13:49 cosine of theta is equal to cosine of theta. I learn much in this channel XD.
@erikkonstas4 жыл бұрын
But not to cosine of Theta... dun dun dun...
@U014B4 жыл бұрын
*Petition To Rename The "±" Symbol As "Plinus"*
@erikkonstas4 жыл бұрын
And the upside-down version would be...?
@anmolbansal26044 жыл бұрын
@@erikkonstas mlus
@tim_tmn Жыл бұрын
@@erikkonstas Mipus
@erikkonstas Жыл бұрын
@@tim_tmn 😂 say that again *slowly* ...
@eboone Жыл бұрын
Legend came back 2 years later to finish the joke
@ahusky44984 жыл бұрын
I like to put my complex number in the Lambert W function
@OriginalSuschi4 жыл бұрын
wait how do you compute the Lambert W function?
@ushasingh62044 жыл бұрын
Using calculators
@That_One_Guy...3 жыл бұрын
Oh wow that's useful
@fsisrael92245 ай бұрын
And then you get the fish back 😊
@plaustrarius4 жыл бұрын
I just saw BPRP has one of the top comments on Herbert Gross's TEDx talk (rewatched it after the recent video the MAA put out) I sent Prof. Gross an email in college to thank him for his lectures. I was so surprised when I actually got a reply, honestly touches my heart when I think about it. Reminds me I have to say more often, Thank you so much for these videos! In the same way Prof. Gross got me excited about all the math I wasn't learning in school, this channel and Dr. Peyam's channel, all of the amazing math outreach you do, your social presence in the community, I never had that in school. It means the world to me seeing the affect you have each day, thank you from the heart Prof. Steve!!!
@blackpenredpen4 жыл бұрын
Thank you very much Carter for your nice comment. Yes, Professor Gross is truly amazing! We actually exchanged a few emails a few years ago! His lectures are pure gold!
@virajagr4 жыл бұрын
@@blackpenredpen just came here to say thanks for the heart on other comment :p Also I don't know why youtube is not notifying me all the replies I m getting....
@virajagr4 жыл бұрын
@@blackpenredpen damn it bprp!! You just gave me a heart on a reply where i talked about getting a heart 😂 that's the first time I have seen a youtuber having half a million subscribers and still replies and chat with people in comments 👍👍
@blackpenredpen4 жыл бұрын
If you want to view this video in 1080P 60FRQ (and don't mind my small mistakes), then it is here: kzbin.info/www/bejne/gnecpnSDd7V5mtE
@karthiksriram46324 жыл бұрын
Hi. Can you like my reply? I am a huge fan of you, your math, your pens and your mics. This will make my day ☺️. Please like my reply
@Rahulsingh-lw8hk4 жыл бұрын
yes we want it and, thanks for the video we loved it!! my support with you.
@sudiptaroyarts38614 жыл бұрын
Hi i am your fan and i also called u junior Euler
@Mystery_Biscuits4 жыл бұрын
60FRQ? Frames Per Question? lmao.
@karthiksriram46324 жыл бұрын
blackpenredpen thank you so much ☺️
@erforscher4 жыл бұрын
I think Euler form has most applications, like in nth roots.
@RedRad19904 жыл бұрын
Well, you have also de Moivre's formula. I, for example, had literally one lecture on complex numbers, and we never used Euler's formula. We did use de Moivre's, though. Wikipedia says that it's a bit older than Euler's identity.
@skylardeslypere99094 жыл бұрын
@@RedRad1990 DeMoivre is also a bit easier imo, I also think that the trigonometric identity of a complex number is more intuitive than the exponential one
@MathemaTeach4 жыл бұрын
I like your mic, you always surprise us, i think your mic is where the complex number is. This is very informative, thanks.. 👍👍👍👍
@geosalatast57154 жыл бұрын
Hahaha 13:45 of course cosθ=cosθ hahaha I cried a lot xD maths are always fun with you as a teacher! Greets from Greece!! PS I do remember that video you mistook the Re with R and Im with C hahaha but dont worry bcs teacher's mistakes are never students' mistakes ;)
@blackpenredpen4 жыл бұрын
Geo Salatast loll thanks!!! I am glad to hear!
@cottonman1324 жыл бұрын
7:51 The video where you made that mistake was the sin(z) = 2 video, but the only reason I remembered is because I remember how great that video was. Keep up the great content :)
@Jiglias15 күн бұрын
damn, was just about to comment the same thing four years later
@treasureberry71544 жыл бұрын
Mike Bison the real MVP
@muttleycrew4 жыл бұрын
As a side note... I like to put my _i_ before any other variable to denote an imaginary component of a complex number since _i_ is behaving like a constant. So I write a + ib instead of a + bi. This seems consistent to me. II don't write b2 for "two times b," I write 2b like pretty much everyone else on the planet so too, for me, _i_ gets treated like any other constant in these expressions and it is _i_ times b and not b times _i_
@divisix024 Жыл бұрын
i is also behaving as the unit vector along the positive direction of the imaginary axis of the complex plane, which is two-dimensional over R. When we write a vector in R^n we take a basis v1,…,vn, real numbers a1,…an, and express the vector as a1v1+…+anvn. So for the complex plane the standard basis would consist of 1 and i, and then since every point on the complex plane can be seen as a vector from the origin to that point, it would then be expressed as a1+bi, but then of course omitting that 1 hurts no one, resulting in a+bi. The usual way is consistent with the vector interpretation but not the constant interpretation, just like a+ib wouldn’t be consistent with the vector interpretation.
@muttleycrew4 жыл бұрын
Next up, "where do you like to put your quaternions"
@byronvega82984 жыл бұрын
Is that an edited minus sign? 😂
@erikkonstas4 жыл бұрын
What is funny with that?
@byronvega82984 жыл бұрын
@@erikkonstas idk I just found it quite amusing, maybe because I would've done the same mistake atleast π times
@dominicellis18672 жыл бұрын
I like to put it in an exponential turning it into an oscillation in time and then adding a frequency term you can generate musical notes using Fourier series. Once you have pitch and rhythm, you can couple them using inner products, outer products, normalization and distance formulas to generate harmonic progressions by tracking the derivative of the intervals of the points on the unit circle of 4ths. This ends up being the jerk of the “quantum note” whereby a half step lasting for one beat is the basis and a tritone has an interval of pi radians on the Bloch Sphere of Harmony.
@rajakannant5724Ай бұрын
I like to put in the following Sin(a+ib) Cos(a+ib) Sinh(a+ib) Cosh(a+ib) tanh(a+ib) tan(a+ib) Sin‐¹(a+ib) Cos‐¹(a+ib) tan-¹(a+ib) Sinh‐¹(a+ib) Cosh‐¹(a+ib) tanh-¹(a+ib) Γ(a+ib) ζ(a+ib) η(a+ib)
@pierre-louisbourgeois11244 жыл бұрын
I've always a complex number in my pocket. We never know...
@blackpenredpen4 жыл бұрын
Loll
@stephanebiwole1034 жыл бұрын
I really appreciated your video! I learnt a lot
@blackpenredpen4 жыл бұрын
Glad it was helpful!
@stephanebiwole1034 жыл бұрын
But I have some question? Is the sin or the cos of a complex number
@julianstoller58834 жыл бұрын
What about complex factorials?
@shishirmaharana40224 жыл бұрын
I love the way how you hold something in your left hand while solving math.
@erikkonstas4 жыл бұрын
Did you just call Mike Byson a "something"??? For shame... 😂😂😂
@interesting62624 жыл бұрын
Today is my birthday and this is my BEST GIFT.....THX bprp
@blackpenredpen4 жыл бұрын
Glad to hear! Happy Birthday!!!!
@interesting62624 жыл бұрын
@@blackpenredpen Thank you so much :) :)
@dominicellis18674 жыл бұрын
tan(a + bi)
@xwtek35054 жыл бұрын
I'd like to put a complex number on riemann zeta function and gamma function.
@區光3 жыл бұрын
For the first one, why can you put the power bi in front of the natural log? I mean you cannot do (ln2)^bi = b*i*(ln2), we can only do ln[2^(bi)] = b*i*(ln2)
@Apollorion4 жыл бұрын
I prefer to put my complex number in the complex plane.
@raghavgarg55384 жыл бұрын
what will happen if we put complex no. in miscellaneous functions like greatest integer function, fractional part function etc.
@invisibleperson8904 жыл бұрын
hello from Turkmenistan and thanks for all amazing videos :)
@気分によって面積が変わる-b9j4 жыл бұрын
What time is it there now? It is very interesting. Please continue to solve interesting math problems
@hadriencrassous21624 жыл бұрын
I want my complex number to stay home.
@virajagr4 жыл бұрын
How about integral of (x dx) from say 2 to 3i? Yes, it is solvable as answer would come out to be [(3i)²-2²]/2 = -13/2 but what does it actually mean? In real plane it would be area under the curve but what about in complex plane?
@angelmendez-rivera3514 жыл бұрын
I hate being the nitpicky person, but what you are stating is wrong. The problem is that the fundamental theorem of calculus, as you are stating it, is not valid for functions that map a nonreal subset of the complex plane to another subset of the complex plane. You need a generalization that holds, and it involves a generalization of the integral operator, known as the contour integral operator, in which the integral depends not only on the boundaries of integration and on the function being integrated, but also on the contour of integration. A few integrals happen to be contour-independent, but in the most general of cases, this is false. Therefore, if you want to calculate the integral z |-> z from first principles between any two points, you need to choose some contour, and then prove that if you choose any other contour, the result is the same. This is how you integrate complex functions, and this is one of a few foundations of what complex analysis is today. In this case, we want a contour that connects the points (2, 0) and (0, 3) in the complex plane. The simplest such contour is a line containing these points. Define a function γ(t) : R -> C such that γ(t) = At + B, γ(0) = 2, and γ(1) = 3i. Therefore, B = 2 and 3i = A + B, hence A = -2 + 3i. Therefore, γ(t) = (-2 + 3i)t + 2. Okay, now we can actually proceed to integrate. The function we want to integrate is f(z) = z, and the boundaries of integration are (2, 3i), with contour γ(t). By the definition of a contour integral, this is the same as the integral from 0 to 1 of f[γ(t)]·γ'(t) with respect to t. The reason the boundaries of integration are 0 and 1 is because γ(0) = 2, which our lower boundary in z, and γ(1) = 3i, which is our upper boundary in z. Of course, if I choose a different contour, then the boundaries in t will be different, but let us not complicate it any further and just work with what we have. Anyway, γ'(t) = -2 + 3i, so f[γ(t)]·γ'(t) = [(-2 + 3i)t + 2](-2 + 3i) = (-2 + 3i)^2·t + (-4 + 6i) = [(-2)^2 + (3i)^2 + 2(-2)(3i)]t + (-4 + 6i) = (-5 - 12i)t + (-4 + 6i). Integrating this over the interval (0, 1) is pretty trivial, because the antiderivative, when evaluated at 0, is equal to 0 plus an arbitrary constant of integration, which then gets subtracted from the antiderivative evaluated at 1. Therefore, the subtraction yields 0. Meanwhile, evaluated at 1 without the constant of integration that we subtracted off, the antiderivative equals -(5 + 12i)/2 + (-4 + 6i) = -(5/2 + 4) - 12i/2 + 6i = -(5/2 + 8/2) = -13/2, which is the same result you obtained. So yes, the integral is indeed -13/2 as you claimed, but the way you went about calculating it is invalid, since you have yet to prove that the integrals over z |-> z are contour-independent, which is not obvious to most readers.
@angelmendez-rivera3514 жыл бұрын
Also, the contour integral is just equivalent to the area under the curve of the function evaluated along the contour.
@virajagr4 жыл бұрын
@@angelmendez-rivera351 so does that mean integrals in complex plane is actually a path function and not a state function? So an integral could give different values (infact maybe infinite)? If so then which contour do we take by default. Also, does taking the simplest contour that is the straight line, would it always give the same result as plugging in the limits in anti derivative?
@angelmendez-rivera3514 жыл бұрын
Viraj *so does that mean integrals in the complex plane is actually a path function and not a state function?* Yes. In fact, the concept of integrating with respect to a line element in multivariable calculus was motivated by the concept of contour integrals, because the study of multivariable calculus, in its modern formalism, came from the study of complex analysis and quaternionic analysis. Line integrals in multivariable calculus are path functions because contour integrals are path functions, although contour integrals can result in state functions more often than line integrals. Whether an integral is contour-dependent or not has a lot to do with the discontinuities of a function and what not. *So an integral could give different values (in fact, maybe infinite)?* Yes. In fact, one of the reasons you can show the complex logarithm is multivalued is by integrating 1/z on the complex plane. If I want the integral of 1/z from z = 1 to z = -1, it is impossible to choose a line that connects those two points, because that line also contains a discontinuity at z = 0. So you need to get more creative. One contour you can choose is a semi-circle of radius 1 whose endpoints are 1 and -1. So you let γ(t) = e^(it), and then you integrate 1/e^(it)·ie^(it) from t = 0 to t = π. This results in πi. However, I can choose any other contour to integrate over. I can choose the contour γ(t) = e^(3it), and then the integral will result in 3π instead. In fact, I can choose any contour γ(t) = e^[(2n + 1)it], and the integral will result in (2n + 1)πi, for some integer n. Hence giving infinitely many different values. And if you want to define log(-1) as being the integral of 1/z from 1 to -1, which logically follows from defining log(x) to be the integral from 1 to x, then you run into this issue that log(-1) is equal to infinitely many different values. In fact, you can even do this positive real x, and you still get infinitely many values. This happens to be true when integrating many other functions in the complex plane as well. *If so, then which contour do we take by default?* Depends on the function and the context of the applications you are working with. I know that is not a very satisfactory answer, but it is what it is. In the case of the integral of 1/z, it's easy to see that the contours γ(2n + 1, t) = e^[(2n + 1)it] are all just re-scalings of the contour γ(1, t) = e^(it), since γ(2n + 1, t) = γ(1, (2n + 1)t), and because we typically only care about angles t in the range (-π, π), we tend to take the contour γ(1, t) to be the most natural contour to integrate 1/z over, and this motivates the definition of the principal branch of the complex logarithm, which is that log(z) = log(|z|) + arg(z)·i, where arg(z) has the restriction that it lies in the interval (-π, π]. This gives you the principal value out of infinitely many, much in the same way that, for example, 16 has four 4th roots, but 2 is the principal value, it is the principal fourth root of 16. But like I said, for other functions, it really depends. *Also, does taking the simplest contour that is the straight line would it always give the same result as plugging in the limits in anti-derivative* Depends a lot on the function. There are some situations in which the simplest contour actually isn't a straight line, and there are situations in which a straight line would have been desired, but is impossible due to the behavior of the function. If you take at face value the idea that the anti-derivative of 1/x is log(|x|), for example, and you plug in 1 and -1, you get 0 for an answer, which disagrees with the answer that you get by contour integrating. Hopefully that answers your questions.
@virajagr4 жыл бұрын
@@angelmendez-rivera351 thanks a lot! That really answered all questions and thanks again for spending so much time in answering :)
@F1U7R2Y94 жыл бұрын
D u know , Grant suggested your video of i^i. ☺️
@brownwater62124 жыл бұрын
As soon as I started the video, I thought you were holding a rabbit. lol
@ronandodds4 жыл бұрын
seamless editing, sir
@muttleycrew4 жыл бұрын
I like to put my complex numbers in my tea with a drop of lemon juice and some honey to sweeten them. So delicious.
@eliardosoarescoelho43334 жыл бұрын
Nota mil cara, parabéns pelo trabalho. Mas algum brasileiro aí?
@megalucasg8018 Жыл бұрын
tem eu aq
@gileadedetogni9054 Жыл бұрын
Tropa dos br fã do bprp
@Shadow-by6xb4 жыл бұрын
I just have a question involving something that is a lot more basic, but how do you write a formula in which it is a value, for example: J. Then have it multiply to (J-1) an n number of times. This is for the ncR because I watched the video about winning the lottery chances. Here would be an example: assume J is 3 and n is 2. It would look like (3*2)/2*1. So I'm trying to write a formula for the top part that has n in it as I already have the bottom part which is simply n!. Any help would be appreciated, especially since I'm probably not explaining this well
@rajatrajmnnit20224 жыл бұрын
I would like to put it inside cosh(a+ib)
@brownwater62124 жыл бұрын
Mic Byson
@shashwatdwivedi81844 жыл бұрын
Hi bprp ! Love from india 😊
@prateekmourya95674 жыл бұрын
In the zeta function
@sueyibaslanli35194 жыл бұрын
Can you proof Euler's formula
@aasimraiyan50014 жыл бұрын
Iota order of derivative of function d^if/dx^i
@sergioh55154 жыл бұрын
Is that a Supreme sweater, BPRP? 😳
@noobiegamer90802 жыл бұрын
sgn(i) sgn == signum function
@cristianv28504 жыл бұрын
(a+bi) tetrated to (a+bi) 😶
@GameJam2304 жыл бұрын
Log base e of the (a+bi)th root of negative pi is where I would put it
@RedRad19904 жыл бұрын
Also, the title of this video be like: *bprp:* Where do you like to put your... *Me:* DAFUQ *bprp:* ...complex number?
@danuttall4 жыл бұрын
Hmm, complex cosine and complex sine remind me of the hyperbolic cosine and hyperbolic sin. Is there an actual connection, other than names and that connection at the end of the video?
Where do I like to put my complex number? i can't imagine!
@DrMaxPlank4 жыл бұрын
@blackpenredpen where is your big whiteboard? what happened?
@stem61094 жыл бұрын
I would like to put that in my head
@claireli884 жыл бұрын
I like to put all my complex numbers in the recycle bin.
@alkaverma59744 жыл бұрын
I want to put my complex number inside factorial
@mateussouza39794 жыл бұрын
Look up the gamma function
@DavesMathVideos4 жыл бұрын
Well, I tried to come here first, but I wasn't first. I like to put my complex numbers to the power of e or in the argument of tranacendental functions like you did. Or maybe in a factorial.
@drpkmath123454 жыл бұрын
Dave's Math Videos wow we share the same preference~
@idontwantanamethx4 жыл бұрын
Can you try out the Trachtenberg System?
@kajarihaldar60614 жыл бұрын
hey. i have been really struggling with a math(INTEGRAL) lately. I dunno know if its right to put it up here under a complex no video but guess leaving it here will grab your(or any other viewers) attention. Will be glad if you or anyone else leave here some hint for solving it. so the question goes: " IF p goes to infinity, then find the value of the integral dx/(1+x^p), limit of the integral goes from '0 to infinity' . It looks like an improper integral, but im not able to solve it, any help is appreciated. p.s: love your videos very much!
@rafaeldubois81924 жыл бұрын
Here's a video from Dr Peyam solving the general case: kzbin.info/www/bejne/rqPSaJZjoqmeZ8k
@kajarihaldar60614 жыл бұрын
Thank you a lot.
@angelmendez-rivera3514 жыл бұрын
kajari haldar Yes, in the video that Rafael Dubois linked, it was proven that integral of 1/(x^n + 1) from 0 to ♾ is equal to (π/n)/sin(π/n). The limit of this as n -> ♾ is equal to the limit of πm/sin(πm) as m -> 0+, which is equal to the limit of p/sin(p) as p -> 0, which is equal to 1.
@mairc92284 жыл бұрын
I like to put my complex number inside an n-th root. Try it and you will get n complex values. Could you imagine the solutions you can get from x^12 = 1?
@angelmendez-rivera3514 жыл бұрын
Yes, the solutions are cos(nπ/6) + i·sin(nπ/6), for n in the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.
@mairc92284 жыл бұрын
@@angelmendez-rivera351 No, I mean the exact simplified solutions. The ones with square roots and stuff.
@saharhaimyaccov49774 жыл бұрын
14:25 'plainus' 👍😂
@我想想-e5d3 жыл бұрын
8:25 than the next lesson could be the structure factor in powder method of x-ray.
@rajatrajmnnit20224 жыл бұрын
I would like to put it inside factorial function
@leoj98544 жыл бұрын
how about sinh (a+bi)?
@valeriobertoncello18094 жыл бұрын
I like to put em in my Gamma Function for sure
@rainbowbloom5754 жыл бұрын
I would put my complex number in the super square root lol
@saradehimi47914 жыл бұрын
Thank you soooo much sir Allah blesses you
@Мартынов-х3ъ4 жыл бұрын
I have little doubts that sin(z) and cos(z) formulas were used in Euler’s formula proof. If it was than it all makes no sense since we literally say a = a. Am I right? PS would be nice if you proof Euler’s formula for a video)
@angelmendez-rivera3514 жыл бұрын
What are you even saying? Of course they were used. And no, it would not be equivalent to saying a = a. You do it via power series.
@Мартынов-х3ъ4 жыл бұрын
Angel Mendez-Rivera Well I wanted to say that if you use A to proof B than you can’t use B to proof A. It’s pretty clear, isn’t it? But the answer with power series is satisfying
I like to square my complex numbers, because i😉 don't like fractions and square roots.☺
@高木清治4 жыл бұрын
θ=arc tan(b/a)
@Be-lo_da_fluff9 ай бұрын
personally i like putting my complex numbers into my integral (i forgot how to do it tho)
@alexismisselyn39164 жыл бұрын
exp() my complex number cuz exp(a+bi) becomes exp(a)*cos(b) + iexp(a)*sin(b)
@srpenguinbr4 жыл бұрын
do W(a+bi)
@naivedyam26754 жыл бұрын
Put it in the domain of arctan.
@Questiala124 Жыл бұрын
I put complex number in the integral limits.
@guest_of_randomness4 жыл бұрын
e^(a+bi) is much more better 😄
@CharIie834 жыл бұрын
I dont like i, does that make me bad at maths? Im down with xy both being negative and the product being negative, but
@-ANaveenBagade4 жыл бұрын
Is the neagtive sign wierd on my screen at the start
@erikkonstas4 жыл бұрын
BPRP fixes the mistakes he catches before uploading like that.
@anandhuh78874 жыл бұрын
*Can we draw the graph of sin(a+ib) ???*
@janami-dharmam4 жыл бұрын
You can use a mesh; use three d plot function on the complex plane
@tomctutor4 жыл бұрын
Hopefully not the last: BPRP I challenge you> try *i^i* ? Or even better 2 to the *base-i* ? Or even better than that *i* to the base_2 so that my computer can store *i* as a number?
@sueyibaslanli35194 жыл бұрын
This is in this channel
@blackpenredpen4 жыл бұрын
I have this (a+bi)^(c+di): kzbin.info/www/bejne/kGWzpH-qbsSer5o
@tomctutor4 жыл бұрын
Ok fair does, just trying to catch you out. Is it real *i^i =exp(-pi/2)* ?
@tomctutor4 жыл бұрын
I think 2 base-i = -i[ (2ln(2))/pi ] and i base_2 = -i [ (pi.ln(2))/2 ] both imaginary so cant really store as a bit number on computer, need someone to check these please?
@muttleycrew4 жыл бұрын
@@tomctutor i^i =exp(-pi/2) is *one* possible answer. i^i has infinitely many solutions because of the polar form of a complex number being equivalent to a rotation of a vector. Consider for instance exp(-5pi/2) which adds a rotation of -2pi to your solution. Another solution is exp(3*pi/2) which has added a rotation of +2pi to your original solution. More generally i^i = exp(-(4n+1)*pi/2) for all integer values of n and yes those results for i^i are _all_ Real-valued.
@alishaaban1554 жыл бұрын
Hi, i have smol question about parametric equation
@blackpenredpen4 жыл бұрын
Yes?
@Darel0074 жыл бұрын
19:16 - "i don't like to be on the bottom, i like to be on the top". OK 😁
@nathanrcoe11324 жыл бұрын
how about complex number double up arrow complex number :P
@nimmira4 жыл бұрын
cosa = zucchini (in Arabic)
@muttleycrew4 жыл бұрын
Which leads to the famous identity, zucchini squared plus sin a squared equals 1.
@atome-ro9yj4 жыл бұрын
I look this video just after the one you speek that the sin(z)=2 video
@iakanoe4 жыл бұрын
Plinus
@WindowsXP_YT4 жыл бұрын
What about 3^(a+bi)?
@Alians01084 жыл бұрын
x^(a+bi) = x^a*x^(bi) = x^a*e^(bi*ln(x)) = x^acos(blnx) + ix^asin(blnx) x cannot be 0 I assume. Plug in 3 for x
@mateussouza39794 жыл бұрын
Same thing, but replace every 2 with a 3 in the formula. You can do that for any positive real number n.
@angelmendez-rivera3514 жыл бұрын
Osu! Top Replays x can be 0 if and only if a > 0.
@hww31364 жыл бұрын
i like to put my complex number into inverse sine
@hww31364 жыл бұрын
well, I tried myself with arcsin(z)=-i*ln(i*z+sqrt(1-z^2)), though the equation became just too long
@angelmendez-rivera3514 жыл бұрын
hww Yeah, that is how you would do it.
@sujanrajakannan9 ай бұрын
In cubrt
@nathanaelmoses79774 жыл бұрын
Plus or minus,ah too long Just use pluinus 14:23 : )
@sueyibaslanli35194 жыл бұрын
😅pink is good
@guest_of_randomness4 жыл бұрын
how many mic do u have?...
@blackpenredpen4 жыл бұрын
WenKang Chiam : ))
@lightyagami66474 жыл бұрын
@@blackpenredpen where can we ask you questions
@sueyibaslanli35194 жыл бұрын
Hi from Azerbaijan
@headlibrarian19964 жыл бұрын
I thought this was black pen - red pen not blue pen - red pen 😀