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@AG-mohamed

17 күн бұрын

لو انني كنت افهم الإنجليزية لكان الأمر سهلا معك

@BajrangJat-mr1gu

16 күн бұрын

I can take this course

@hiyayahiyaya5645

16 күн бұрын

I want to say that obviously , 8>√5>2 so [5-√(5)]/8

@bozydarziemniak1853

11 күн бұрын

Hello blackpenredpen! I have found a pattern for a prime numbers.

- Can't be negative - Can't be 0 - 72° > 60° so the sin has to be greater than √3/2, but √((5-√5)/8) is clearly less than √3/2 - Therefore A is the answer

@gagadaddy8713

18 күн бұрын

Oops! You got bprp this time 😆

@z000ey

18 күн бұрын

Same line of thought. Had the proposed answers had more diversity, twould be much much tougher to evaluate

@vashista4583

18 күн бұрын

We actually have to remember angles like 18,36,54,72,7½, and other angles🥲🥲

@orangee_blox

18 күн бұрын

process of elimination is always simpler 😂

@josepherhardt164

18 күн бұрын

Exactly. You can eliminate three answers off the bat, and the (d) value looked suspiciously low to me (did a bit of mental back-of-the-envelope calculation) and I settled on (a).

Nah, ain't no way. I just did this EXACT question at school today. I come home and, boom, there it is

@ShreshyaR

12 күн бұрын

U from India?? Just asking cuz we were also taught this in 11th grade

@rudradutta2741

10 күн бұрын

​@@ShreshyaRyes. I also did in 11

@BroomieHERE

4 күн бұрын

We all study this in 11th not just indians​@@ShreshyaR

@4fgaming925

3 күн бұрын

@@BroomieHERE I studied in 10th in india...

@StudyOn-yv4wj

6 сағат бұрын

​@@4fgaming925i m studying in 9th... basic maths in physics and vectors.

Commenting before watching the video, I eliminated the negative because sin 72 is near 1 than zero, so the square root has to be near 1 so we must choose the bigger alternative

@formaniclv1046

18 күн бұрын

True

@gagadaddy8713

18 күн бұрын

Yes! If this question is really a multiple choice, we don't even need to solve the equation. Just simple elimination can figure out which one is the real answer. That makes our life, may be an Oxford's life, more easy! 😂

@barrym5310

18 күн бұрын

My thoughts, as well.

@mlgswagman6002

18 күн бұрын

Does not necessarily mean we must choose the bigger of the remaining two. However, note that 5 - sqrt(5)

@anigami01

15 күн бұрын

let's suppose the values given are a. 0 b. 1 c. .973 d. .987 now you can't solve like by eliminating them

6:55 37 appears everywhere, even in your jokes :)

@DodgerX

16 күн бұрын

Fellow veritasium follower

@sowndolphin5386

14 күн бұрын

veritasium was here

@utvikrama

12 күн бұрын

Funfact : 37 is the only number in the entire universe which is spelt as Thirty Seven

@lanye2708

12 күн бұрын

@@utvikrama wow... i can't believe it. do you have a source for that??

I solved this question knowing that in a pentagram, if you cut a point, it will form an isosceles triangle with angles 72º, 72º and 36º. But, in a pentagram the aurum ratio is a rule, so if the base is equal to x, the equal sides are equal to x((5^0.5-1)/2). Applying the cos rule, you can find that cos72º = ((5^0.5+1)/4). Therefore, the fundamental law of trigonometry garantees the sin72º = ((5+5^0.5)/8)^0.5, as sinx >0 for 0>x>90º

@trueriver1950

16 күн бұрын

Aurum ratio? I've not heard it called that before, so initially I was puzzled till I recognised the quantity you gave. Then I realised: Aurum = pertaining to gold, as in the chemical element. Yes, this is just a posher name for what I was taught as the golden ratio. Thanks for your comment: which turned out to teach me something about language as well as maths :)

@oliveirapoli1

15 күн бұрын

Thanks for the clarification. In my native tongue it's called "razão áurea", as in spanish "proporción áurea", in french "nombre d'or" or in latim "aurea ratio"... I forgot that in english it doesn't follow the same pattern. All those words come from the hebrew "aur", which means divine light, hence the creationists idea that this ratio is present in all creation (which of course is just a point of view).

Indian student do this is 11th class maths as using sin (90-12) =cos (12) and x=12 5x=90 3x=90-2x Cos 3x = cos (90-2x) 4cos^3 x -3cosx = sin 2x 4cos^3 x -3cosx =2.sinx.cosx 4cos^2 x -3 = 2sinx Covert cos^2 to sin^2 then use the quadratic formula u have your answer

@justnowi8967

8 сағат бұрын

It should be (90-18)

golden triangle has 72°, 72° and 36° and sides a golden ratio (phi) larger than the base. Split the base in half to obtain the right triangle with 72° having a base of 1 and hypotenuse of 2 phi. The remaining far side of 72° will be phi*√(phi +2). Sin 72 = √(phi +2)/2 where phi is the positive golden ratio = (√5+1)÷2 √((√5+1)÷2+2)/2 = 0.951056

I love these kinds of admission test math problems. Would love to see more of your videos on this

It's not necessary to notice that 36° is also a solution. By inspection sin(72°) is pretty close to 1, approximately 0.95, so (sin(72°))^2 must be approximately 0.9. Now (5 - sqrt(5))/8 is much too small and (5 + sqrt(5))/8 is the correct answer.

@KaiserBob99

17 күн бұрын

You can't compute the actual value of the sin. You have to figure it out yourself. You can't say it's 0.95 therefore it's that thing

@borstenpinsel

16 күн бұрын

Of course you can. If the 5 solutions were reasonably close, your have to it the hard way. But if 3 options eliminate themselves right away and you have a rough estimate of the ball park and one solution is way off, tadaaa. ​@@KaiserBob99

@ghostboi71

3 күн бұрын

​@@KaiserBob99 At least guessing is ok.

Nice - I sorta got stuck on the last step, but your explanation made perfect sense.

Beautiful math excercise. ❤

I'm learning to like maths. It's slow but i'm beginning to appreciate it. That's after having my potential love of maths beaten out of me by a less than ideal experience of being taught it at school. My algebra is at the point where I followed everything that was done here without scratching my head. Although I probably could not have solved it on my own. Slow and steady. I'll get there in the end.

Alternatively, consider sin(3x) + sin(2x) = 0, cancle one sin x after expanding and replace sin^2 by 1 - cos^2. We have a quadratic is cosine which is simpler to solve. Side note : Depending on what kind of education system you are in, you may have some trig values of 18 and 36 degrees memorized, in which case the question is trivialized. There also probably is a complex number solution but I am too lazy to find it.

Recall that sin(θ) is the perpendicular height to the x-axis from the point of intersection of the unit circle and the line making the angle θ. Clearly, this height is not 0, and also it is above the x-axis, so it is positive. We can therefore exclude B, C and E. We know that sin(45°)=sqrt(2)/2≈0.7, so sin(72°) must be greater that 0.7 But sqrt(5)≈2.2 So option A ≈ sqrt(7.2/8), and option B ≈ sqrt(2.8/8) From A and B, only A is greater than 0.7. As a multiple choice question, this is a one minute question.

So it doesn't change the fact that I don't like minus and I remove it, and I am still correct

Very insightful

Normal question: 95% time solving equation, 5% time choosing the correct answer This question: 50% time solving equation, 50% time choosing the correct answer I have never seen a question that takes a long time to filter and reject answers.

At school I just did this question. When I saw 36 degrees, the golden triangle was the first thing that came to my mind. Then using sine rule to find a relationship between sin72, sin36, and (√5-1)/2. I got the same answer. However, I tried to remove the root sign, hoping that 5+√5 could be expressed as the form ()^2. Then one hour was wasted.

It’s worth noting that at 4:55 we arrive at the four choices which are given in the paper. We could have started there, and in the real exam, you should, as it saves time. Nonetheless, I appreciate seeing the derivation of the solution. It would also be interesting to see where the trig identity comes from. It comes from applying de Moivre’s theorem.

Very nice. I would never solve this myself but love seeing the magic happen haha

This is the Best content on KZread in the world 🙏🏿🌍!

I had a similar question where we were supposed to find sin(72) on a complex analysis final.

Facinating question

Teaching skill Love it

I used the same approach you did to calculate sin72 = x, obviously x cant be negative so we igonore that solution, now to determine if it sqrt((5 +/- sqrt5)/8, I set up a condition where sin72 lies between sin 60 and sin90 since sin is increasing from 0 to 90 degrees, we know that sin 60 is sqrt3/2 ~ 0.866 and sin 90 = 1 so when we approximate the two solutions, the one with + sign is ~0.951 and the one with negative sign is ~ 0.56 which does not satisfy our condition, there fore the one with + sign is the answer which is option A

do more calculus 3 question on you channel, i really like them

Exelente Ejercicio. 😃

Tried and also done 👍🏻

Sir videos help me a lot to make my concept very strong in mathematics love from India❤❤❤❤

From scratch it will be using similar and isosceles triangles Angle is acute so we can immediately eliminate (b) , (c) , (e) Sine is increasing in first quadrant so after comparing (a) and (d) , (a) is closer to one than (d) so we can suspect that (a) is correct but at this moment we can not to be sure that (a) is correct (In fact (a) = sin(72 degrees) and (d) = sin(36 degrees) )

@Misteribel

17 күн бұрын

Yes, you can be sure, as it needs to be quite close to 1, and (a) is the only one close to one. Or, since sin(60) is sqrt(3/2), it's clear out cannot be any other value.

I love your teaching so much ❤ form Cambodia teacher

Thanks for improving my "trigonometry" skills

You could also use a simple complex numbers strategy. Let z = exp(2πi/5) which is a fifth root of unity, therefore it satisfies z^4 + z^3 + z^2 + z + 1 = 0. We know that z^4 = 1/z^2 and z^3 = 1/z so this equation becomes (z^2 + 1/z^2) + (z+1/z) + 1 = 0. We can turn this into [(z + 1/z)^2 - 2] + (z+1/z) + 1 = 0. Setting x = z+1/z gives x^2 + x - 1 = 0, which has solutions x = (-1± √5)/2. We reject the negative solution, giving us that z + 1/z = (-1+√5)/2. We know that z + 1/z = 2cos(2π/5), so cos(2π/5) = (-1+√5)/4. We use the Pythagorean identity to find that sin(2π/5) = √[(5 + √5)/8]

I think we ought to just compare the answers to sin(60) BUT to perform it differently i just did: sin(72)=sin(45+27) with addition formulae, and then again i splitted (27) as (30-3) and used same formulae. Then i approximated sin(3)~0 and cos(3)~1, last thing i did few simple divisions to compare and A is the answer :)

Very good video ! I have a question I would love you to answer : for any ineger n, what is the integral going from O to 1 of (lnx)^n

Omg u just took me back to trigonometric algebra thnx 💀

All the simple right triangles come from the √ 2, √3 and golden triangles. Triangles with sides 1:1:√2, 1:,√3,,2 and 1:2phi:X Which came from dividing the square, equilateral and golden triangle. Knowing this, then no trig formula is required.

@MonkeyDLuffy-ed4fj

17 күн бұрын

Triangles have 3 sides, that’s all I know

There are a lot of ways to approach this that I took in a vastly different way and still got the same results :3

very nice....i understand it, only i wouldn't know where to start

I actually remember doing this question 2 years ago for the MAT

I guess you can also write the equation as 4^2.x^4-4.5.x^2+5 and transform the equation a quadratic in terms of 4. then you can use the quadratic formula to find out what equals 4 and pull out the x value from that. a lot more work but its kinda fun that you can write a quadratic in terms of 4

@MonkeyDLuffy-ed4fj

17 күн бұрын

Those are for sure numbers

For the finaly answer, can I say that: Since Sin(45)=root(2)/2, and sin(72)>sin(45), so the answer should be lager than root(2)/2. And since root(5) > root(4) > 2, and 5-root(5)

In order to choose the right value between the + vs -sqrt5, you can compare to the value of sin45

sir please make video for calculus 3 for multivariable case

I've been your fan for so long.Could you please help me with this integral (Infinite Integral of xlnx/x+1)? Thanks in advance ❤

Please do integral of e^x/x dx, without Taylor or MacLaurin Series expansion. Thank you very muchA

can you do Singapore H3 A level math? been suffering through those although the threshold for distinction isn't that high

Can you do integral of 1/(x^i)?

It would be so cool if you actually did a maths Oxford Admission. Other KZreadrs have done it with Tom Rocks Maths.

The 37 joke was cute 😂

Master, try some ITA and IME from Brazil!

Can we just eliminate the minus because we dont like it? I laughed so hard 🤣

Before you showed the answer, I squeezed the answer to be greater than root 3 / 4, and used inequalities to prove that +-root 5 cannot be negative 😂 What a way of solving the issue - understand the properties of the value you are dealing with - in this instance sin 72 was simply greater than sin 60 ...

You can approximate which one it is numerically, without using the identity they give you. And yes, you can do this without a calculator, remembering sin(60 deg) is approx 0.866. First, we can rule out the two negative choices, and zero. From our special angle sin(60 deg) = sqrt(3)/2, which is approx 0.866, we know the answer must be between this and 1. This narrows it down to option A and option D. sqrt(5) is approx. 2.25, which we can find by linearizing sqrt(x) at x=4 to get L4(x) = 1/4*(x-4) + 2. Plug in x=5 to L4(x). (5 + 2.25)/8 = 7.25/8 = 0.875 + 0.03125 = 0.90625. Linearize sqrt(x) at 0.81, to get L_0.81 (x) = 5/9*(x - 0.81) + 0.9. Plug in 0.9 to approximate the result. 5/9*0.09 + 0.9 = 0.95. This confirms option A meets the range where we expect sin(72 deg) to be. Now try option D: (5 - 2.25)/8 = 2.75 = 2/8 + 3/32 = 11/32 is approx 0.34 Use nearest number with rational square root of 0.36, to linearize. L_0.36 (x) = 5/6*(x - 0.36) + 0.6 L_0.36 (x) = 5/6*(-0.02) + 0.6 = 1/6 + 0.6, is approx 0.76667. This rules out option D, and allows us to conclude option A is correct.

I can calculate value of trigo ratios degree... 5,105,18,20,22.5,5,,.25........

You should try the 2011 IMO question number 2 it is very fun

one with the greater valve as sin is an increasing function

Based on the fact that it's multiple choice, I wonder if you could've cut straight to the analysis at the end, without doing the algebra. Like you could throw out all the non-positives right off the bat; then show that a and d both satisfies the given equation for theta = 72 deg; and then do a little geometry to show that answers a and d are actually talking about congruent triangles (a^2 + d^2 = 1); and the correct triangle is a.

I saw the blue pen 😂

I finally got something right in one of these videos :D

This is so useful

How often does real life come in multiple choice? As an engineer, I'll say sometimes it does, and it really saves time and effort to eliminate the answers that are clearly incorrect and go from there. In this case, one can find the right answer without doing any of the work.

@carultch

14 күн бұрын

One place that real life problems come in the form of multiple choice, is with product selection. You may not need an exact answer, but you just need a best-enough answer. You have a finite number of options to choose from, since manufacturers only make certain sizes. As an example, consider the sizing of a water pipe. Suppose you are given the flow rate, the allowable pressure drop, the length and type of the pipe, and need to determine the diameter of the pipe. This is a difficult problem to solve directly, because turbulent flow (which water usually has) doesn't have an analytic solution, and the standard equations require you to know diameter and velocity in advance. It also isn't possible to algebraically invert the combination of equations to solve for these values. But you don't start with diameter and velocity; you start with flow rate and pressure drop. So, what you can do in practice, is use the options for diameter as inputs to the calculation, and find the corresponding flow velocity. From then on, the problem is ready for you to directly solve for pressure drop, and match it to your given data. The smallest pipe that produces a pressure drop less than the allowable pressure drop you were given, will validate selecting that pipe diameter.

@carultch

14 күн бұрын

To put specific numbers on my example: Flow rate m_dot = 8 kg/s Max allowed pressure drop deltaP = 60 kPa Length: 20 meters PVC pipe sizes considered: 25 mm, 35 mm, 50 mm, and 75 mm, internal diameter. Corresponding speeds at each diameter, using m_dot = rho*v*A, solved for v: 25 mm: v = 16.3 m/s 35 mm: v = 8.31 m/s 50 mm: v = 4.07 m/s 75 mm: v = 1.81 m/s Reynolds number at each size: 25 mm: Re = 407436 35 mm: Re = 291026 50 mm: Re = 203718 75 mm: Re = 135812 Friction factors at each size, from the Colebrook equation: 25 mm: f = 0.013 35 mm: f = 0.014 50 mm: f = 0.015 75 mm: f = 0.016 Pressure drops, from deltaP = 1/2*rho*f*(L/d)*v^2: 25 mm: deltaP = 1373 kPa 35 mm: deltaP = 273 kPa 50 mm: deltaP = 49.1 kPa 75 mm: deltaP = 6.995 kPa Of these options, the 50 mm is the smallest choice meets the criteria. 0.479 mm would be closer, but we only care about which of the options given we'd recommend.

@user-gs6lp9ko1c

14 күн бұрын

@@carultch Great example!

Wow I'm speechless you are legendry so MAT for who ?

All 5 solutions of the original equation (0, two positive and two negative) are valid. For angles: { 0, 36, 72, 108, 144 } + k * 180 for integer k including zero. For an even k they are 0 or positive and for an odd k they are 0 or negative.

I got right until the last step rigorously, but I didn’t spot the sin(36) trick, so I had to approximate some square roots in my head which led to me having a good guess that the larger one is greater than sqrt(3)/2 and the lesser one is less than sqrt(3)/2. Since sqrt(3)/2 = sin(60), the answer follows.

U know already that the sinx function is increasing in the interval (0,π/2) so since 72° = 2π/5 > π/3 we have sin π/5 > sin π/3 = √3/2. And you can check that √ (5 -√5)/8 < √3/2 by calculation. So it is automatically dropped.

Hi bprp, I have "found" an interesting thing, if you compute the integral of the function x^(ln(x)/ln(1/2)) from 0 to infinity i dont' know why but you will get the square root of pi in the result. I will be really happy if you can make a video about this, thanks.👍👍

@omkarjoshi9137

12 күн бұрын

It must turn into the Gaussian integral after replacing x with e^t

∆ABC, AB=AC=1 and ∠A=36° Let D is on AC such that AD=BD then ∠BDC=∠C=72° and ∆ABC~∆BCD (AA) Let AD=BC=a ( cos72°=√[1-(a/2)²] ) then 1:a=a:(1-a) => a²=1-a => a=(-1+√5)/2 Thus cos72°=√(a+3)/2 =√[2(5+√5)]/4

Pretty sure there is an exact value for sin(36°), could use the sin(2x) formula

Somewhere around 24/25

Can you do the same for sin(73) ?

You can see you're going to get a quadratic formula for s², so take the positive square root. Then 72° is in a positive interval for sin, so take the positive square root again. That gives A.

@MonkeyDLuffy-ed4fj

17 күн бұрын

Those are for sure a bunch of words, do I know what they mean, no

You have to go with the greater positive answer, simply because 72 is approaching 90

My idea to solve this even without the given equation would be to draw a unique 72/18/90 deg right triangle. The only thing we need to do now is to scale it to the proper size. It is obvious the answer is either A or D so we know side/hypotenuse ratio and can pick them to fit the answer. At this point i dont have a pen and paper to actually go forward. Any thoughts?

@MATHS_FOR_FUN

18 күн бұрын

That's nice but the proper angle doesn't drawable so can only approximate the value .

@maxrs07

18 күн бұрын

@@MATHS_FOR_FUN u dont draw to look at it u draw just to visualize the relations for angles and side lengths lol

@trueriver1950

16 күн бұрын

​@@MATHS_FOR_FUN the proper angle is drawable: you just need more than a ruler and compass...

Nobody ever remembers that 5x angle formula and it takes a while to derive. I’d stop at 3x for 54 and 2x at 36 and know that 54 + 36 = 90 so sin of one is cos of another. You’ll find sin18 in two minutes.

It is another way of asking about pi/5. We will have to solve a 5th degree poĺynomial.

👍 job

Pls make the proof for sqrt of a+sqrt of a - sqrt of a so on

i got this question for 11th grade

Can you solve this high school integral ∫(1+xcosx)dx/x(1-x^(2)*e^(2sinx))

For me in India we were taught sin,cos,tan of 18,72,7½ so on angles and i actually remember sin72 as =√10+2√5÷2√2

@andromeda16384

17 күн бұрын

Wait WHAT

@vashista4583

17 күн бұрын

@@andromeda16384 yes and more than these it's not necessary to remember all of them but we need to remember multiples of 36

@NadiehFan

17 күн бұрын

Actually it is sin 72° = ¼√(10 + 2√5)) sin 36° = ¼√(10 − 2√5)) and you also have sin 18° = ¼(√5 − 1) sin 54° = ¼(√5 + 1) I also know all this by heart and no, I'm not from India.

@MonkeyDLuffy-ed4fj

17 күн бұрын

I don’t know what any of that stuff means :)

@vashista4583

17 күн бұрын

@@NadiehFan oh yeah it's ¼ for all and damn nice bro

In India we were asked to memorise special angles (18,72,36,54)..

@asd-wd5bj

15 күн бұрын

Most schools do that, that's why they specifically asked you to find it given the equation at hand. "I remember sin(72) from memory" would not be a valid answer

@ghostboi71

3 күн бұрын

​@@asd-wd5bj When there are multiple choices use memories. But not in the description.

please solve this limit for me, I am trying to figure out this from long time lim x - 0 ((1+x)^1/x-r+ex/2)/x²

When the question is way easier if you just use logic and common sense instead of full math lmao

Come to India these are the basics

@divinecreation7429

13 күн бұрын

nope

Why can't you just draw a sin graph to determine sin(72)>sin(36)

@blackpenredpen

9 күн бұрын

You could.

72*5 is 360 and we know sin360 is 0 and in the end it becomes a matter of getting the correct zero

В Оксфорде нет таблиц Брадиса, что ли?

Day 1 of asking u to solve 1/(cos(x)+x²)

You shoudl try the joint enterance exam advanced 2016 paper. Its a tough pill!

@cdkw2

18 күн бұрын

Agreed

@fibby7069

18 күн бұрын

You mean a tough suppository?

@Sa-adMP

18 күн бұрын

Please send the paper

@cdkw8254

18 күн бұрын

jeeadv.ac.in/past_qps/2016_2.pdf

@cdkw8254

18 күн бұрын

This ones for shift 2

Looking at the problem, lol, I'm guessing you can take theta = 12...we know sin(60), and that is sin(5 theta)...replace all the theta's with 12's, etc, lol...then, of course, one has to solve an equation of degree 5, etc, not sure about that...the moment one has sin(12), knowing sin(60) (I don't actually remember what sin(60) was, lol, it was either 1/2 or sqrt(3)/2, but it's something that is usually known), one can easily find sin(5theta + theta) using the sum formula, something like sin(x+y)=sin(x)cos(y) + sin(y) cos(x) or something like that, don't quote me on that one, lol...that is, we would know sin(72)=sin(5x12 + 12), etc...perhaps that's the trick, the only part I don't see is how to solve the degree 5 equation, maybe the fact that there are no 4th nor 2nd degrees will help, lol...if it weren't for that part, I would consider I basically solved it, lol...I mean, of course, it's not like I actually know the sum formula or the sin(60), etc...72 might also be a multiple of 18?...Ah, yeah, lol...perhaps that would be an easier way?...But it's not like we know what the sin(18) is, traditionally, etc...we could use sin(90) = sin(18 + 72), etc, using the sum formula and the formula provided in the video, etc...at any rate, maybe that would avoid solving a quintic...at any rate, lol, my first thoughts on how one could try to solve this problem...

@archangecamilien1879

18 күн бұрын

Maybe 18 would indeed be better, lol...I mean...90=18x5, 72=18x4...we know what sin(90) is, lol, that would give us an equation involving sin(18), but a quintic again...hmm...it's the quintic I'm not certain about...

@archangecamilien1879

18 күн бұрын

Ok, maybe one can reduce the 5th of the equation to something else, using some square angle formulas or something...I think there was a formula somewhere that involved sin(2theta) = some function of [sin^2 (theta)]...that might reduce sin^5(theat)=sin^2(2theta) * sin(theta), etc, we can perhaps use some identity to get rid of the 5th, and even the third of the polynomial...in other words, it would probably be manageable if I remembered more trigonometry, lol...I don't remember any identities...

Plzzz Solve a^a=a

without solving: zero is obv not the answer it cant be negative (value is in the range [0,1]) and the one which is bigger is the answer (since it is close to sin 90, i can infer that the answer is closer to 1, so, a, yup) did this in literal 1 minute lol

@UmairMalik-rj3in

18 күн бұрын

yh its literally Q1 F, these are questions you are supposed to run thought in less than 3 minutes each. The actually harder questions come up later

@loonaloonaloonaloona

18 күн бұрын

@@UmairMalik-rj3in ooh i didnt know that, thanks

@ghostboi71

3 күн бұрын

​@@UmairMalik-rj3in Wait, you are saying, you have to prove that "The answer is 'A'"?

Plzzz Solve the eqn a^a=a

@Gordy-io8sb

17 күн бұрын

1 seems to be the only solution, even considering rings with characteristics a (assuming a is prime). So yeah, a=1. 1^1=1. This is extremely trivial. I don't think bprp will solve this.

Thank you for the upload! This method can be extended for proving constructability of polygons, and I have a nice visualization of proving with just ruler and compass! kzread.info/dash/bejne/h2GBq7qnf8bHZ7w.html

No radians?

@Gordy-io8sb

17 күн бұрын

I thought one of the most prestigious universities in the US or otherwise would know to use radians lmao

@asd-wd5bj

15 күн бұрын

@@Gordy-io8sb Degrees are more common in secondary education due to being considered more intuitive. The test is for highschoolers, so it makes sense to use the notation that the kids would be more familiar with. Radians are awesome for uni math, but here it literally doesn't matter

@asd-wd5bj

15 күн бұрын

@@Gordy-io8sb Also Oxford is in the UK

adding 0 as a choice B is the dumbest idea.. sin0 is 0.

Its easy

F**k that I'll just guess! The graph of f(x)=sin(x) looks like that way it's bigger than sin(60 degrees aka pi/3). Therefore it's between sqrt(3)/2 and 1. A is the answer. And that sine wave is common sense in any of high school students.