Search results

Like commentators have said on your other thread Cambridge is the king and honestly is the only textbook you need.

wdym 99.95 thats an atar? As for scaling a 94 scales to rougly 97-98 looking at previous years so many try to get low to mid 80s for 4U since moderation bumps their mark up significantly.

Let there be 5-2z stars and 1 bar (since we want to split 5-2z into two parts the number of stars in each part is x and y). We can arrange these in {{6-2z}\choose{1}}=6-2z ways (including the extra bar we have 6-2z objects needed to be arranged). Now all we have to do is sum this up from z=0-2...

To be fair the whole thread was based around recommendations for jkjkjk so it would be odd for him to advertise to a random commenter rather than the guy who made the thread.

I read in another thread that the reason delta is so expensive is because they pay their tutors top dollar because they all state ranked?

Direction is zero? Thats clearly wrong which textbook are you using?

Even easier as the vector points 5 units up at an angle of 90 degrees to the positive x-axis.

Its quite easy C_n outputs C_n

Just a pedantic correction, it may be better to name your constants C_1 and C_2 rather than both C as this implies they are both the same.

Nope no need for AP and GP, but you definitely need to understand what a 'sequence,' and what a 'series' is. It may also be wise to understand how the number of terms of a series and the last term of a series can both depend on a variable. For example the series: 1^2+2^2+3^2+...+n^2 has n terms...

Using the identity: \sin\alpha\cos\beta=\frac{1}{2}\left(\sin{(\alpha+\beta)}+sin{(\alpha-\beta)}\right) \int_{\frac{1}{4}}^{\frac{1}{2}}\sin\left(x^{x}+\ln x\right)\cos\left(x^{x}-\ln x\right)\log_{2}\left(ex\right)dx...

So you have learnt this? Im confused.

Yup why?

Yup sure now prove this identity: \log(ab)=\log(a)+\log(b) for a,b positive numbers and also \log(a^b)=b\log(a)

^ Try this next. You said you have learnt it.

Ok prove the log laws?

Even then that isnt equivalent... (and you didn't write that in your working)

Thats one of the definitions of the log function usually your meant to learn that at the start...

e^x=5 is not equivalent to \log(xe)=5?

I would suggest revising logarithms as a whole.