"Tom has a rectangular underground swimming pool 15 feet by 30 feet and he wants to put a uniform border of designer cement blocks around it. If the total area for the border is to be 424 square feet, find the length of the border. If each cement block is 2 feet by 2 feet, and costs $1.24 per block, how many blocks will be needed to make the border and what will the cost be, including a 7% sales tax?"
215020
The area of the swimming pool is 15 times 30 which is 450.\nThe area of the pool and the border is\n450+424\n= 874\nLe the width of the border be x.\nSince it is uniform, the width of the pool with the border is 2x+15 and the width is 2x+30\nThe area of the pool and border is\n(2x+15)(2x+30)\nWe know the area is 874\nThus,\n(2x+15)(2x+30) = 874\nFOIL left side\n4x^2 + 90x + 450 = 874  Subtract 874 from each side\n4x^2 + 90x - 424 = 0\nDivide bot sides by 2\n2x^2 + 45x - 212 = 0\nFactor\n(x-4)(2x+53) = 0\nx = 4 or x = -53/2\nx cannot be negative.\nThus, x = 4\nThe length of the border is 4\n\nNow a cement block is 2ft by 2ft.  The area of the block is 2 times 2 which is 4\nTake the area of the cement border and divide by 4.  This will give you how many blocks you need.\n424/4 = 108\nThus, you need 108 blocks.\nEach one costs 1.24\nTotal cost = 1.24(108) = 133.92\nAdding 7% sales tax gives you\n133.92 + 133.92(.07)\n= 143.29
18194
You can find how many square feet the kitchen is by multiplying 18*15. This equals 270 sq. feet.  (The * means multiplication)\n\nSince each tile is one square foot, you will need 270 tiles.  Multiply the number of tiles by $18, and thats the cost to cover the floor.\n\nEquations:  \nLength * Width\n\n(Number of Tiles) * (cost per tile)
380304
Thinking about this intuitively, if the cost for all 4 walls was the same, what design would give you the smallest perimeter of a rectangle where the product of the sides is 14,400?  That would be a square.  Now, you can see that because that one wall has a greater cost, you want to make that wall's length less (which means making the corresponding back wall less, and the sides grow longer).  But at some point we lose so much efficiency (because the side walls get SO much longer that the total price goes up, that we start paying more. We need to figure out at what point the cost of the front + back walls equals the cost of the side walls. \n\nLet's develop a couple of formulas.\n\nlet c = the cost/linear foot, x = length of one wall (we will arbitrarily assume that x and it's partner x' will contain the glass), y = length of an adjacent wall.\n\nThen total cost = c*x + 1.88*c*x +2*c*y or simply 2.88*c*x + 2*c*y\n\nThe total square footage of 14400 is equal to x * y:  x * y = 14400\n\nSince we already observed that the greatest economy is found when the cost of the front and back walls (2.88*c*x) is equal to the cost of the side walls (2*c*y) we can solve this equation:\n\n2.88*c*x = 2*c*y; or 1.44*x = y\n\nsubstitute for y in: x * y = 14400 yields:\n\nx * 1.44*x = 14400, which simplifies as:\n\n1.44*(x^2) = 14400, which when solved yields:\n\nx = 100 feet\n\nThis implies that y = 144 feet.
463991
Width = x\nLength = 2x+3\n\n2(2x+3) + 2(x) = 72\n\n4x+6+2x = 72\n\n6x = 66\n\nx = 11 and  2x+3 = 25\n\nSo, the width is 11 and the length is 25.\n\n\n\n\n140 + commission = 340\n\ncommission = 200\n\n200 / .2 = 1000\n\nShe earned commission on $1000 but she doesn't start earning until she sell in excess of $300.  Therefore, her total sales must be $1,300.
430215
perimeter = 2w + 2l = 440\nw + l = 220\n\nA = lw = w(220-w) = 220w - w^2\ndA/dw= 220-2w\nmaximum point occurs at dA/dw=0\n2w=220\nw=110\n\nMaximum area= 110(220-110) = 12100sq. ft\n\nI'm not quite sure whts the string of numbers about
350042
First let us calculate the net area of the 4 walls:\n2(36*5)+2(24*5) = 600sq m\nCalculate the area of doors and windows:\n2(3*3) + 4(1.2*1) = 22.8 sq m\nThen to get the area of the walls to be painted:\n600 - 22.8 = 577.2 sq m\nthe rate of white washing the walls is 7$/sq m therefore the cost will be:\n577.2 * 7 = 1177.3 $
90498
Your rectangle (in feet) comes out to 8.2' x 19.7'.  Each row needs 8 tiles plus one strip of 1x6 tiles (if your tile sheet is 2" tiles in a 6x6 configuration).  You will have 19 full rows of tile in the other direction, plus 1/2 sheets of 3x6 tiles.  All in all, you will use 152 FULL tiles, plus 1.3 sheets for the 1x6 strips and 9.5 sheets for the 3x6 sheets.  All in all, you will have 12 tiles left over (2 1x6 strips) if everything lays out properly.\n\nTotal number of tile sheets needed: 163\nTotal left over: 12 2" tiles.
345428
set up 2 equations with 2 unknowns and solve\n\none equation will involve the total poundage\nx + y = 9\n\nthe other equation will  involve your  costs with oh say the Right Hand Side having 9 * .96 as its value\n\n.... i want you to think on your own so this is as far as i go :)
219757
this is very easy, just use the pythagorean therom (i might spell it wrong) the formula is: a^2 + b^2 = c^2 (c is the length of the pipe)\n\n15^2 + 8^2 = c^2\n\n225 + 64 = c^2\n\n289 = c^2\n\n17 = c\n\nyou need 17 feet of pipe
172793
let x = number of 3 cent stamps, and y = number of 2 cent stamps.\n\nThen x + y = 345.\n\nthe total cost will be ($0.03 * x) + ($0.02 * y) = $8.90\n\nYou have two equations with two unknowns. Solve by substituting \n\ny = 345 - x\n\n($0.03 * x) + (( $0.02 * (345 - x)) = $8.90\n\nmultiply by 100\n\n3x + 690 - 2x = 890\n\nx + 690 = 890\n\nx = 200\n\ny = 345 - 200 = 145
432041
I hope you meant to type "what is the width of the rectangle."  If there really is a triangle in this problem I'll need more info.\n\nAssuming it was the rectangle you meant to ask about, the first thing to do with a word problem is write equations for the data.  We'll call the width W, the length L and the perimeter P.  We're told that one side is 15 cm longer than the other.  We aren't told which is which, so I'll say it's longer than it is wide.  Therefore L = W + 15.  We're also told that the perimeter is 94 cm.  The perimeter is the sum of the lengths of all four sides, so it's 2L + 2W.\n\nSo our equations are:\nL = W + 15\n2L + 2W = 94.\n\nNow, substitute the first equation into the second:  2(W + 15) + 2W = 94.\n\nDistribute the 2:  2W + 30 + 2W = 94.\n\nSubtract 30 from each side: 2W + 2W = 64.\n\nCombine the Ws: 4W = 64\n\nDivide by 4: W = 16.\n\nNow go back to the first equation:  L = W + 15.  So L = 16 + 15 = 31.\n\nThe rectangle is 16x31 cm.  I don't know which is the length and which is the width, I just arbitrarily picked, so the answer could be either 16 or 31.  Hopefully you have additional information that will allow you to decide which it is.
156278
Right.  I'm not sure what the question is, but let's assume that the space we are talking about is a rectangle, and you have to figure out what the dimensions are.  Well, the perimeter, 26, is the distance around the space which has the area of 36, which for the sake of argument, we will assume is square feet.  What times what equals 36?  9 x 4, right?  So, it's possible, but not certain that we are looking at a rectangle whose dimensions are 9 feet by 4 feet.  To prove this, remember that a rectangle has four sides.  If our assumptions are correct, it has two sides of 9 feet and two of 4 feet, right? So then,  9 + 9 + 4 + 4 = 26.  It would seem that this is the right answer.  \n\nOf course 12 times 3 also equals 36.  But this time, if we substitute the 12 and the 3 as the dimensions, we would get a perimeter of 30 feet, obviously the wrong answer.  Looks like you have a rectangular space that is 9 by 4 feet.\n\nI hope this helps and that I haven't just muddied the water further.
349087
add the parts together 4 + 1 = 5\ndivide the 100 by 5 = 20\ntherefore one part = 20\nyou need one part of cement = 20kg\ni hope that helps
236531
(a)\nC = mx + b\nC = 10x + 40\n(where fixed cost could be due to the cost of the workers or rental of the factory...)\n(The variable cost is the cost required to make 1 item)\n\n(b)\nR = 50x\n(The revenue, money collected, from selling x items is 50x.\nSo if you were to buy 1 item, you would pay $50\nand if you were to buy 3 items, you would pay $150.)\n\nRevenue = Cost\nR = C\n50x = 10x + 40\n50x - 10x = 40\n40x = 40\nx = 40/40\nx = 1\n\nThe selling of 1 item would result in the break-even point.
561294
Let the width be x ft and the length be y ft.\nSo perimeter is 2x + 2y = 74\nor x+y=37........(1)\n\nCalculating cost of lengths....\n2y*1 = 2y\n\nCalculating cost of widths....\n2x*4.5 = 9x\n\nTotal cost = 2y +9x = 159........(2)\n\nSolving (1) and (2) simultaneously you get x(width) = 12.14 ft and y(length) = 24.86 ft
378060
D+P=500\n300D+225P=217500\ndividing by 75\n4D+3P=2900\n-3D-3P=-1500  (1)*-3\nadding\nD=1400\nsomething wrong with the problem\neven if all the trees were douglas the total price will be only 150,000
404215
let the length and width be 'x' and 'y' reply.\ngiven xy=512\nso x=512/y\nthe perimeter=2x+y\n=2x+512/x\nfor optimising\ndP/dx=0\n2-512/x^2=0\nmultiplying by x^2\n2x^2-512=0\n2x^2=512\ndividing by 2\nx^2=256\nx=rt256=16\ny=512/x=512/16=32\nso the dimensions for min perimeter\n=16'*32'\nperimeter=2x+y=64'\ncost=64*2=$128