GYanggg

Let there be three regular polygons with \(x,y,\) and \(z\) as their number of sides. Their internal angles must sum to \(360º\) and individually be supplement to their respective external angle. Therefore, we can express the sum of the three angles as:

\((180º-\frac{360º}{x})+(180º-\frac{360º}{y})+(180º-\frac{360º}{z})=360º\\ 540º-\frac{360º}{x}-\frac{360º}{y}-\frac{360º}{z}=360º\\ 180º=\frac{360º}{x}+\frac{360º}{y}+\frac{360º}{z}\\ \frac2x+\frac2y+\frac2z=1\) 

\(3\le x \le 6\), because three sides is the least number of sides a regular polygon can have, and if \(x,y,\) or \(z\) is greater than 6, the sum will be less than one. 

Casework: 

\(x=6\Rightarrow y=6, z=6\\ x=5\Rightarrow y=5, z=10\\ x=4\Rightarrow (8,8);(6,12);(5,20)\\ x=3\Rightarrow (10,15);(9,18);(8,24);(7,42)\\\)

Therefore, there are 9 groups of regular polygons that can surrond a point.

I hope this helped,

Gavin. 

\(A_2=A_1+2\\ A_3=A_2\cdot(-4)\\ A_4=A_3+2\\ A_5=A_4\cdot(-4)\\ \)

For the next term, you take 120 and add two to get 122. 

Then, you multiply by -4 to get -488. 

Finally, you add two again to get -486. 

The next three terms in order are 122, -488, -486. 

If \(|x|=3x+1\), find \((64x^2+48x+9)^{2018}\).

Rewriting expression:

\((64x^2+48x+9)^{2018},\\ =[(8x+3)^2]^{2018},\\ =(8x+3)^{4036}.\)

Solving for \(x\):

\(|x|=3x+1\\ x^2=(3x+1)^2\\ x^2=9x^2+6x+1\\ 8x^2+6x+1=0\\ (2x+1)(4x+1)=0\\ x_1=-0.5, x_2 = -0.25\)

After plugging our values into the equation to check if they work, we will see that only \(-0.25 \) satisfies the equation. 

\((8\cdot(-0.25)+3)^{4036}=(-2+3)^{4036}=1^{4036}=\boxed1\)

If \(x^2 + y^2 +\frac12 = x + y\), find \(x^y+y^x\).

Now plugging our values into the expression: 

\(x^y+y^x\Rightarrow \frac12^\frac12+\frac12^\frac12\\ =\sqrt{\frac12}+\sqrt{\frac12}\\ =\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\\ =\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\boxed{\sqrt2}\)

a)

You just plug in the values, and evaluate the expression. 

b) 

\(x+y=\sqrt{x^2+y^2}\\ (x+y)^2=x^2+y^2\\ x^2+2xy+y^2=x^2+y^2\\ 2xy=0\)

Therefore, either x, y, or both variables have to be 0 for the expression to be equal. 

This works in reverse. If \(x+y\ne\sqrt{x^2+y^2}\), then \(2xy\ne0\).

I hope this helped,

Gavin. 

1. 

\(5ax^2 + x^3 - 5a - x\\ =(5ax^2-5a)+(x^3-x)\\ =5a(x^2-1)+x(x^2-1)\\ =(x^2-1)(5a+x)\\ =(x-1)(x+1)(5a+x)\\\)

2. 

\(a^2 - b^2 + 12b - 36\\ =a^2 - (b^2 - 12b + 36)\\ =a^2 - (b - 6)^2\\ =[a - (b - 6)][a + (b - 6)]\\ =(a - b + 6)(a + b - 6)\)

3.

\(3x^3 - x^2y + 6x^2y - 2xy^2 + 3xy^2 - y^3\\ =x^2(3x - y) + 2xy(3x - y) + y^2(3x-y)\\ =(3x - y)(x^2 + 2xy + y^2)\\ =(3x - y)(x + y)^2\)

I hope this helped,

Gavin

\(2a^4-a^3-6a^2-a+2\\ a^3(2a-1)-[(2a-1)(3a+2)]\\ (2a-1)(a^3-3a-2)\\ (2a-1)(a^3+1-3a-3)\\ (2a-1)[(a+1)(a^2-a+1)-3(a+1)]\\ (2a-1)(a+1)[(a^2-a+1)-3]\\ (2a-1)(a+1)(a^2-a-2)\\ (2a-1)(a+1)(a-2)(a+1)\\\)

I hope this helped,

Gavin. 

Consecutive terms of an arithemtic sequence have a common difference. 

We can write the equaiton:

\((x-1)-\frac12=3x-(x-1) \\ x-1.5=2x+1\\ x=-2.5\)

Then, we can plug this value into the terms to check if they have a common difference. 

The sequence goes: \(\frac12, -2.5-1, 3\cdot(-2.5) \Rightarrow0.5, -3.5, -7.5\)

The common difference is 4, so the value works!

I hope this helped,

Gavin. 

Point \(P\) is inside equilateral triangle \(ABC\) such that the altitudes from \(P\) to \(\overline{AB}, \overline{BC}, \text{and} \ \overline{CA}\) have lengths 5, 6, and 7 respectively. What is the area of triangle \(ABC\). 

Connect \(P\) to \(A,B, \text{and } C\), to form \(\triangle ABP, \triangle ACP, \triangle BCP.\)

\([ABC]=[ABP]+[ACP]+[BCP]=\frac12x(5+6+7).\)

The area of \(\triangle ABC\) can also be represented as \(\frac{\sqrt3}{4}x^2\).

Therefore: \(\frac{\sqrt3}{4}x^2=\frac12x(5+6+7)\Rightarrow x=12\sqrt3\).

Plugging \(x\) into one of our area formulas, \(\frac{\sqrt3}{4}x^2=\frac12x(5+6+7)=108\sqrt3\).

I hope this helped,

Gavin.