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Using gradients: For YX\bot XZ we have m_{YX}\times m_{ZX}=-1 where m means the gradient. So \frac{\lambda-0}{1-(-1)} \times \frac{2-0}{\lambda-(-1)}=-1 \frac{2\lambda}{2(\lambda+1)}=-1 \lambda=-(\lambda+1) 2\lambda=-1 \lambda=\frac{-1}{2} Oh just realised username already did that.

What is your typical day like?

Do what Idkddi did. 2 combinations diagonally, 2 combinations vertically. Then horizontally 5 dots need to choose 3 5C3 so in total:Ways=4+5C3=10+4=14

Another cool way of doing the poly q: sub \alpha, -\alpha into the original equation to get: (\alpha)^3+(\alpha)^2+c(\alpha)-10=0 and (-\alpha)^3+(-\alpha)^2+c(-\alpha)-10=0 Add these two to get:2(\alpha)^2-20=0 i.e. \alpha=\sqrt{10} WLOG. Making c the subject...

The saying correlation does not equal causation is kinda true. Not only does Du get ruse kids (who are already the top students) he gets the top ruse kids.

Arent most agriculture state ranks from ruse? Also math is never used in medicine yet pretty much every ruse kid does 4u?

I think mechanical is more specialised in the mechanical aspects so the former is true.

I think mechatronics adds electronics and robotics on top of mechanical making it more broad.

Another way is to note that: x^4+4x^3+4x^2=(x^2)^2+2(x^2)(2x)+(2x)^2=(x^2+2x)^2 So p(x)=(x^2+2x)^2+7x^2+14x+10. From here we will try getting a quadratic in x^2+2x and so p(x)=(x^2+2x)^2+7(x^2+2x)+10. Then use quadratic factoring techniques to obtain: p(x)=((x^2+2x)+5)((x^2+2x)+2).

Same idea as akshat but another way is: Let x^4+4x^3+11x^2+14x+10=(x^2+bx+5)(x^2+cx+2) (5 and 2 since constant terms multiply to 10). Now compare coefficients of x^3 and x to get 4=b+c and 14=2b+5c solving these simultaneous equations:b=2, c=2 So p(x)=(x^2+2x+5)(x^2+2x+2)...

If you want to know how he came up with that factorisation: Notice that 1+w+w^2+...+w^6 is a GP so using the GP sum: 1+w+w^2+...+w^6=\frac{w^7-1}{w-1} now we know w^7=1 and w \neq 1 so 1+w+w^2+...+w^6=0 subtract 1 from both sides you get w+w^2+...+w^6=-1

For 2: w^n=1 by definition. dividing by w on both sides: w^{n-1}=\frac{1}{w} Also note that |w|=1 so w^{n-1}=\frac{\overline{w}}{w\overline{w}}=\overline{w} since w\overline{w}=|w|^2

A lot of laws are based on morality. For example indecent public exposure (especially for women) is seen as immoral by western society and many religions whereas in other cultures its perfectly ok. This law is almost entirely based on 'personal feelings,'' i.e. women should dress modestly.

I assumed it was obvious guess I should have explained more.

I'm a night person :)

There really isn't a need for induction here. All you have to note is that 1>\frac{1}{2} and \frac{1}{3}>\frac{1}{5} and so on until (-1)^{n-1}\frac{1}{n-1}>(-1)^n\frac{1}{n} add all of these inequalities and you get your desired result.