Alan

I did use brackets - Math (Input=Result) removed them!  Try it!

10x - 3y = 69   ...(1)

2x - 9y = 39    ...(2)

Multiply (1) throughout by 3

30x - 9y = 207   ...(3)  Rewrite (2) below it

2x - 9y = 39      ...(2)

Subtract (2) from (3)

28x - 0 = 207-39 or

28x = 168

Divide both sides by 28

x = 168/28 or

x = 6

Put this back into equation (2)  (or equation (1), whichever you prefer);

2*6 - 9y = 39 or

12 - 9y = 39

Subtract 12 from both sides

-9y = 27

multiply both sides by -1

9y = -27

Divide both sides by 9

y = -27/9

y = -3

so x = 6 and y = -3

It's R₃-R₁ followed by a row switch R₃ <-> R₄

The word "first" obviously allows two operations!

4/(√2*a) = √2*4/(√2*√2*a) = 4√2/(2a) = 2√2/a = 2√2a^-1

Careful with b²! 

$${\mathtt{b}} = {log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right) \Rightarrow {\mathtt{b}} = {\mathtt{0.619\: \!331\: \!048\: \!066\: \!094\: \!5}}$$

$${\mathtt{bsquared}} = {{log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}^{\,{\mathtt{2}}} \Rightarrow {\mathtt{bsquared}} = {\mathtt{0.383\: \!570\: \!947\: \!098\: \!647\: \!1}}$$

$${\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right) = {\mathtt{1.238\: \!662\: \!096\: \!132\: \!189}}$$

What information do you know about the pyramid?  Height? Slant height? ??

If Elk City is at the midpoint of the other two, its coordinates are ((8+10)/2, (-5-3)/2) = (9, -4).

Its distance, r, from Cheyenne in coordinate units is given by Pythagoras theorem:

r² = (9-8)² + (-4 -(-5))² = 1² + 1² = 2  so that r = √2

i.e. the distance from Elk City to Cheyenne is √2 coordinate units.

(At least, I think that's what you are after!)

If you are referring to the five Platonic solids then there are only two of them, the cube and the dodecahedron, that do not have triangular faces.

To be doubly clear, let me do the calculation with the numbers:

Sum = 22 + 23 + 24 + ... + 39 + 40 + 41

Sum = 41 + 40 + 39 + ... + 24 + 23 + 22

Add up these two:

2*Sum = 63 + 63 + 63 + ... + 63 + 63 + 63

There are 20 lots of 63 on the right-hand side, so

2*Sum = 20*63

Sum = 20*63/2 = 10*63 = 630

Note that 63 can be written as 22 + 41 (i.e. first + last).

So we could write Sum = 20*(22+41)/2

The goal is to add up all the numbers from 22 to 41 inclusive.  Sum = 22 + 23 + ... + 40 + 41.

This is an arithmetic progression.  More generally, we might have:

Sum = a + (a+d) + (a+2d) + ...  (a+(n-3)d) +(a+(n-2)d)+(a+(n-1)d)  

where there are n terms, the first of which is "a" and the same difference, "d", is added on to each successive term.

Let's rewrite the sum and below it write the same thing, but with the terms written in reverse order:

Sum =     a +            (a+d) +             (a+2d) + ... + (a+(n-3)d) +(a+(n-2)d)+(a+(n-1)d)  

Sum = (a+(n-1)d)+ (a+(n-2)d) +   (a+(n-3)d) +... +     (a+2d)   +  (a+d)      + a

Now add up the two series term by term:

2*Sum = (2a+(n-1)d) + (2a+(n-1)d)+(2a+(n-1)d)+ ... +(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)

or, setting b = a+(n-1)d, the last term in the original series, we have:

2*Sum = (a+b) + (a+b) + (a+b) + ... + (a+b) + (a+b) + (a+b) 

There are n terms on the right-hand side, all equal to a+b, so

2*Sum = n(a+b)

so that

Sum = n(a+b)/2

In the original question, n = 20, a = 22 and b = 41.

Does this make it clearer?