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Rewrite it as (15x-5y)^2 + (5y-3z)^2 + (15x-3z)^2 = 0
implies 15x = 5y = 3z = k say
implies 2y = 2k/5 = k/3 + k/15 = x+z
Hence proved!

To make it more simpler

Continuing my solution Mr. Green

$\text{TPT} 3^3 < (2.33)^4$

Now $(2.33)^4 > (2.3)^4 = (5.29)^2 > (5.25)^2 = \left(\frac {21}{4}\right)^2 = \frac {441}{16} = 27.5\cdots > 27 = 3^3$

Hence Proved

Sorry Feynmann, didnt see your start :)

PS: Ignore the extra TPT at the bottom!

$\sin 1 < \sin 60^{\circ} = \frac {\sqrt{3}}{2}$

TPT $\sin 1 < \log_{3}{\sqrt{7}} \leftrightarrow \frac {\sqrt{3}}{2} < \log_{3}{\sqrt{7}} \implies \text{TPT} \sqrt{3} < \log_{3} {7} \implies \text{TPT} 3^{\sqrt{3}} < 7 \implies \text{TPT} 3^{\sqrt{3}-1} < 2.33 < \frac {7}{3} \implies \text{TPT} 3^{\sqrt{3}-1} < 3^{\frac {3}{4}} < 2.33 \implies \text{TPT} 3^3 < (2.33)^4$ which is true Smile

Hence proved Mr. Green Razz

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Is it necessary for you to fill in the off campus counselling forms or can we directly attend the on campus counselling?

Put x+3 = t.

You mean alkyne Huh?

Well. add HgSO4/H2SO4. Also be more specific about the alkene and ketone you want because the question is quite ambiguous.

1. HgSO4/H2SO4
2. alpha bromination
3. KCN

Its a direct consequence of modulos! We all know that a number can be represented by the form $\sum_{i=0}^{n} 10^{i} \cdot a_{i}$ where $a_{i} \in \mathbb{R^{+}}$ say

Ofc $10 \equiv 1 \mod 9$

Hence the result Razz Mr. Green

Id advise you to take bits!

bits
Dont ask why because you said "i'm asking your personal choice........" :P :P

I'd advise you to go for bits!

There may be other factors like OBC quota which influence it.

The best way is to keep in pace with your professor (If you are going to one).