Alan

Like so:

Two lines pass through the point (–6, 7) and each is a distance of 2 from the origin at their closest to the origin. What is the sum of the slopes of these lines? 

Rewrite l_1 in the form  y = -(3/4)x - 14/4  Its slope is -3/4, which means the slope of a line perpendicular to it has slope m = +4/3.

Thus, line L_2 can be written as  y = 4x/3 + b

If it goes through point (-5, 7) then we must have  7 = 4*(-5)/3 + b.

Can you take it from here?

The question implies the result is true for all such x and y, so we could choose x = 2, which has remainder 2 when divided by 5, and y = 4, which has remainder 4 when divided by 5.  Hence x + y = 6, which has remainder 1 when divided by 5, so z = 1.  Therefore (2z - 5)/3 = -1.

There's a difference between working with two decimal places and displaying with two decimal places.  It's dangerous to do the former!  Errors build up quickly.  To display x to two decimal places you can do:  round(100x)/100.

Compute (1^2 + 2^2 + 3^2 + ... + 10^2)/770.

In general

\(\Sigma_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}\)

When n = 10 we have: 

\(\Sigma_{k=1}^{10}k^2=\frac{10.11.21}{6}=385\)

And 385/770 = 1/2

There is missing information here!

3 and 5 make 8, which in base 6 is 2.

As follows:

1*3+2*5+3*7+4*9+5*11+6*13+7*15+8*17+9*19+10*21+11*23+12*2 = 1102

i.e. there are 3 ones, 5 twos, 7 threes, ... 23 elevens and 2 twelves.

I suspect "strictly inside" excludes those on the boundary.