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.
Consider the volume of the triangular pyramid TABC.
\(\dfrac{6\cdot 12^2}{2\cdot 3} = \dfrac{\text{Area of }\triangle ABC\cdot \text{Distance from T to }\triangle ABC}{3}\\ \text{Distance} = \dfrac{432}{\text{Area of }\triangle ABC}\\ \text{Half perimeter of }\triangle ABC = \dfrac{2\sqrt{6^2+12^2} + \sqrt{12^2+12^2}}{2} = 6(\sqrt 5 + \sqrt 2)\\ \text{Area} = \sqrt{(6(\sqrt5+\sqrt2))\cdot (6(\sqrt5+\sqrt2) - 6\sqrt 5)^2 \cdot (6(\sqrt5+\sqrt2) - 12\sqrt 2)} = 36\sqrt6 \text{ unit}^2\\ \text{Distance} = \dfrac{432}{36\sqrt6} = 2\sqrt6 \text{ unit}\)
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 \(a+b=80?\)
Length
= 3 (120/360 * 2(pi)(10)) + 6 (10)
= 60 + 20 pi
a + b = 80.
Picture, please?
As shown in what?
There are no pictures!
(1/2*BE*AB)+(1/2*CE*BC)+(1/2*ED*CD), and the answer is thus \((\frac{1}{2}*12*12\sqrt{3})+(\frac{1}{2}*6*6\sqrt{3})+(\frac{1}{2}*3*3\sqrt{3})=\boxed{\frac{189\sqrt{3}}{2}}.\)
:O ok.
Area
= (24^2 + 12^2 + 6^2)/2 * (sin 60 cos 60)
= sqrt(107163)/2
= 189sqrt(3) / 2 unit^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.
Sum of areas = 6(3 ft)(6 in) = 6(36)(6) in^2 = 1296 in^2
\(|5 - 12i| + |3 - 4i|\\ = \sqrt{5^2 + 12^2} + \sqrt{3^2 + 4^2}\\ = 13 + 5\\ = 18\)
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.
Let C1 be the center of the circle with radius 10 ft, C2 be the center of the circle with radius 17 ft, A and B be the endpoints of the common chord.
\(\angle \text{C}_1\text{AB} = \arccos\left(\dfrac{\dfrac{16}{2}}{10}\right) = \arccos\left(\dfrac{4}{5}\right)\\ \angle \text{C}_2\text{AB} = \arccos\left(\dfrac{\dfrac{16}{2}}{17}\right) = \arccos\left(\dfrac{8}{17}\right)\\ \text{Distance} = \sqrt{10^2 + 17^2 - 2(10)(17)\cos\left(\arccos\left(\dfrac{4}{5}\right)+\arccos\left(\dfrac{8}{17}\right)\right)} = 21\text{ ft.}\)
What is m and n? The problem is still not complete!
Please mind what you are posting, do not just copy from some website and throw it here.
At least check if there are any mistakes in the problem before posting... *facepalm*
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.
(As you edited your question, I will edit my answer.)
Theorem: If the denominator of a simplest fraction is 2a 5b, then the fraction is terminating.
2010 = 2 * 3 * 5 * 67
We need to eliminate 3 and 67 from the denominator.
Therefore, \(n = 3\cdot 67 = 201\).
The question is not complete.
(1 + i)(1 + 2i)(1 + 3i)
= (1 + 2i - i) (1 + 2i + i) (1 + 2i)
= ((1 + 2i)^2 + 1)(1 + 2i)
= (-3 + 4i + 1)(1 + 2i)
= (-2 + 4i)(1 + 2i)
= -2(1 - 2i)(1 + 2i)
= -2 (5)
= -10
Expanding, gives us \(\left(1+i\right)\left(1+2i\right)=-1+3i\). Then, expanding again gives us \(\left(-1+3i\right)\left(1+3i\right)=\boxed{-10}.\)
\(\text{DF} = \sqrt{4^2+4^2} = 4\sqrt2\)(Pythagorean theorem)
\(\text{Let }l \text{ cm be the side length of the equilateral triangle.}\\ \dfrac{\sqrt3}{4} l^2 = 64\sqrt3\\ l^2 = 256\\ l = 16\\ \text{New length} = 12\text{ cm}\\ \text{New area} = \dfrac{\sqrt3}{4} (12^2) = 36\sqrt3\text{ cm}^2\\ \text{Area decrease} = 64\sqrt3 - 36\sqrt3= 28\sqrt3\text{ cm}^2\)
