tertre

I will assume a drawing...The best thing to do is draw an equilateral triangle in the middle, and connect the centers of the circles with lines going to the bends and corners of the belts. This makes a circle with a radius of ten incnhes and a circumference of $2*10\pi=20\pi$ inches. $60+20\pi$, thus the answer, and $a+b=80?$

Picture, please?

(1/2*BE*AB)+(1/2*CE*BC)+(1/2*ED*CD), and the answer is thus $(\frac{1}{2}*12*12\sqrt{3})+(\frac{1}{2}*6*6\sqrt{3})+(\frac{1}{2}*3*3\sqrt{3})=\boxed{\frac{189\sqrt{3}}{2}}.$

There are six "non-hexagonal faces" with a height of three feet and a width of six inches. Six inches is the same as 1/2 feet, and the answer is thus $6*(3*1/2)=\boxed{9}$ square feet.

We just use the distance formula: For lines, YA and AB, draw a diagram and label all the points!

Thus, we have $\sqrt{{10^2}-{8^2}}=\sqrt{100-64}=\sqrt{36}=6$ feet.

Similarly, we do it for XA and XB, and since this is $17$ feet, we get $\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$ feet.

Thus, the answer is $6+15=\boxed{21}$ feet.

30-60-90 triangles override this problem. If AE=24, then AB=$\frac{24}{2}\sqrt{3}=12\sqrt{3}.$ BE is equal to twelve inches, and this is opposite to the ninety(90) degrees angle in triangle BCE. And, CE=6 inches, while BC is equal to $6\sqrt{3}$ inches. Thus, opposite to the ninety-degrees in the triangle CDE, means that ED=three inches, and CD equals $3\sqrt{3}$ inches. Thus, the perimeter of quadrilateral ABCD is $12\sqrt{3}+6\sqrt{3}+3\sqrt{3}+24+3=21\sqrt{3}+27$ inches.

Expanding, gives us $\left(1+i\right)\left(1+2i\right)=-1+3i$. Then, expanding again gives us $\left(-1+3i\right)\left(1+3i\right)=\boxed{-10}.$

The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle. Consequently, setting this formula and expression equal to $64\sqrt{3}$, we get $s^2=256$, and $s=16$ centimeters. Since each side of the triangle decreases by four centimeters, the new side length of the equilateral triangle is twelve centimeters. Thus, the new area is $36\sqrt{3}$ centimeters, and it has decreased by $64\sqrt{3}-36\sqrt{3}=\boxed{28\sqrt{3}}$centimeters.

-tertre

The best thing to do here is to check if the determinant is equal to or not equal to zero in the three systems of equations.

Thus, we have the matrix $\begin{bmatrix} 1 & 1 & 2\\ -2 & 4 & 2\\ 0 & 1 & 1 \end{bmatrix}$, and it's determinant, which is $D$, is thus $1\begin{vmatrix} 4 & 2\\ 1 & 1 \end{vmatrix} - 1\begin{vmatrix} -2 & 2\\ 0 & 1 \end{vmatrix} + 2\begin{vmatrix} -2 & 4 \\ 0 & 1 \end{vmatrix}$, and this is equal to $0$, so the three systems of equations have no solutions.

-tertre

Maximum Value: A=B=C=60 degrees, 1/2+1/2+1/2=3/2.