Melody

Anonymous has got the right idea too

10%3 = 1 ==>  10mod3 is 1 because 10 divided by 3 has a REMAINDER OF 1

11%3 = 2 ==> 11MOD3 is 2

12%3 = 0 ==> 12 mod 3 is 0

That would be terrible zegroes.

A place with no maths - DEFFINITELY NOT GOOD!

Ok My turn

the roots have to be factors of 6 so I'd look at 1 first

f(1)=1+4+1-6=0 therefore (x-1) is a factor    [I am using factor theorem but it is just logical really]

I could now do algebraic division or synthetic division to find its cofactor but I'll just try easy possibilities first.

f(-2)=-8+16+-2-6=0 therefore (x+2) is another factor   

(x-1)(x+2)=x²+x-2

$$(x^3+4x^2+x-6) \div (x^2+x-2)
=x+3$$

I did the division but I don't know how to set it out in latex.  SO

$$(x^3+4x^2+x-6)=(x-1)(x+2)(x+3)$$

And that will work every time so long as the roots are integers. 

I don't know -we don't have tornados here.  

Doesn't sound good though zegroes.  

$$\sqrt{(4+16)}=\sqrt{20}=\sqrt4\times\sqrt5=2\sqrt5$$

sunday 0.5

Mon 0.5*1.5

Tues 0.5*1.5*1.5

Day 5 (friday): 0.5*1.5^5=

$${\mathtt{0.5}}{\mathtt{\,\times\,}}{{\mathtt{1.5}}}^{{\mathtt{5}}} = {\frac{{\mathtt{243}}}{{\mathtt{64}}}} = {\mathtt{3.796\: \!875}}$$

which is 3.8 miles

Yes the search function is problematic but the site in general is not too bad - you just have to get used to it.

I think we should call today.

"Mutual Admiration Day"  MAD for short.

Hi Kitty!