rectangular box problem?
492961
Yes 152 cm is correct.\n\nTo solve the problem factor 2002 = 2x7x11x13\n\nThe closer the box is to a perfect cube the lower the ratio of the surface area to the volume.\n\nThis occurs when the sides are  11, 13 and 2x7=14\n\nIn the cube, each edge appears 4 times and so the sum of the edges will be 4 times (11+13+14) = 4x38 = 152 cm
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/
882047
I can't draw a diagram, but that would help.  So, picture the box and the corners.  Label the amount cut off (like the size) from each side x.  The new length on each side is going to be 11-2x or 14-2x (as there are two squares on each side).  These multiplied together (like a rectangle) form the area of 80.  Let's solve.\n\n(11-2x)(14-2x)=80\n\nExpand.\n\n154-50x+4x^2=80\n\nBring the 80 over\n\n4x^2-50x+74=0\n\nBring out the two\n\n2(x^2-25x+37)=0\n\nx is about 1.715 solving this equation.\n\nThere is actually another solution, somewhere around 10.8.  However, this is an unnecessary solution:  10.8*2 is obviously larger than both 11 and 14.  We can throw it out.  \n\nCheck that:\n\n(11-2(1.715))(14-2(1.715))=80\n\nThat's your answer.\n\nHope this helps.
913838
A rectangular prism is like a box. So it has 6 sides in 3 matching pairs: top and bottom, right and left sides, and front and back. You can find the area of each pair and add them together:\nABxBC+ABxBC+BCxCD+BCxCD+CAxCD+CAxCD = PA (Prism Area)\n\nOr you could get the areas of one of each pair and double it:\n2x(ABxBC+BCxCD+CAxCD)=PA (Prism Area)\n\n;-D AB is the length of the front left side, with A at the bottom left corner, B at the top left corner. BC is the length of the front top with C at the top right corner, and CD is the length of the depth, with D at the back right corner.
783613
Area = length x width
97536
BOX!!!!!!!!!!!!!!!!!!!!!!!!
376854
I think practice is the only way. I use several different methods. But usually, I start with the obvious ones. Look within the boxes first to fit in the easy ones. Do that once again in case you missed any, then look at each row and column. Sometimes, if there looks like a few possibilities, you can work out if there is a number which can only go in one square which you might not see straight away. Sometimes, there will be squares where only 2 numbers are possible. If there are 2 squares in the same row/column/area which have the same 2 possiblities and then there is only one square remaining then you know the last square must be the other number. Some people don't think of that.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
844185
square-ish in shape\ncome in different sizes\nmany varieties
504740
After several attempts, it seems that it is rather impossible.  If you start in the bottom corner, you can get all but 3 squares rather easily.  If you're careful, you can have 1 remaining if you start diagonal to the corner.\n\nWhat are you trying to do this for?  Is it a website?  I'd love to try it again.
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
306191
a) cube...final answer
465093
Just search on yahoo for vb tutorials. If you are specific enough with your problem, it will display many forums, that people have had the same problems with, and others have answered.
359093
Square and Rectangle= lenght*height\ntriangle=base*height divided by 2
954958
Top line little square box with green inverted arrows in it
578272
Try downloading a I-mesh.\n\nThey may have it.
756458
w1=width of 1st rect.\nL1=length of 1st rect\nA1=area of 1st rect\nw2=width of 2nd rect\nL2=length of 2nd rect\nA2=area of 2nd rect\nw1=x\nL1=4x\nA1=length*width\n=4*(square of x)\nw2=x+2\nL2=4x+5\nA2=(x+2)(4x+5)\n=x(4x+5)+2(4x+5)\n=4*(square of x)+5x+8x+10\n=4*(square of x)+13x+10\n\nAccording to the given condition\nA1+270=A2\n4*(square of x)+270=4*(square of x)+13x+10\n270=13x+10                //4*(square of x) gets cancelled\n13x=260\nx=20\ntherefore width of 1st rect=20cm\nlenght of 1st rect=20*4=80cm