The area of a soccer field is 5000 yd^2. The length of the field is twice the width. Find the dimensions of the field.\nI do not know how to do this so if u do not know ethier please do NOT post.
200219
l=2w, and length times width is area, so...\nl*w=A...use l=2w, w=w, A=5000\n2w*w=5000, or\n2w^2=5000...divide by two and solve for w!
534142
Being adjacent they share a common side. Assuming the corrals are identical, then if we let x be the length of the common side and y be the length of the other side, the amount of fencing required is 3x+4y = 240ft. The area of each corral is xy =1000sq ft. So y = 1000/x. Substitute this value in the first equation and you have:\n\n3x+4(1000/x)=240\n\n3x^2 +4000 =240x\n\n3x^2 - 240x+4000=0\n\nFrom the quadratic formula, x= 56.33ft. or x= 23.67ft.\n\nIn the first case the corral would have dimensions:\n\n56.33ft and 1000/56.33 = 17.75ft (approximately)\n\nIn the 2nd case, the dimensions are:\n\n23.67ft. and 1000/23.67 = 42.25ft (apprx)\n\nIn both cases 240ft of fencing would be used. Check:\n\n3x+4y=240ft\n\nCase 1: 3(56.33)+4(17.75)= 239.99 = 240\n\nCase 2: 3(23.67)+4(42.25) = 240.01 = 240\n\nIf you wanted the corrals to be closer to squares you would choose the 2nd dimensions.
159959
Part a)\n\nx = width\ny = length\n2x + 2y = 1000, so y = 500 - x\n\nA = xy = x(500 - x) = 500x - x^2\n\nb)  x should be positive (and small enough to fit the plot of land that he is planting the garden in)\n\nc)  If one corner is at the coordinates (0,0), the opposite corner is at (x,y) as defined above.\n\nd)  A = 500x - x^2\n\nCalculus Solution:\n\ndA/dx = 500 - 2x\nWhen 500 - 2x = 0, the area is a maximum.  Here, x = 250 and y = 250, so the maximum area is 62,500 square yards.\n\nNon-calculus Solution:\n\nGraph the equation A = 500x - x^2.  You can see that the parabola crosses the x-axis at x = 0 and at x = 500.  By symmetry, you know that the peak of the parabola must be half-way between 0 and 500 at x = 250.  When x = 250, A = 62,500, which is the maximum area.
100505
Maybe I'm not understanding the problem correctly, but if "There is only space for the pature 2 be @ most 100ft" means that no side can be longer than 100 feet then I think you may have written the question down incorrectly.\n\nIf each side has to be less than or equal to 100 feet, then let's say that each side *is* 100 feet in length. There are 4 sides, so the perimeter is 4 times 100 feet = 400 feet. You say you have 500 feet of fencing. This is more than enough to surround the maximum-sized pasture allowed. So what's the problem?\n\nThe only two linear equations you need to worry about in this case are:\nwidth <= 100\nlength <= 100\n\nChoose any values for the width and length that satisfy these constraints,\nsay the width = 75 and the length = 50 or the width = 2 and the length = 89\nand you have solutions to your question.\n\nOften these questions ask for the maximum area that it is possible to enclose. In this case there is only one solution: width = 100, length = 100 therefore the area enclosed = 10,000 square feet and the amount of fencing used is 400 feet.
226627
A full size adult soccer field is 60 yards x 100 yards.  This means the perimeter is 320 yards which is 0.18 miles.\n\nso if you just run around the perimeter (in a rectangular path), you're doing 0.18 miles per lap.\n\nif you ran around a 400m track that encircled the soccer field, you're doing 0.25 miles per lap.
366864
You could look up the stuff on google...I know the football field if thats what you want but a football has about 4 different sizes...there are peewee, Junior, College, and NFL..\nThe field is 100 yds long plus 10 yds on each side for endzones...and about 53 yards wide..
282380
To find area you multiply the length times the width. It's pretty easy once you know
33674
1.)The area of a carpet is 308sq.m. If its length is 2m less and its width 6m more, the carpetwill be square. Find the dimension of the carpet...\n\nSo we can see the length of the carpet is 8m longer than the width\n\nL B = Area\nL (L-8) = 308\nL^2 - 8L - 308 = 0\n(L-22)(L+14) = 0\n\ntherefore L=22m or L= -14m\n\nNow a carpet that was -14m long wouldn't be very useful, so we discard that result as not representing the real world.\n\nL = 22 ; B = 308/L = 14m\n\n2.) Gerry wants to fence a rectangular field whose area is 1200sq. m. He has only 100m of fencing so he decided to fence only the three sides of the rectangle letting the wall be the fourth side. How wide should the rectanglebe??? \n\nLet a and b  be the length of sides of the field\na.b = 1200       ------(1)\n2a+b = 100       ------(2)\nb = 100 - 2a     ------(2a)\n(100-2a)(a) = 1200     (substituting into eqn 1)\n2a^2-100a+1200 = 0     (expanding and collecting terms)\na^2-50a+600 = 0        (divide by 2)\n(a-20)(a-30) = 0\na = 20 or a =30\nb = 100-2a\nif a = 20, b = 60      (Check L=20+20+60,A=20*60 correct)\nif a = 30, b = 40      (Check L=30+30+40, A=30*40 correct)\n\nSo two feasible answers.
599383
No, it's not.\n\nOne acre is slightly less than 91 yards on an American Football field, with the full field, including the end zones, it would cover approximately 1.32 acres.
324783
1.  Area of rt triangle whose leg is 6ft and whose\n    hypotenuse is 10ft?\n\n    Area = 24 sq. ft.  \n\n  Explanation:\n\n    First we need to find the other leg's length...\n    Using Pythagorean Theorem,\n    \n    c**2=a**2+b**2 \n       \n    where c is the hypotenuse, a is one leg, and \n    b is the other leg.\n\n    So, 100 = 36 + b**2;  b**2 = 100-36 = 64; therefore b=8.\n    Now, Area = 1/2(base*height) since the base is one leg\n    (we'll say it is a) and the height is the other leg \n    (therefore we'll name it b) \n    the Area = 1/2(8*6) = 1/2(48)= 24 sq. ft.\n\n2.   The hypotenuse is approx. 4.9 meters.\n    \n  Explanation:\n     \n     Using Pythagorean Theorem: c**2=a**2+b**2 we find that\n     c**2 = (3.5)(3.5) + (3.5)(3.5)\n     c**2 = 12.25 + 12.25\n     c**2 = 24.25\n     c = sq. rt. (24.25)\n     c = 4.9244 meters\n     hypotenuse is approx. 4.9 meters.\n\n3. hypotenuse is approx. 20 inches long.\n\n  Explanation: Area = 1/2(x)**2 \n               where x is length of one leg.\n\n     x**2 = 2*Area;\n     x**2 = 2*98\n     x**2 = 196\n     x = sq. rt. (196)\n     x = 14 inches\n\n     Now, to find length of hypotenuse, we use the\n     Pythagorean Theorem:\n\n     c**2 = a**2 + b**2; since a=b this can be rewritten so\n     c**2 = 2(a)**2\n     c**2 = 2(14)**2 -- Remember 14 is the length of a leg.\n     c**2 = 2(196)\n     c**2 = 392\n     c    = sq. rt. (392)\n     c = 19.7989\n     c is approx. 20 inches long.\n\n\n4.   The researcher will climb approx. 10 meters \n     to reach the crack.\n\n   Explanation:\n      \n      Using the Pythagorean Theorem: c**2 = a**2 + b**2\n      If we set a = 24, then c**2 = (24)**2 + b**2,\n      but we also know that c + b = 36 \n      which means c = 36 - b.\n\n      So to solve for this we need to set\n      (36-b)**2 = (24)**2 + b**2\n      1296-72b + b**2 = (24)**2 + b**2\n      1296 - 72b = (24)**2\n      1296 - 72b = 576\n      -72b = 576 - 1296\n      -72b = -720\n         b = -720/-72\n         b = 10 \n\n      And now we have our 10 meters the researcher \n      must climb to reach the crack in the tree.\n\n\n6.    Approximately a 42 inch stick is the longest stick\n      that will fit in a box that is 24 in X 30 in X 18 in.\n\n  Explanation:\n\n      First use Pythagorean Theorem to find the diagonal\n      inside the rectangle of 24 in X 30 in.\n\n      c**2 = (24)**2 + (30)**2\n      c**2 = 576 + 900\n      c**2 = 1476\n      c = approx. 38.418 in. -- we will call it 38 in.\n\n      Now, use Pythagorean Theorem to find the diagonal\n      inside the "new and invisible" rectangle that is \n      38 in. X 18 in.\n\n      c**2 = (38)**2 + (18)**2\n      c**2 = 1476 + 324\n      c**2 = 1800\n      c = sq. rt. (1800)\n      c is approx. 42.43 in.  \n      \n      We will call "c" 42 inches long.
517952
this would be easier to see if we could draw on here, so bear with me.\n\nThe sidewalk is a series of 4 rectangles, each 4' wide and of varying lenghts, to surround the plot of 75' x 80'\n\nthe four rectangles then are of lenthgs 75, 75, 80, 80.\n\nThis would leave 4' x 4' squares at each corner.\n\nSo the area would be:\n\n75*4 + 75*4 + 80*4 + 80*4 + 4*4 + 4*4 + 4*4 + 4*4 = 1304 sq ft
562281
A soccer field does not have a standard size. A tennis court is 78 feet x 27 feet.\n\nIf you did want to try to put a standard soccer field size, it should be in the range of 100-130yds x 50-100yds
616857
area = pi times the radius squared\n\nThe diameter is twice the radius.\n\nSo divide the diameter in two, square that and multiply by pi.
589842
Always try to find the open space. Pay attention to the off side rules. Learn to kick the ball in a way that the ball ends up were you want to send it..  The same heading the ball. Practice until you can do that. The ball is the tool of the game.
508281
Its really hard to say because all courses are different lengths and widths, but a fair average would be in the area of 6,500,000 sq yards.\n\nA golf course would consume about the same area as a football stadium with surrounding parking lots.\n\nThis would make a golf course about 2 sq miles.
457468
Word problems like this can be hard.  Everybody has problems with them sometimes.\n\nIt does help to draw it out.  So you draw a little rectangle like this:\n ______________\n|                            |\n|                            |\n|______________|\n\n2 of the sides are 120 m and the other 2 are 150 m.  So you take 2 times each of these numbers and add them together:\n2(120) + 2(150)\n\nThat's your perimeter, in meters.  Now if your poles are 2 m apart, you don't need one for every meter you get in your answer, you only need 1 every 2 meters.  So you divide your answer by 2, and that gets your number of poles you need to put around the field.
323445
yup your answers are correct.\n\nhowever your units are not, length and width are measured in yards not sq yards.