Kara wants to buy fencing for her rectangular garden. The garden is 24 feet long and half as wide.?
544478
Length = 24 ft\nWidth = (1/2)*24 = 12 ft\n\nPerimeter = (24 + 12)*2 = 72 ft\n\nNumber of boxes = 72/25 = 2.88 boxes.\n\nSo Kara has to buy 3 boxes.
23529
that is one of the answers, if the length of one side is 5. Use the Phytagorous theorem.
199571
8,364 ft (2,550 m) or 2,788 yards
840518
Yvonne Mystic found a shape that produces an area three times the original area. But she did not add the same width on all sides.\n\nIf you assume you are adding x meters to each side, the new shape is (6+2x) by (8 + x), and the area is (6+2x) (8+x) = 144 (i.e., it equals three times the original area, which was 48 sq. ft.).\n\nIf you take the equation from the above paragraph (the equation that ends "= 144"), multiply it out, and rearrange it into standard quadratic form, you will be able to solve for x, which is the width of the strip.
191768
r u sure its not 40 m less, if it is here:\nLet w=width\n    w+40=length\nw(w+40)=21^2\nw^2+40w=441\nw^2+40w-441=0\n(w+49)(w-9)=0\nw=-49 or w=9\nthe width is 9 m and the length is 13 m
487271
suppose the size of the square need to be cut is x^2, (it means that the length of each side is x).\nwe will have an equation:\n   80= (14-2x)(11-2x)\nthen you will get the answer.\n\ni am from china, i not sure i fully expressed my solution:)
800206
The 400 feet of fencing forms the perimeter P of the rectangular area.  The perimeter of a rectangle is P = 2*W + 2*L.  The area of a rectangle is A = L*W.  Take either the W or L from the perimeter equation and solve for it in terms of the other.  For example. 2*W = P - 2*L, or W = (P - 2*L)/2.  Put this expression for W into the area formula\n\nA = L*(P - 2*L)/2\n\nA = P*L/2 - L^2\n\nTo maximize area, take the derivative of A with respect to L to get\n\nA' = P/2 - 2*L  and set it equal to zero\n\nP/2 - 2*L = 0, or  L = P/4;  This gives you the length of the area.  You can get the width from the perimiter formula W = (P-2*L)/2;\nPutting in L = P/4 gives W = (P - P/2)/2 = P*(1-.5)/2 = P/4.\n\nNotice that L = W = P/4; in other words the maximum area rectangle is a square.
832240
We know that the width is 5 yards, and that the length is double that, we can deduce that the length is 5x2=10.  So the rectangle is 5x10.\n\nThis means that the perimeter is 5x2+10x2=30 yards.  Since there are two sides length wise, and two width wise.\n\nThe area is the length times the width, 5x10=50 square yards.\n\nAgain, if you want to grow as a person you need to solve these yourself.  The time it took you to copy all these questions onto Yahoo! you should have been able to answer them yourself.  If you don't know how, you should be asking your teacher to explain them and give you extra tuition.
847943
24.  The perimeter is just the sum of all the sides
268407
Actually it depends on the layout of the meters.  If it is a true square, it would be 2x2 meters... so 2 meters wide.  It is also possible to have four square meters be 4x1 meters.\n\nMore information is needed.
880678
Area = length times width\nA = L time W\nReplace L with 32 and W with 22\nA = 32 times 22\nA = 704 square feet
300586
Adding. example if the lenght of rectangle is 4 and width is 2 then to work out the perimeter you do this: 4+2+4+2= 12 = perimeter.
1077978
23 on the small side.\n\nIf the small side is x, then the long side is 5x.  That makes the perimeter x+x+5x+5x or 12x.\n\n12x=276\nx=276/12\nx=23\n\nThere for the rectangle is 23cmX115cm.
309337
3.2 in an acre I think\n\nwill modify later if I find otherwise
394628
P = 2l +2w\nP>54 = 2(12) + 2x\n54 = 24 +2x\n54-24 = 2x\n30 = 2x\nx= 30/2\nx =15\n\nso at least 15 inches such that perimeter should be equal to greater than 54
423490
since 24 inches equals 2 feet divide 44 by 2 which would give you 22 strawberry plants needed
1073112
First convert everything to feet. The perimeter of 25 1/3 yards x 3 = 76 feet.\n\nThe length of the house is 22 feet. So the two lengths add up to 44 feet of the perimeter. That leaves\n\n76 - 44 = 32 feet for the two widths. So one of those widths would be:\n\n32 / 2 = 16 feet.