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Loved this video. Great work.

100% = 1 sqrt(100%) = sqrt(1) = 1 = 100%, so its c

Its so easy 1+2=3 2+3=5 3+5=8 8+5=13 Why are you making it so complicated

W(2ln11) or - like in the video - W(4.7958) is approx. equal to 1.303 instead of 1.409. The correct solution differs by ~0.32 since e^W(4.7958)=3.6804...

Sorry, but that answer is completely incorrect, obviously so, as 4^4=256, already too large to be the answer, the answer is actually about 3.68.

Somewhere between 3 and 4 for the rational answer.

Sometimes "guess-and-check" and a bit of insight can be quicker than an analytic method. Guessing that an integer solution is available, we know that integers squared are mostly bigger than the integer, so we are probably going to find a solution when a^2 is a bit bigger than 43. Try a = ±7, which makes b = 49 - 43 = 6. Then b^2 = 36, making a = 36 - 43 = -7. Yup, that is a solution. So a^2 + b^2 = 49 + 36 = 85.

Isn't that the square of 11

the x must have been between the interval (3;4) i dont think the answer is correct

It's nice

Hello, this is a good video! And you sound like a good person

This seems way more complicated than necessary? It's simply a substitution problem Step 1: Set equations equal to find b, in terms of a a2-b=b2-a a2-b2=b-a (rearrange terms) (a+b)(a-b)=b-a (left side is difference of squares) a+b=-1 (after dividing both sides by a-b, which isn't 0 because it's given that a/=b) b=-a-1 Step 2: Substitute b=-a-1 in the first given equation and solve the quadratic. a2-(-a-1)=43 (substituting b=-a-1) a2+a-42=0 (arithmetic, shuffle terms) (a+7)(a-6)=0 (factor) Thus a=-7 or a=6 Step 3: You now have a-=7, b=6 (or vice versa), thus you have the answer: a2+b2=49+36=85

I tell you the answer value for the big X and small x are 11

But, x=4.093 does not satisfy the equation : x^x = 121 because (4.093)^(4.093) = 319.95....

√(100%) = √1 = 1 = 100% √(100%) = √(1 00/100) = √(100) / √(100) = 1 = 100%

sqrt(100%) = sqrt(1) = 1 = 100%

stupid

I don't like the notation to be honest

Thanks alot for your handwork sir❤

By the thumbnail, I solved it like this (I havent watched the vid, YET): 2a^2 - 2b^2 = 98 divide both sides by two a^2-b^2=49 (a+b)(a-b)=49 Set up a system of equations: a-b=1 a+b=49 a=1+b 1+b+b=49 2b+1=49 b=24 a=1+b a=1+24 a=25 Edit: he did the same thing as me

Isn't a=7 and b=0 also a solution?

Thanks alot sir

X=1

6! times 7! is obviously 43! Case closed

wow

I'm guessing you find the square roots of 48 and 49 and subtract them and then find the square root of that?

X=1

13, 21, 34, 55, 89

13

This is hardly a 15 sec ans. 7! * 6! Just elaborate 6! 6*5*4*3*2 Form next numbers of 7! i.e 8.9.10... 8=4*2 9=3*3 Left with 2*5 = 10 So 10!

-> Assume answer exists in form integer factorial -> realise answer must be larger than 6! -> realise answer must have a second factor 5 on top of the 5th index, hence must be at least 10! -> realise answer cannot have any prime numbers larger than 7 as factors, hence must be less than 11! answer is 10!

3 million factorial, is astronomical.

Wtf am i doing here

So simple X=4 Y=3 X²+XY=28 Y²+XY=21 So 4²+(4*3)=28 16+12=28 3²+(4*3)=21 9+12=21