First, draw a diagram. Label the information that has been presented to you in the problem, then male the steps and moves.

Drawing an altitude from the top of the base to the bottom, we should cleverly use the six units and eight units. By the Pythagorean Theorem, the diagonal length is ten inches. Now, take half of the diagonal, which is five, and point it towards the corner. Here, we should see a 5-12-13 triangle, and thus the height is twelve inches. The volume of the pyramid is \(\frac{B*h}{3}=\frac{48*12}{3}=48*4=\boxed{192}\) squared units(units squared).

Label the square bases' area as \(x.\) The area of one of the triangular face is thus \(\frac{1}{2}x\), and we have \(4*(\frac{1}{2}x)+x=432\),, and we have \(2x+x=432\), and thus \(x=144\) square units. The side length of the square is the square root of 144 or twelve inches. And, the height of the triangle is twelve inches. The Pythagorean theorem gives us \(12^2-6^2=144-36=72\longrightarrow\sqrt{72}\). Thus, the answer is \(\frac{144*\sqrt{72}}{3}=\boxed{48\sqrt{72}}\) cubic units.

We have the Vectors: \(AB=(-12, 12, 0), AB=<-12, 12,0>\) and \(BC=( 0, -12, 6), BC=<0, -12, 6>.\)

Using the cross-product, and the equation of the plane, we get(heft calculations), but this simplifies to \(\frac{12\sqrt{6}}{6}=\boxed{2\sqrt{6}}.\)

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.

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