Maths year 11

Nooo:*(

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Multiply both sides by 3, then square root both sides.

That would indeed be the radius, but you should get used to working like this in the exams:

$27=\frac{1}{3}r^2$ (27 comes from 270 over 10 as you might expect)
$\rightarrow 81=r^2$
$\rightarrow 9=r$

Yepp

Then

Posted from TSR Mobile

Yes, but my layout would be better for the examiner/whoever is marking it. You can move on nonetheless

Yepp

Then

Posted from TSR Mobile

sorry im at work so quick typing this.

∫ 1/Cosx dx. This is a standard integral but you can use the quotient rule to prove it. This is just workings but if you dont get any part of it then ask.

∫ 1/Cosx dx = ((Cosx)(0) - (1)(-Sinx)) / (Cosx)^2 +C

= Sinx / Cos^2x +C

= (Sinx /Cosx)*(1/Cosx) +C

=SecxTanx +C

sorry im at work so quick typing this.

∫ 1/Cosx dx. This is a standard integral but you can use the quotient rule to prove it. This is just workings but if you dont get any part of it then ask.

∫ 1/Cosx dx = ((Cosx)(0) - (1)(-Sinx)) / (Cosx)^2 +C

= Sinx / Cos^2x +C

= (Sinx /Cosx)*(1/Cosx) +C

=SecxTanx +C

Wrong.

Can you do it?

Indeed, but only because I remember what to multiply the integral by. If I didn't, I would probably struggle without help.

Integrate, not differentiate.

Sorry I'm hangover and at work so my brain no work ahah yeah but all you do is 1/cosx = secx and then use data sheet as its on there,the proof is difficult

Indeed, but only because I remember what to multiply the integral by. If I didn't, I would probably struggle without help.

So the radius is 9?

Posted from TSR Mobile

yes

Why are you working out the volume of the cone if you're given it in the question? Also that wouldn't give you the right volume anyway because $9^2 \not= 18$ and not to mention you're missing $\pi$