BuilderBoi

Hi Shockfish, nice to see you here 

Make a CPhill diagram: 

                        A 

                       \(36^ \circ\)

                  1          1

               B               C

Using the Law of Cosines, we have: \(BC = \sqrt{1^1 + 1^2 - 2 \times 1 \times 1 \cos(36)} \approx \color{brown}\boxed{0.61803}\)

Let \(x = 0. \overline{72}\)

This means \(100x = 72.\overline{72}\)

Subtracting the 2 gives \(99x = 72\), meaning \(x = {72 \over 99} ={ 8 \over 11}\).

The reciprocal of this, as Guest found, is \(\color{brown}\boxed{11 \over 8}\)

The area of the total region is 6 x 6 = 36. 

The area of the 2 squares is 1 + 1 = 2. 

The area of the triangle outside of DYES but inside the big square is 2.5 ((5 x 1) /2)

There are 4 of these, so the total area of the triangles is 10. 

Can you tell me the area now?

Set equal: \(2x^2-7x+1 =8x^2+6x+1\)

Simplify: \(6x^2 + 13x = 0\)

Factor: \(x(6x + 13) = 0\)

If \(6x + 13 = 0\), \(x = -{13 \over 6}\), else \(x = 0\)

Substituting this into the quadratic, we find the points of intersection are \(\color{brown}\boxed{(-{13 \over 6},{ 43 \over 9}) \ \text{and} \ ({0, 1})}\)

Use the formula \(180(s-2)\) to find the sum of all the angles in a polygon with s sides. 

The prime factorization of 60 is: \(2^2 \times 5 \times 3 \).

The largest single-digit factor of 60 is 6. This will be our unit digit. 

This means the remaining numbers must multiply out to 10, which means we need a 2 and 5. 

Now, what's the smallest number possible with these 3 digits?

Sorry Melody! I won't do it again. 

Recall vertex form: \(a(x-h)^2 + k = y\)

Substituting the vertex, we have: \(a(x + 4)^2 = y\)

Plugging in the point, we have: \(25a = -7\).

Can you take it from here?

Here, I'll make your life a little bit easier:

We have: \((2x+10)(x+3)<(3x+9)(x-8)\)

Now, factor \(3x + 9\): \((2x + 10)( x+3) < 3(x + 3)(x-8)\)

Note that because both sides have an \(x + 3\), we can cancel them, giving us \(2x + 10 < 3(x-8)\).

To finish, solve it like you would a regular equation, but make sure the sign( < or >) satisfies the equation.