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These are yr 11 cambridge 14B Q22-24

This is yr11 cambridge 14B Q19

This is yr11 cambridge 14B Q18

It won't end in 0. The digits are 1,2,3,..,9 which does not include 0.

It is cambridge y11 14B Q15

So what would a better rule be? One that relates to the position on the number line. In the middle or above, round up. Below the middle, round down. And this can be illustrated on the fx-92B Secondaire (but not fx-8200 AU or most of the other emulators): \text{And from }this\text{ rule we...

Here's a thing you can do with the fx-8200 AU for recurring decimals. First let's do it without the calculator. Here are the rules in the Year 7 Cambridge: \text{So applying these rules to }0.14\dot{9}\text{ rounded to 1 decimal place we get } 0.1 \text{It's an example where these rules don't...

I think the really big improvements are in statistics and probability distributions. For example, - All the stats at once instead of finding each one individually definitely saves time - Ability to do inverse normal distribution isn't even a thing on the older calculators I get that as with...

Maybe we can look at other things like this: Prove or disprove: \text{For any set }\Bbb X\text{ such that }\Bbb Z \subset\Bbb X \subset\Bbb R\text{ there is either a 1-1 correspondence \text{between }\Bbb Z\text{ and }\Bbb X\text{ or there is a 1-1 correspondence between }\Bbb X\text{ and...

Douglas Adams may have thought it was 42, but nope it is ½.

Cambridge Ext.1 Year 11 Ex. 16A Q11. Here is e and f solutions

Maybe this one. Prove the Riemann hypothesis: The Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part ½: https://en.wikipedia.org/wiki/Riemann_hypothesis

It is Cambridge Ext. 1 Yr 11 Ex. 16A Q9 Here is the solution to b:

I found the rest. It is Cambridge Ext. 1 Year 11 Ex. 17G Q7 and here is the solution:

There are of course other methods, some listed here: https://math.stackexchange.com/questions/2991542/proving-cos-frac2-pi7-cos-frac4-pi7-cos-frac6-pi7-frac12 I suspect there is more to the question yellowhighlighterr didn't put there. It is unlikely such question would be asked unscaffolded.

Note this method can be generalised to establish \sum\limits_{k=1}^n\cos\left(\frac{2k\pi}{2n+1}\right)=-\frac12

Here is one way...

\text{Well you know }\cos(x)=-\cos(\pi-x) \begin{aligned}\text{So...

\text{They are simplfying the inequality so that it is easier to solve.} \text{Starting with }\frac{x^2+x-2}{x^2(x-2)}\le0,\text{ multiplying numerator and denominator} \text{by }x-2\text{ like this: }\frac{x^2+x-2}{x^2(x-2)}\cdot\frac{x-2}{x-2}\text{ is the same as multiplying by 1} \text{but...

Here is a quicker way: \textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)\ \!dx=\frac{i}{2}[\text{Li}_2(e^{2ix})]_0^\pi=0\text{ where Li}_2\text{ is the dilogarithm function} \text{However} \begin{aligned}\textstyle\int_0^\pi\ln(-2ie^{ix}\sin x)dx&=\textstyle\int_0^\pi(\ln2+\ln(-i)+ix+\ln(\sin x))\...