Відео

What about 13500

2³=26-3•2(³√(x²-13²)) x²=(⅙(26-8))³+169=196 x=±14

Friend may I ask: I thought you made a similiar question somewhere and it was from Olympiad but I cannot find it on your site here! Where is it? Also friend - is the key to understanding this, to learn about symmetric polynomials?

It had already been done in 2009 by someone first name Jason I forgot the last name. People are just trying to make a big deal over it because the two who supposedly figured it out are black. FACT

Hmm...looks like the prisoner is sentenced to death

A perfect video

Excellent! Thank you ❤ I searched for the derivation of this formula for a long time and got it only in your video!

How did you get y= r/h x z

by the way the other girl had a different proof not this one. She used a circle

Very helpful great work 👍

It's a shame that he did not pursue the fact that x=(-1+sqrt(3)i)/2 or x=(-1-sqrt(3)i)/2, because from this it immediately follows that x^(3n+k)=x^k and that x^n+x^(n+1)+x(n+2)=0. Hence, x^77+x^76+x^75+x^74+x^73=x^2+x. But x^2+x+1=0, hence x^2+x=-1.

n=9216 n=6156

OMG! THANK YOU x100!!!

a=3, b=1 I can't believe I solved it! I'm going to give myself a break from maths for a week 😊

Excellent!

I was in 6th grade a year or so ago and I learned basic calculus and trigonometry very easily on UA-cam.

PLEASE DO THIS FOR THE OTHER FIVE EXPONENT LAWS!!!! I AM BEGGING YOU!

I so much like the technology you employ to make illustrations easy. Please, what software do you use in making your videos?

ohshit

The title of the video is missing a word.

f(x)x⅔+(3.3-x²)½.sin(aπx)

3,6,9,3,9,6,

Shout out to St Mary’s high school.

Why do you use the inverse tangent rather than tangent?

I don't think the gravity of this proof has been truly felt...

Bakwass

Great Mind Thank you very much 🎉❤😊

Wat

I am sorry I still don’t understand

I still don't understand.. how come the denominator can represent all the possible outcomes? C(N,n) only tells us how to pick up 5 elements out from a set of cardinality of 25 , it doesn't matter if they are m or f.. So how come this can represent all possibile cases, if this number contains NO information about the "sex" ?

The problem with this is "knows". Albert knowing Bernard doesn't know only eliminates 18 and 19th. If the birthday was May 16th, statement 1 shouldn’t eliminate May and also June. The problem is how the information between B and A are communicated.

Brilliant very underrated

Yea the foil method was now already answered on the notebook

Fantastic and aaasam bro thank you so infinity ♾️ it is use ful for me❤

oh wow, its easier than it looks!

49x7=343

I found this phillipine map on 10

This is just a preexisting geometric series that they found and they put a triangle on top. I know because I found an older version of the exact same series with completed calculations. This is more the product of dumb luck than any real intellectual insight. Look up: "math.stackexchange Is this series representation of the hypotenuse symmetric with respect to the sides of a right triangle?"

How is their proof anything other than a copy of John Arioni's proof at cut-the-knot (proof 100)?

Surprisingly, it was easier to solve the Pythagorean theorem problem than to prove that a woman with a penis is not a woman!!

simple and beautiful, thanks for the explanation.

thank u sm

Brilliant! Proud of their hard work and success!

Plz make a video on how find cube root

What if its the square pf a 3 digit number or more?

It is combination of trigonometry and algebra. So we may say it a trigonometric -algebraic proof.

Simply assume x = 2 Thus square root both side you get the answer... Which is x = √2 I.e 1.414 approximately

By inspection x³+x²=12 can be written as x³+x²=2³+2². Hence x=2. To find other root, if any, note that the last equation can be written as x³-2³+x²-2²=0 (x-2)(x²+2x+4)+(x+2)(x-2)=0 (x-2)(x²+3x+4)=0 x=2, x=½[-6±isqrt(3)] For real root x=2 is the only one. Other root is a complex one.

Sir i have one doubt x4 is always zero . Sir please clarify my doubt or you make a videos on this topic .please

hi