The breadth of a rectangle is 3cm less than its length...?
928715
Say the length in centimeters is l\nThen the bredth(b) is 3 less than that, \nb = l - 3\n\nNow the bredth, length and diagonal(d) form a right triangle, \nso we can say d^2 = b^2 + l^2 by pythagorus.\nd^2 = (l-3)^2 + l^2\n15^2 = (l-3)(l-3) + l^2\n225 = l^2 - 6l + 9 + l^2\n216 = 2l^2 - 6l          div by 2\n108 = l^2 - 3l            move over the 108\n0 = l^2 - 3l - 108\nYou could use the quadratic formula, or factor.\n0 = (l-12)(l+9)\nl = 12\n\nb = 12 - 3 = 9\nThe bredth is 9cm and the length is 12cm.\n\nSince the other diagonal is just the diagonal flipped along the center, it is the same length: 15 cm.
1065648
square inches doesn't matter. 3/4" is thicker than 1/2"
217916
the diameter cannot be 9cm when the radius is 12cm because Raduis (r) = Diameter (d)/ 2 = 9/2 = 4.5 .\nRadius is half of diameter
923158
Thinner simple work its cause the 3 is greater than the one other than drwing the grph there is really no work. And if u need the graph then just make an X Y table using values -3 to 3 then solve for y and gragh them
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
745313
Not enough information. If the rectangle is also a square, then the length of each side would be 32/4 = 8 cm. Letting x be the length of one of an arbitrary pair of sides and y represent the length of a side from the other pair, you do know of course that x and y must satisfy the equation 2x + 2y = 32 cm --> x + y = 16 cm. Since you have two unknowns but just one equation, there are uncountably infinite many possibilities for (x, y).
1073502
Once again you will see that this is a rectangle.\nA = length times width\nA = 7*2\nA = 14
592688
assuming the board is also of uniform dimensions the CM is at 2.325m from left end
1089271
I don't know how these could be the sides of a rectangle since rectangles have two dimensions, not 3.  And if you mean a rectangular prism (box) none of these are invalid as dimensions.\n\nI think you mean triangle and the answers are 3,4 and 7 as well as 3,5 and 9 because that would result in triangles with no area.  \n\nThe sum of the shorter sides should always be more than the third side.  The only one that meets those criteria is #1.  \n\n#1:  2+3 > 5 --> good triangle\n#2:  3+4 = 7 --> bad triangle (flat)\n#3:  3+5 < 9 --> bad triangle (impossible)\n\nSo #2 and #3 are not the sides of a valid triangle.
65434
The question is WRONG. If you have length =6 and width =7, P=26 and A=42.
976501
You mean "one less."\n\nCall the base x.\nThen each leg is 2x-1.\nAnd the perimeter is x + (2x-1) + (2x-1), which equals 18.\n\nJust solve for x and you can calculate the lengths of the sides.
917141
the radius is half the diameter.
123925
the same way you figure small radius.  i'm not sure i see the dilemma. ???
705154
9cm3\n\n9x9x9=729!!!!!
282503
Let's break the area into many vertical rectangular strips between x=1 and x=2, each strip so narrow that we can assume that the top of the strip is horizontal. Thus each strip will be a very narrow and tall rectangle.\n\nLet's say the number of vertical strips we use is n. The first strip will have its left-hand side at x = 1, the next will have its left-hand side at x = 1 + 1/n, the third will have its left-hand side at x = 1 + 2/n, and in general the kth strip will have its left-hand side at x = 1 + (k-1)/n.\n\nNow, what is the y value at each of these x points? Using f(x)=x² we get, for the kth point,\n\ny = (1 + (k-1)/n)² = 1 + 2(k-1)/n + ((k-1)/n)² \n\nNow the area of each strip is equal to its base times its height. The base of each strip is 1/n. So the area of the kth strip is\n\nA = 1/n * (1 + 2(k-1)/n + ((k-1)/n)²)\n\nNow we have to add the areas of all these strips from k = 1 to k = n and rearrange the terms.\n\n.....k=n\nA = Σ (1/n + 2(k-1)/n² + (k-1)²/n^3)\n.....k=1   \n\n.....k=n\nA = Σ (1/n + (2k - 2)/n² + (k² - 2k + 1)/n^3)\n.....k=1\n\nGiving all fractions the common denominator of n^3,\n\n.....k=n\nA = Σ ((n² + 2kn - 2n + k² - 2k + 1)/n^3)\n.....k=1\n\n.....k=n\nA = Σ ((n² - 2n + 1)/n^3  + (2kn + k² - 2k)/n^3)\n.....k=1\n\nThe first term in the summation sign appears n times in the sum, so we can take it out by multiplying it by n. \n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ (2kn + k² - 2k)/n^3\n\nNow let's clear the parentheses in the summation:\n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ 2k/n² + Σ k²/n^3 - Σ 2k/n^3)\n\nThe first summation contains k=1, 2, 3, ... n, so we use the formula for the sum of the arithmetic series:\n\nS = n(n+1)/2\n\nso the first summation becomes\n\nn(n+1)/n² = (n² + n)/n² = 1 + 1/n\n\nThe last summation also contains k=1, 2, 3, ... n, so we use the same formula for the sum of a series, and we get\n\nn(n+1)/n^3 = (n² + n)/n^3 = 1/n + 1/n²\n\nFor the middle summation, we have to compute the sum of k² for k=1, 2, 3, ... n, or that is the sum of 1, 4, 9, 16, ... n².\n\nThe site\n\nhttp://mathworld.wolfram.com/PowerSum.html\n\ngives the formula for the sum of the squares of a series as\n\nS = (2n^3 + 3n² + n)/6\n\nso the middle summation becomes\n\n1/3 + 1/2n + 1/6n²\n\nNow, we are almost done.\n\nLet's let the number of strips get very large, go to infinity as a limit. Let's see what each term becomes:\n\n(n² - 2n + 1)/n² becomes 1 as n gets very large,\n1 + 1/n becomes 1 as n gets very large,\n1/n + 1/n² becomes 0 as n gets very large, and\n1/3 + 1/2n + 1/6n² becomes 1/3 as n gets very large.\n\nThe area under the curve is 1 + 1 + 1/3, or 7/3.\n\nYou can see why integral calculus was invented. Using integral calculus, this problem can be solved in two steps. The integral of f(x)=x² is x^3/3. Inserting the limits of x=1 and x=2 gives\n\nA = 2^3/3 - 1^3/3 = 7/3
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.
965465
It is related to the length.