As of May 31, 2023, we have updated our Code of Conduct.
Mathematics
SmarthBansal's user avatar
SmarthBansal's user avatar
SmarthBansal's user avatar
SmarthBansal
  • Member for 5 years, 3 months
  • Last seen more than a week ago
  • India
Profile Activity
Stats
704
reputation
17k
reached
5
answers
27
questions
Communities
View all
About

Apparently I like to keep an air of mystery about myself

Badges
View all badges
This user doesn’t have any gold badges yet.
6
silver badges
21
bronze badges
Top tags
View all tags
3
Score
7
Posts
22
Posts %
2
Score
3
Posts
9
Posts %
0
Score
5
Posts
16
Posts %
0
Score
5
Posts
16
Posts %
0
Score
3
Posts
9
Posts %
0
Score
3
Posts
9
Posts %
Top posts
View all questions and answers
All Questions Answers
Score Newest
question
6
What is the value of $\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $?
May 4, 2018
question
5
If the complex number Z satisfies $ |Z^2 - 9| + |Z^2| = 41 $ then the true statements among the following are?
Apr 30, 2018
question
4
For rotation matrix $A$, find $B = A^4- A^3 + A^2 - A$
Apr 16, 2018
question
4
If $aZ^2 + bZ + 1 = 0$ and $|a|=0.5$. What is the value of $|ab' - b|$?
Apr 30, 2018
question
4
> Let $X$, $Y$, and $Z$ be three independent uniform random variables on $[0, 1]$. What is $P(XY < Z^2)$?
Sep 29, 2019
question
3
Solving for $k$ in $\cos \frac{\pi}k - \cos \frac{2\pi}k = P$
Apr 11, 2018
question
2
For how many $k$ do the lines $3x-2y-12=0$ and $x+ky+3=0$ meet the coordinate axes in four concyclic points?
Apr 2, 2018
question
2
What is the image of P in BC?
Apr 3, 2018
question
2
In a triangle ABC if $B-C = \frac\pi 4$ and .......... What will be the value of $\frac {a+b+c}{a}$
May 4, 2018
question
2
Why isn't $\lim_{x \to 0^+} x \left[ \frac px \right] = 0$?
May 8, 2018
Top network posts
View all network posts
7
Minimal velocity to throw an object to the Sun
7
A $3\times3$ ColorSaw
7
If you have to ask what the answer is, you'll never know!
6
Grooves on water tank
5
Why can't we see the satellites?

Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.