 OOPS that is true toooooo
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
So k =1/6 also satisifies your inequality
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
What if z=0?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Well if z = 0 then go for it: k < 1/2
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
@aajugdar I don't mind if you check my matrix work above. I can't find a unique solution to the system.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Well the solutions given by you and shubhamsrg have many cases left out .....
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
But when z=0, the inequality fails.
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
put them in equation dude,the values are correct ok hold on,i wil tell ya
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
perform R3 R2/2 on R3 and then again arrange them in equation form,u will get x+2z =1 and then 3/2x+z =3/2
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Well ok in that case we get (by putting in my quadratic inequality in terms of z and k), k < 1/2 What about that ?
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
solve them,u will get values of x and z,from that values,find value of y by putting it in equation 2
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
@mathslover That is not even the answer @aajugdar Check this dw:1363447547599:dw
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
what operation did you perform on 3rd row????
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
2 x R1 + R3 > R3
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
@mathlover When I learnt inequality, my teacher told me that if (1) ax^2+bx+c>0, where a>0, then the solution would be like (alpha)<x or x>(beta) (2) ax^2+bx+c<0, where a>0, then the solution would be like (alpha)<x<(beta)
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
nope man you didn't solve it as it should be solved,you see by makng complete 1 row 0 you wil never get solution you shouldn't keep a row NULL its a rule
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@Callisto very correct
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
remember 3 2 unknowns,2 equations,3 unknowns 3 eqautions.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
(4k+1)z^2  4kz  2k>0 I used that (please have a look at my solution)
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I am not a man This system has no unique solution, that's what the determinant of coefficient matrix tell us.
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
well,solve it correctly,it has a solution I found it
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
@aajugdar You can try using WolframAlpha, you can't get a unique solution. @mathslover Can you guarantee that (4k+1)>0?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
k > (x1)^2 /(4x^2 12) the derivatives are 0 at x=3 and 1 respective values of y are 1/6 (local maxima) and 0(local minima) so k > 1/6 and k<0
 one year ago

aajugdar Group TitleBest ResponseYou've already chosen the best response.0
facewall
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
only in that range x>f(x)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
I mean k>f(x)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Case  1 : 4k + 1 > 0 => k>1/4 Now, that means the inequality has a graph upward. Now for the inequality >0. b^2  4ac < 0 => k(6k+1) < 0 => If k< 0 then 6k+1>0 i.e. k>1/6 And if k>0 then 6k+1 < 0 i.e. k<1/6 (This option is not possible) So, k<0 ; k>1/6 ; and k>1/4 Take intersection: k belongs to (1/6,0) @Callisto I am thus taking cases
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Case 2 : (4k+1)<0 => k <1/4 i .e. the graph is bent downwards. Now for inequality> 0. b^2  4ac > 0 => k(6k+1) >0 => k> 0 and k>1/6 or k<0 and k<1/6 => k>0 or k<1/6 But we took k<1/4 So, k>0 case is eliminated. So, k belongs to (infinity, 1/4)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Case  3: 4k+1 = 0 => k = 1/4 So, inequality reduces to z>1/2
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Which obviously consitutes part of real values of z.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
So, I said answer should be: k belongs to (infinity, 1/4] union (1/6,0)........ Kindly please tell if I am correct or If I went wrong somewere
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Well I agree with you that k = 1/4 and 1/6 do not constitute the solution so modify it k belongs to (infinity, 1/4) union (1/6,0) But I still don't know that how will I then deal with 4k+1 = 0 case
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I'm thinking.. for the inequality, (4k+1)z^2  4kz  2k>0, it doesn't imply that (4k+1) >0 or (4k+1)<0. It can be 4k+1 =0 Oh, you've pointed the problem out.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
Think about that
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
I hope I didn't annoy you?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Nope, it just the exchange of ideas.
 one year ago

stamp Group TitleBest ResponseYou've already chosen the best response.0
it seems the solutions are not numbers, but rather variables (see attachment) also, http://tutorial.math.lamar.edu/Classes/Alg/AugmentedMatrixII.aspx
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
The solutions to the system are not numbers, that is it has no unique solution. The link doesn't wok.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@stamp read the question again ;)
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
The system is easy to solve for x, y, z. It's just a little tedious. I am in a rush as it's nearly 4 am so I'll give you a brief solution to this problem. First rearrange x + 2z = 1 to z = (1  x) / 2 Now you have z in terms of x. Find the value of y in terms of x from the second equation and plug in (1  x)/2 for z in it. Then plug these y and z values in terms of x in the third equation and solve for x. Then solve for z and then y. You can of course use GaussianElimination Method but that can be annoying at times. Anyway, after you got the solution for x,y,z, just plug the values in to the inequality and rearrange it to k > yz/(x^2  3). And that will give you the range for k. G2g now. @Callisto
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
1. There is not unique solution to the system. I have solved it using Gaussian elimination and have already put what I have got above. 2. I have put x, y, z into the inequality (in terms of z), but it just leaves me a range with unknown z. 3. I have tried differentiate z to get min/max, but the values I got (z=0/z=1) does not match with the answer.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
So does the system have infinite solutions or a finite number of solutions?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Infinitely many solution.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
If it has infinitely many solutions then which solution gives the smallest value for yz / x^23?
 one year ago

some_someone Group TitleBest ResponseYou've already chosen the best response.0
I would say range is all real numbers?
 one year ago

some_someone Group TitleBest ResponseYou've already chosen the best response.0
no? do you know the answer then?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I know the answer.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I don't know how to get the answer though.
 one year ago

some_someone Group TitleBest ResponseYou've already chosen the best response.0
oh ok got you
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Do you know what the smallest solution is for x,y,z such that yz / x^2  3 is the smallest possible value? Try to find such a solution.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
@genius12 Would you mind checking me steps of solving the system?dw:1363507811134:dw
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Did you do this through gaussian elimination?
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Have you tried solving this algebraically? It will be tedious, but could be promising. It's easy to mess up on matrices, but not as much when you solve algebraically.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
I know for sure that you will be able to solve this algebraically, and I am pretty confident that you won't get "infinitely many solutions", hopefully. 4:16 AM. Dark room, people sleeping in here, I think I'll head to sleep as well. Good luck Callisto.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
x + 2z =1 (1) x+2y+4z =1  (2) 2xy+3z=2 (3) (1)+(2) 2y + 2z = 0 => y = z  (4) Sub. (4) into (3) 2x  (z) + 3z =2 x+2z = 1  (5) , which is also (1) Pretty obvious, huh?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
If it is not infinitely many solutions, then it will be no solution.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
You'd better sleep now. Good night.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Then there is infinitely many solutions.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
What's next?
 one year ago

some_someone Group TitleBest ResponseYou've already chosen the best response.0
Wouldn't the system of equations be inconsistent?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
The system should be consistent.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Yes, for you to solve the inequality and get the range for k, it should be consistent which it isn't.
 one year ago

some_someone Group TitleBest ResponseYou've already chosen the best response.0
@genius12 yes i agree it is inconsistent.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Please point out my mistakes then.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
FYI: http://www.wolframalpha.com/input/?i=x+%2B+2z+%3D1+and+x%2B2y%2B4z+%3D1+and+2xy%2B3z%3D2
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
I could pick any value for z and get the same three equations. This gives us no way to come up with a range for k. =.=
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
That's why we need to think how to get a way to solve it.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Something wrong with the question? lol
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
No way! It is a past exam question.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
What unit does this question cover?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
Matrix and determinant.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
If it's a past exam, isn't there a given solution?
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
The final answer  Yes.
 one year ago

genius12 Group TitleBest ResponseYou've already chosen the best response.0
Check it? lol
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
I know what the final answer is. I just don't know how to get the final answer.
 one year ago

Callisto Group TitleBest ResponseYou've already chosen the best response.5
When z=1, (max) => k=min (x, y, z) = (3, 1, 1) \[k>\frac{yz}{x^23}\]\[k>\frac{(1)(1)}{3^23}\]\[k>\frac{1}{6}\] When z = 0, (min) => k = max (x, y, z) = (1, 0, 0) \[k<\frac{yz}{x^23}\]\[k<\frac{(0)(0)}{1^23}\]\[k<0\] @shubhamsrg Your method works!! Too intuitive for me though :(
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
I am not too sure how to put my words more properly. : Hey, why don't you try to understand this with the graph, you may use wolfram for the graphing.
 one year ago
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