It's also can be 65?

Because of this, i had dobe wrong !

i assuming since Br2 reacts w it that it does not exhibit aromaticity? but it certainly fits Huckel's rule...? OH but i suppose those 4 substituents on the middle ring act as activating groups... that must be it

SIR why didn't it happn through SNae the carbon is sp2 in rcocl SNae occur why not here

thanks for this, very helpful

The final equation has a positive sign after Pr

When u > u min. , the dumbell will have some velocity at top point When u < u min. Dumbell will not complete vcm coz it will not reach the top point

HYDROLYSIS OF AMIDE!!📢🗣️🗣️🥵🔥🤬🥶👺👹GIVES AMMONIA

With accurate drawing , it’s becomes feasible to tell R by eyeballing hence making it easier to find shaded area

RESPECTED Sir if WE use formula ∆H=E@(f) - E@(b) . We are getting directly answer 5 which is correct. PLZ tell me deep analysis about this method.

Thanks

Very nice explanation. Still, prying force is not available on you tube, a single video found which explained

Better solution than physics galaxy itself No comparison or hate But sir u solved this problem in detail so that an avg student can also understand it In every other places , the solution is a little bit of hard to understand Thank you so much sir❤️

man you are underrated

10:30 how is dy/dx = tan(theta/2)

(5)^2=25 36°B+36°A+90°C}= =162°BAC {162°BAC ➖ 180°}= 18°BAC 25/18°BAC=1.7°BAC .1^1.7^1 7^1 (BAC ➖ 7BAC+1)

a=2, b=4, r=(ab)/(√a + √b)^2

3*2^998

(2N)^2= 4N^2 (4M)^2^=16M^2 {4N^2+16M^2}= 20NM^4 360P/20NM=18NMP^4 3^6NMP^4 3^3^2NMP^2^2. 3^13^2^1^NMP2^1^2 1^1^31^1NMP^1^1^2 32 (NMP ➖ 3NMP+2)

beautiful question

THANK YOU BETA

This is awesome, many thanks! φ = 30°; ∆ APO → OPA = 2φ/3; PAO = 3φ → AOP = 7φ/3 → AOB = 14φ/3 → 12φ - 14φ/3 = 22φ/3 arc AB/arc BC = 3/2 → AC = 3k → BC = 2k → 5k = 22φ/3 → k = 22φ/15 → BOC = 2k → BAC = BOC/2 = k = 22φ/15 = 44°

Nice

X=5

PRE-MATH know how to take tan of an angle? BS

No need all these calculations. Subtract the area of a 3/2 circle of radius 2 from the area of a quarter circle of radius 6. That's it.

Laal Phool Neela Phool Sir k videos beautiful😍😍

I looked at it as 328 being a sum of 3 numbers all power of two. 256, 64 and 8 so 2^8, 2^6 and 2^3 then quickly match a b and c but all as powers of 2 so 1a, 2b and 3c then matching form there as C can only be 1 or 2 etc

Your explanation was clear, succinct, and well-paced. The figure was not to scale which some people would find distracting.

Sir the time period formula is 2π√l cosA÷g

Simpler argument not using theorem other than Pythagoras. Let BO = R, the semicircle radius: ACO is a right angled triangle. CO = R - 3. OA = R - 4. AC = 3. So AC² + AO² = CO² (Pythagoras). SO (R-4)² + 3² = (R-3)² and hence R = 8, so area subtraction of the semicircle and circle yields 23π.

Reflect ∆PSQ about QS so that P falls on QR at P'. ∠QPS= ∠QP'S and QP= QP', and P'R= QR-QP' = PS. Hence, ∆SP'R is Isosceles and ∠P'SR = ∠SRP' =θ This implies, ∠QPS= 2θ= ∠QPS In the ∆QPR, 60+θ+2θ=180 θ= 20°

इस विषय पर बहुत अच्छा स्पष्टीकरण वीडियो, रूसी गणित ओलंपियाड बहुत उल्लेखनीय और बहुत दिलचस्प हैं जब आपको लागू करने की तकनीकों का पता चलता है।

It is at P😂

Can you at least try to center your radius line to the center of the circle?

Six circles of radius r=2 inside one circle of radius R=6. (π*6^2-6*π*2^2)/4=3π

W.L.O.G. Let P = (-1,0), S = (0,0) and R = (1,0). angle(PQR) = theta + (180° - theta - 45°) = 135° Therefore Q lies on intersection of the circle y = sqrt(2 - x^2) - 1 and the line y = -x This solves to: x = 1/2(1-sqrt(3)) , y = 1/2(sqrt(3) - 1) theta = atan( 1/2(sqrt(3) - 1) / (1/2(1-sqrt(3)) - -1) = atan(1/3*sqrt(3)) = 30°

Who is here from 2024… 😅 Thank you sir This is extremely helpful

r^2=4^2+(1/2)^2=16+1/4=65/4; r=√65/2

First transfer 8l into the 5l and 3l beaker, now the 8l is empty and other both are full. Now transfer 1l from 5l into the 3l so 5-1 = 3+1 = 4

sqrt(2i) is also a solution not arrived at with your method.

The vid's 11 years old, and yet is very efficient ❤

256/3

Already done in pg , pls bring more questions (unique)

Because we have already calculated the work done by the weight

Great sir

Great

WELL EXPLAINED 😍

How net force= 0 when it have acceleration

Who the hell ar you man ? Thank you for so much effort