jfan17

Let's look at the squares between 100 and 200.

First is obviously 100, which is equal to 10²

^{\(100 = 10^2\)}

^{\(121 = 11^2\)}

\(144 = 12^2\)

\(169 = 13^2\)

\(196 = 14^2\)

\(15^2 = 225\), which would be too large, making our answer

5 square numbers inclusive between 100-200. 

Hey guest! Personally, I've been just grinding some math lol(i know, sounds boring :P). I've also been reading some books on the side and also some work for school(they actually assigned some to us). What about you?  

Hey guest!

1. Let's start by expanding out first.

We get:

\(-4(x+4) > x+ 7 = -4x-16 >x+7\)

Adding 4x on both sides and subtracting 7, we get:

\(-23>5x\)

Dividing by 5, we get:

\(-23/5>x\)

That gives us an interval of:

\((-\infty,-23/5)\)

2. Similar to the last problem, let's just first add like terms to get a constraint on x.

We're given:

\(2x-5\leq -x + 12\)

Adding 5 and x on both sides, we get:

\(3x \leq 17\)

Dividing by 3 on both sides, we get:

\(x \leq 17/3\)

This gives us the interval of:

\((-\infty,17/3]\) with a square bracket on 17/3 to indicate its inclusion in the interval(remember, round parentheses are exclusive!)

Hey qwerty, glad to see you again!

This problem is just some expanding and then rearranging. 

Let's start with an understanding of the word "constant". the constant of any polynomial is defined as the term that has an x coefficient of "0"; in other words, it has no "x" coefficient. 

Now that we know this, it's a simple matter of distributing and multiplying out everything and then combining like terms. We get:

\(3(x - 4) + 2(x^2 - x + 7) - 5(x - 1) = 3x-12 + 2x^2-2x+14-5x+5\)

Combining like terms, we get:

\(2x^2-4x+7\)

Because the problem asks for the constant coefficient, we then get:

7 as our final answer

Sorry, I think I may have messed up my answer. The answer should be:

\(g(x) = -8x^4+7x^2-7x+8\)

So sorry for my mistake!(On subtraction too :P)

I think it means the unit square can't be shaped like a "diamond"; in otherwords, diagonally oriented

In the question, does it mention whether or not Larry and Laura's squares must be different? What if they choose the same unit square? I think that part of the question is very unclear.

Hey there qwerty! First, this is just some basic rearranging. We subtract on both sides, to get:

\(g(x) = x+ 1 -8x^4 +7x^2 -8x +7\)

Next, let's collect like terms and rearrange it in order of decreasing degree. What that means is that the highest "power" goes to the leftmost of the equation, with the lowest(the 0th power) degree going to the right. We then get:

\(g(x) = -8x^4+7x^2-7x+8\) as our final answer!

Following up on CalTheGreat's tidbit on the exterior angle sum(the sum is important, as exterior angle theorem is something different) theorem, if you don't understand it, here's a proof:

Start by realizing that the interior angles of an n-gon have an angle sum of 180(n-2). This is derived from the fact that an n-gon can be divided into n-2 triangles, which all have an angle sum of 180 degrees(there are other derivations of it, but I think I'll use this one for now). Now if you look at an exterior angle, it's pair(the interior angle) forms a 180 degree angle, so the total sum of these pairs would be 180n(the exterior angles + the interior angles). Realize that because the interior angle sum is 180(n-2), the exterior angles sum is just:

\(180n-180(n-2) = 180n-180n+360 = 360\). That's just a short proof of why the exterior angles of a polygon ALWAYS add up to 360. 

No problem!