Topic, Offtopic post

  UK, Canada, Korea, France, Czech Republic, and Belgium. (consecutive multiples of 5)
UK, Canada, Korea, and France. (consecutive multiples of 10)

CG - Neusis Construction
GIS - Euclidean Construction

 

  Best Buy, Walmart, Target, Kohl's, Shoe Carnival and Marshall's

  Complete 5 world names
Some letters were given, esay riddle.

S___m A__
Renai_______ A__
F_i__Tale F___st
S__ C_ty
K___ F_ World

  It would be better to create world-themed levels with puzzle it & jellify rather than bb 1 & 2. because both have a lot of colors.

jellify only has 10 colors outside of gray, which is a flaw. It should made into puzzle it rather than jellify.

still bad point: missing pink, typical blue, salmon, bright orange, lavender, navy, olive, light yellow, pale green, rose red, royal blue in puzzle it.

GeoGebra supports #RRGGBB.

Hexagon, there 19 world colors I made , one of which is #6ee700. if you want to get those colors, try it. esay. no descriptions.

 

  Answer those World_Theme Names;

, , : ?

, : ?

, , , : ?

  What is that? A riddle?

  , , =?

, =?

, , , , =?

  Binders, backpacks, rulers, protracters, desks and chairs.

  Sketching, writing, marking, erasing, colouring, measuring, cutting, and folding.

  Pencils, pens, markers, erasers, crayons, rulers, scissors and paper.

  That was a brilliant method Nelson, I never really thought of it that way. I forgot the term polynomial and square root in terms of solving equations. There are many different solutions and possibilities.

  @nelson Nice solution! I tried to reverse engineer my way and it will involve guessing solutions anyway so I think your way is better. That was the way I expected you to solve that question. I came up with it by the method in the spoiler at the bottom (because it reveals a way to solve it).

As for the general procedure for solving a quartic, I think that’s a little crazy to suggest a problem that requires it, because the solutions are far too large to write out. I used to think that the ‘guess and factor/divide’ methods weren’t as direct but honestly solving that with a general procedure is far too extreme! As far as I understand though it would require solving a general cubic, which itself isn’t commonly known how to do.

@chris Yes, I have heard of inequalities. Linear inequalities are usually almost as simple as linear equations but adding higher powers or other functions makes inequalities far more tricky and require a lot of work.

To generate the problem:

  Have you guys heard of inequalities? It's one of the algebra terms back in high school when I attended in 2009.

  equation x^4-2x^3-6x^2+10x-3=0

The general method to solve this kind of equation is very difficult. I don't remember the whole process but it seems to me that it will be necessary to solve an equation of degree 3.

The simplest method is to find obvious solutions and factor the polynomial:

  Pumpkins, Scarecrows, Ravens, Skeletons, Witches and Goblins

  We're going to the pumpkin patch tonight for a Halloween party. Enjoy this holiday guys. You can get candy from trick-or-treaters.

  Pi, 3.14, #, circumference, radius and circles

  Happy Halloween everyone. Enjoy all the decorations and candy that's passed out by trick-or-treaters. However I'm too old to do that sort of thing now.

  Here’s a potentially trickier to solve equation if you want something even more difficult. I know of two methods that will work here, but it may not be easy to ‘guess’ one method, the other is similar to Nelson’s other solution - which is the typical way to solve this type of equation:

x^4-2x^3-6x^2+10x-3=0

(Again there are four separate solutions!)

  s^2, pi*r^2, pi*r^2*h, s^3, b*h/2, (3*sqrt3/2)*s^2 (regular).

  Squares, Circles, Cylinders, Cubes, Triangles and Hexagons

  Red Riding Hood, Three Little Pigs and the Big Bad Wolf.

  , ,

  4, 8, 12, 16 and 20

  0, 1, e, pi, i

  Grass, dirt, sand, water, ice.

  Rain, thunder, wind, hail, tornados and hurricanes.

  @nelson Very good! Polynomial division is definitely a possible alternative to my substitution, and honestly I didn't think the equation was too hard when I first set it, I guess I should have assumed most people didn't see it.

@suhangha I derived the formulae (there are technically two if you count the + and - versions as separate though for legitimate triangular numbers you should always use the + formula), they are very interesting and can provide a generalisation to the question - 'If 2, -1/4, or whatever were a triangular number, what place would it be?'

I'd be interested to see if I got the correct answer though, I checked it and it seems to work.