Mathmatically, how can I lay out 2 archs that reach a height of 11' starting and ending  at 8'9"?
113113
Think of this as an even power function.  Look at the graph of y=-x^2.  Looks like an arch.  So does y=-x^4.  Now you just need to calculate the other parts of this function to match your initial, final, and maximum.  The expression you will be using is y=-(ax+c)^2+b.  The 'a' will change the "roundness" and the 'b' will change the maximum.  The 'c' will change the horizontal position.
872092
It is always understandable when a student feels overwhelmed by "word problems" because they are often presented in an intimidating manner.  If only they could be FUN again ... if YOU wrote the problem, I believe you might find it fun!  :-)\n\nBut let's allow ourselves some joy once we uncover all the unknowns in this problem.  Perhaps we'll get results from converting "written words" into "mathematical expressions."\n\nWhat's most important is that you understand the shape involved (think "shoe box") and how the "openings" are placed (think "windows" on three faces of the shoe box).  All the openings are understood to be identical.\n\nUsing only the letter X -- which stands for the length of an opening -- we can indeed come up with expressions for every other dimension of the container, as well as the openings.\n\nX = Opening LENGTH (given)\nX-2 = Opening HEIGHT ("length is 2 more than height")\n\n2X+8 = Container LENGTH ("8 more than twice X")\n3 = Container HEIGHT (given)\nX+2 = Container WIDTH ("2 more than length of opening")\n\nThe surface area of the container's sides (minus the three openings) is given as 300 square meters.  The equation is also given as 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings), so let's just set these equal:\n\n300 = 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings)\n\nAll we need to do is to write, and then substitute, expressions for the areas in parentheses.\n\nSide area = length * height = 3 (2X+8) = 6X + 24\nFront or back area = width * height = 3 * (X+2) = 3X + 6\nTop or bottom area = length * width = (2X+8) * (X+2)\n____________________________ = 2X^2 + 12X + 16\nOpening area = length * height = X (X-2) = X^2 - 2X\n\nTime to substitute:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n\nNow that this (long) equation has been written, it is necessary to solve for X, since we need that value in order to calculate what the problem asks for:  The area of each element.\n\nCollect like terms:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n300 = 12X + 48 + 6X + 12 + 4X^2 + 24X + 32 - 3X^2 + 6X\n300 = X^2 + 48X + 92\nX^2 + 48X - 208 = 0\n\nWhat we hope is that the trinomial on the right-hand side of this equation is able to be factored easily!  Does it?\nMy approach is to factor '-208' and see whether I can 'build' the linear term's coefficient ('48') from the factors.\n\n-208 = -2 * 104 = -2 * 2 * 52 = -2 * 2 * 2 * 26 = -2 * 2 * 2 * 2 * 13\n\nCan we group these factors into (one positive, one negative) numbers that produce 48 when added together?  Yes!\n\n(-2*2) + (2*2*13) = -4 + 52 = 48\n\nWe got lucky ... the '-4' and '52' provide the perfect pair of factors needed to create a 48 (through addition), so we can now factor.\n\n0 = (X-4)(X+52)\n\nThings are getting better still, because only one of the zeros (some call them 'roots') from the right-hand side will be positive.  The zeros are 4 and -52, and we will only choose the 4, since it represents a physical measurement of length (of an opening).\n\nWe can go back to the expressions above and substitute X=4.\n\nSide area = 6X + 24 = 6(4)+24 = 48 sq. m\nFront/back area = 3X + 6 = 3(4)+6 = 18 sq. m\nTop/bottom area = 2X^2+12X+16 = 2(4^2)+12(4)+16 = 96 sq. m\nOpening area = X^2 - 2X = 4^2-2(4) = 8 sq. m\n\nI hope this helps!
1183842
find the are  (base x height = area) and time that by 3........   OR  ........ take each of the length of the sides and times it by 3...
975485
H*W= area\nheight times width\n\none side has a height of 8 and its width is 9 thus the area would be 72
414837
The idea is to build up the corners, you want to get a 2x2x2 corner lined up and then work from there.\n\nHere's a very helpful link\nhttp://lar5.com/cube/
352673
Make right triangles with the height as one leg, a radius (segment from center of base to one of its vertices) as the other leg, and lateral edge as the hypotenuse.  The radius will equal the length of the base edge as it's a side of an equilateral triangle.  So 10^2 + 15^2 will equal lateral edge squared.\n\nSlant height would be the hypotenuse of a right triangle with legs of the pyramid's height, and the apothem (distance from center of base to midpoint of a side.)  You can get the apothem from Pythagorean theorem from the right triangle formed by half a base edge, a radius, and the apothem.  Then use the apothem in another Pythagorean theorem to get slant height.\n\nLateral area = 6 times the area of any one triangle.  Each triangle's area is 1/2 times 10 times slant height.\n\nVolume = 1/3 (area of base)(height).  The area of the base is 1/2 (apothem)(perimeter).\n\nThose are the steps.  Hope you can figure them out from this typing!
763822
Divide the right angle into three equal parts, or three 30-degree angles.\n\nSounds like homework to me (which, hey, I'm cool with!), as if you owned this parcel this is NOT the way you'd want to divide it for real -- you wouldn't want three neighbors all bunched up like that at the confluence of the former right angle.\n\n.
322681
For the first, if one side is x, then another side would be 9-x (so that they would add up to 18).  The area is then\n\nA = x(9-x)\n= 9x - x^2\n\nTo find the maximum area, take the first derivative and set it to 0:\n\nA' = 9 - 2x\n0 = 9 - 2x\n2x = 9\nx = 9/2\n\nSince the parabola opens down, the maximum area happens when x = 9/2.\n\nSo if x = 9/2, the other side is 9 - 9/2, or 9/2 - it turns out to be a perfect square!  And the area is then (9/2)^2, or 81/4.\n\nFor the second, you already have an area and you need to find the perimeter.  So, lw = 36, and we're trying to find the smallest 2l + 2w.  Well, l = 36/w, so the perimeter can be expressed as\n\nP = 2l + 2w\n= 2(36/w) + 2w\n= 72/w + 2w\n\nWe can take a derivative and set it to 0 to find the minimum value here:\n\nP' = -72/w^2 + 2\n0 = -72/w^2 + 2\n72/w^2 = 2\nw^2 = 36\nw = 6\n\nSo if w = 6, then l = 6 as well, and the perimeter would then be 24.
830486
..... ^\n... / .. &#92;\n.. / A1 &#92;\n../-------&#92;\n./ .. A2.. &#92;\n/______&#92;\n\nIs this what you mean (where A1/A2 = 1/φ; and φ = golden ratio = ½(1 + √5))?\n\nIt is my understanding.\n\nSo the two triangles are similar \n\nArea = ½ base * height\n\nA1 / A2 = 1/φ\nThus A1/ (A1 + A2) = 1/(1 + φ)\nie A1 /A = 1/(1 + φ)\n\nThus corresponding sides are in the ratio 1/√(1 + φ) \nNow √(1 + φ) = φ \n(since φ - 1 = 1/φ then φ² - φ = 1 so φ² = φ + 1)\nSo the top base = 1/φ * 8 inches ≈ 4.94 inches\nand the top side walls = 1/2 * 8/φ * tan 52°\n≈ 3.16 inches\nThus the bottom side walls are 8 - 1/2 * 8/φ * tan 52°≈ 4.84 inches
909305
you might try a figure 8 knot, or a stopper knot.\n\nhttp://www.chockstone.org/TechTips/Stopper.htm
348345
12 inches + 7 over 8 inches long
1009332
Convert everything to inches, then multiply.  LxHxW.  They need to be the same units, though.
11716
by using a compass, draw two archs of the same degree to intersect each other on each end.  each point of intersection will be your point A & B use a straight edge to connect. Now you have a proven straight line.
856174
You need two right triangles, I've attached a diagram...\nhttp://img247.imageshack.us/img247/7576/bldgny3.gif\n\nThe bottom triangle has an angle of depression of 20°.  We know the opposite side is 30m (same height as the window, assuming a level street).  We want to figure the adjacent side 'A', being the distance across the street.\n\nFrom the SOH-CAH-TOA mnemonic --> Tan(angle) = Opposite/Adjacent.\n\nSo tan(20°) = 30/A\n\nSolving for A:\nA = 30/tan(20°)\nA ≈ 82.4m\n\nNow we look at the top triangle.  The angle of elevation is 50°.  We now know the adjacent side is 82.4m.  We need to determine the opposite side 'O'.\n\nAgain, tan(50°) = O/82.4\n\nSolving for O:\nO = tan(50°) * 82.4\nO ≈ 98.2m\n\nThe total height of the building is the sum of the two heights (30m to the window and 98.2m to the roof).\n\nSo the building is approximately 128.2m tall.
882237
2 ways:\n1. take the width in inches times the length in inches and divide by 144\n2. convert the inches to feet by dividing by 12, then take the width in feet times t he length in feet
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
163044
DIVIDE SQ INCHES BY 12 AND YOU GET SQ FEET