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p and q could be both positive or both negative, so we may assume WLOG they are both positive.

very elegant!

No need to open convert all of them into perfect squares

WHAT WORD DID YOU SAY? “The answer is yes…if we know how to use (#*%}*#) here.”

For sure it's tough ......... but not toughest.....

seems that you just make it up by reverse engineering. as we never know how can you make up all those cosine formula for a, b , c ,d and let e and f work separately. better to study the real geometry and their roots relationship. Though it seems you have find solution by your approach but the initial thinking is not convincing. Is it really the original solution from Gauss, can you provide more information or the related resource for this approach

For the bijection f: (0,1) --> [0,1]... Could I just switch the domain and range? So f(1/2)=0 and f(1/4)=1 to account for the endpoints. Then f(x)=1/2^n (n=1,2,...) if x=1/2^(n+2) and f(x)=x otherwise.

THANK YOU!!!

clear explanation

Bhai Hindi me bhi bata diya karo

How did you find those x, y so quickly? Can you provide some reference? UPD: Got this. It's directly from an unequallity.

The actual solutions are (1,a,-a),(b,1,-b),(c,-c,1) where a,b and c are arbitrary integers and also (1,2,3)(2,4,4)(3,3,3)

Thank you

Plz solve IIT paper

LOW volume pls increase it or change the mic

What about n=0

How did you come up with the sequences for the a^2 + b^2 + c^2 one

Complex numbers solution will be a lot easier

Kaprekar's findings are incorrect There are numbers that do not produce the number 6174 I dare to be tested.

0

Hey what is the the answer for hhtt in a row i am getting 16 is it correct?

1/³a² b²-> ¾->¼->⅓¾->3/3/6/3/3/3->3->13

2 1/in¹-1/in <2

¹¹/16sin²x+cos²x½x² y² sinx³ +cos³x=½+׳->11/16

A¹ a² a³ a⁴ a⁵ a⁶ a=18=a² a³=18->¼

Mnu->-2 -1 0 1

/2×+5>×+1>-2 2 -× 2.5

3x-1ײ->⁶ax⁶ ax⁵ ax⁴ ax½ axⁿ->3-1-1->1-1-3->ax³-axⁿ->ax⁶->f(1)-f(1)

Ax⁶-ax³ ax⁴ ax² ax ¹->3-1-1 1-1-3->f(1)-f(1)

3m-2n

Sin -1 ½ 1pie/¹-

A²+b²/c² sinab²->4t-4i->2i/2/2->2ti

B=a²+b²=c²=Sina sin b/sin²c c²Sina/sinb=sinb->pie/4²-a/2->b->2pie/²

C<a<b>a=0.1e b =⅛+1c=1/9inx²-1=Infinity!/E- k->0!->x¹/ײ! X²/×¹+x³/ײ!->c<a<b

😮

This is wrong. What is your reasoning for ignoring the higher powers? If you put x=0.9 you will find that -ln(1-x) > xe^x so exactly how small does x have to be for your theory to work and why?

Nice Approach

Shldn’t it be 9<= x-y-z<12

US and NATO planes and drones fly from the Black Sea every day for 3 years. What exactly is your problem?

Can other countries student give this exam Pls make video on full detailedinformation about This exam

w

Where available papers pdf in English

Inequality name?

Jaka muda kolega to je znana Bellova enačba

I understand this was posted over a year ago, but I'm certain that your method 1 solution is incorrect. If t=0 and s=1, you would have x=5, y=160 and z=-208, which gives a large negative number, not 1. I think it came from a typo. What you're looking for is the following: x=-5+11t y=(3+7s)(16-35t) z=(-2-5s)(16-35t)

😂😂😂😂😂 this is 😤😤😤

For instance when a=-1, the vertex of the parabola will have a maximum point of x=1, can you explain how this doesn’t show in the graph at the end?

You better darn well be able to answer that if you're going to college. It's sad the way things are going here in Canada. I think my grandpa, who got pulled out of school in grade 3, was smarter than my nephew. My nephew finished high school.

Great explanation

You didnt show how you got the numbers?? Did you brute force it?