5 th circle in centre

I could definitely not pot it.

What if we take r =72 , this rule doesn't work you all dumbo kids believing in him He has told it for normal rates like up to 20% but mathematically this statement is completely impossible

Unless your control is so bad, you give unintentional english every shot

Is it the easiest pool game?

Now I want to buy a mirror.

The ball won't always fall into the hole. Usually an inadvertent sidespin is placed on the ball when it's hit.

🕺🕺🕺🕺🕺🕺🕺🫄

i dunno mate id still find a way to miss it if I hadn't had a pint beforehand

DO NOT REDEEM IT

i would screw this up with torque shot

*Me who hits the ball lightly*

112.5

This is called gerrymandering, deckard is the person i actually learned it from and if you are interested in gerrymandering puzzles i recommend checking put his channel

If someone has solved cengage CoM chapter.....they will know a similar question was there as a numerical. I could not understnad it that time. But this video actually clears my doubt😊

seems like diminishing returns though lol

i havent done math in years but this came naturally to me. guess all those math lessons taught me something that stuck.

Coefficient of restitution left the chat

I can imagine that hitting the ball at a 90 degree angle to the bunker - like exactly back on itself, you could fail to reach the hole. But maybe in two banks it would go in? I almost care enough to investigate...

yeah ok cumin warrior

I can't focus because of the accent

100 square units

I bet I could make it miss Keepo

Thank you

And what if you add sidespin to the ball, angles of reflection mean nothing in billiards

Good bro

u must also be standing a certain distance from the mirror...

Wrong!

Aplying similar method we get that the distance concerning the octagons should be 5. And if someone is intersted in the distance of the other corner this should be 2sqrt(2-sqrt(2))+5/2 sqrt(2+sqrt(2)) ==6.1

what if you sidespin the ball though

😍 Yes. Infinite and balanced INFINITELY 💯♥️👍👍👍👌

So that the other exagon has side length 4 does not play a role at all? Or does it? If the side length of the big exagon were > 6 then such a scheme is not possible.

whats yall's cuemath teachers

6 square unit becacuse Base and Height are equal 😘

½ × d1+d2

This is a classic case of an Indian KZbinr teaching a year's worth of school in a single video.

You can just move one step right first

Some focii just know how to join the dots better than the rest.

This is BS... the table is a sink. He never places the ball anywhere

Wow! I didn't know that its unlimited. I used to think that there must be a limit for this.

❤❤❤

4?

You made it very easy to understand

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LETS HIT THE BALL VEEEEEEERY SOFTLY

My noob azz would fail this miserably 😭

That accent is awful and makes it impossible to take anything he says seriously.

...and I bet I will still be able to miss the hole 🙈😢😅🙈

Bhai yeh toh sure hote hi khatam ho gya 😂😂

Class 10th