12 inches wide by 8 inches high with the lid still being able to be closed?\n\nsteps
121441
The cubes, when stacked, form a brick with sides divisible by 3. \nTwo dimensions of the tin are divisible by 3, while 8, the height, is not. This means we will only be able to put two stacks (height of 6) into the box with 2 inches of air on top. The volume of the stack is 6*12*9 = 6*4*(3^3), so we will only be able to fit 24 cubes inside, unless we cut up the remainder (which will, incidentally, fit inside).
257012
U ripped dis off from da AISSE 2006 sample que paper didn't u. Here's da value points for it\nLet ABCD be the bucket which is the frustrum\nof a cone with vertex O (as in fig.) ½\nLet ON = x\nΔ ONB ~ Δ OMC ∴ = Ü x = 4 1\n∴ height of cone = 8 + 4 = 12 cm ½\nVolume of bucket = [π(9)².12 — π(3)².4] cm³ 1\n= 312 π cm³\nSlant height of cone of radius 9cm = 9² + 12² cm ½\n∴ L = 15 cm ½\nSlant height of cone of radius 3cm = 3² + 4² cm\n = 5 cm ½\nArea of the copper sheet used to form bucket\n= [π(9) (15) - π(3) (5) + π(3)² cm² 1\n129π cm²
469917
It is always understandable when a student feels overwhelmed by "word problems" because they are often presented in an intimidating manner.  If only they could be FUN again ... if YOU wrote the problem, I believe you might find it fun!  :-)\n\nBut let's allow ourselves some joy once we uncover all the unknowns in this problem.  Perhaps we'll get results from converting "written words" into "mathematical expressions."\n\nWhat's most important is that you understand the shape involved (think "shoe box") and how the "openings" are placed (think "windows" on three faces of the shoe box).  All the openings are understood to be identical.\n\nUsing only the letter X -- which stands for the length of an opening -- we can indeed come up with expressions for every other dimension of the container, as well as the openings.\n\nX = Opening LENGTH (given)\nX-2 = Opening HEIGHT ("length is 2 more than height")\n\n2X+8 = Container LENGTH ("8 more than twice X")\n3 = Container HEIGHT (given)\nX+2 = Container WIDTH ("2 more than length of opening")\n\nThe surface area of the container's sides (minus the three openings) is given as 300 square meters.  The equation is also given as 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings), so let's just set these equal:\n\n300 = 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings)\n\nAll we need to do is to write, and then substitute, expressions for the areas in parentheses.\n\nSide area = length * height = 3 (2X+8) = 6X + 24\nFront or back area = width * height = 3 * (X+2) = 3X + 6\nTop or bottom area = length * width = (2X+8) * (X+2)\n____________________________ = 2X^2 + 12X + 16\nOpening area = length * height = X (X-2) = X^2 - 2X\n\nTime to substitute:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n\nNow that this (long) equation has been written, it is necessary to solve for X, since we need that value in order to calculate what the problem asks for:  The area of each element.\n\nCollect like terms:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n300 = 12X + 48 + 6X + 12 + 4X^2 + 24X + 32 - 3X^2 + 6X\n300 = X^2 + 48X + 92\nX^2 + 48X - 208 = 0\n\nWhat we hope is that the trinomial on the right-hand side of this equation is able to be factored easily!  Does it?\nMy approach is to factor '-208' and see whether I can 'build' the linear term's coefficient ('48') from the factors.\n\n-208 = -2 * 104 = -2 * 2 * 52 = -2 * 2 * 2 * 26 = -2 * 2 * 2 * 2 * 13\n\nCan we group these factors into (one positive, one negative) numbers that produce 48 when added together?  Yes!\n\n(-2*2) + (2*2*13) = -4 + 52 = 48\n\nWe got lucky ... the '-4' and '52' provide the perfect pair of factors needed to create a 48 (through addition), so we can now factor.\n\n0 = (X-4)(X+52)\n\nThings are getting better still, because only one of the zeros (some call them 'roots') from the right-hand side will be positive.  The zeros are 4 and -52, and we will only choose the 4, since it represents a physical measurement of length (of an opening).\n\nWe can go back to the expressions above and substitute X=4.\n\nSide area = 6X + 24 = 6(4)+24 = 48 sq. m\nFront/back area = 3X + 6 = 3(4)+6 = 18 sq. m\nTop/bottom area = 2X^2+12X+16 = 2(4^2)+12(4)+16 = 96 sq. m\nOpening area = X^2 - 2X = 4^2-2(4) = 8 sq. m\n\nI hope this helps!
97674
Yes 8x8x8 = 512.\n\nah, puddles_818 .. putting a placeholder answer of just "h", then doing the work, and then coming back to post the answer after half a dozen people answered it correctly .. funny.
38944
With the dimentions of the container, calculate its volume. One cubic foot = 7.48 gallons.
25632
The question is WRONG. If you have length =6 and width =7, P=26 and A=42.
566103
Depends, what kinds of containers they are, for eg. a cylindrical container will hold more than a rectangular container with the same base Perimeter and height.  This question is very unclear.  The answer would probably be the cylindrical one.
417223
Here's the answer: (I'm assuming the top of the box is counted in the area)---------   x = side of square base, h = height, S = surface area, P = perimeter length\n\nS = 2x^2 + 4xh\n\nP = 8x + 4h ---->  h = (P-8x)/4  Substituting,\n\nS = 2x^2 + 4*x*(P-8x)/4 = Px - 6x^2\n\nFor your particular case where P = 120 cm,\n\nS = (120x - 6x^2) cm^2
58722
You all thought way too hard on this one. There is no way to know who has more because the asker did not tell what size the containers were. \nWhoever had the biggest container had more because the numbers he gave are more than 1, you can not fill a container 1 and 2/3 of the way, nor can you fill one 7/4 of the way. Once they get to the top, they are full at 1 each. For example, if you try to pour 1 and 3/4 gallons (which is 7/4) into a one gallon container, you will only have one gallon in the container.\nIf both containers were the same size they both have the same amount.
645989
There's a control.......we know the box remains the same........but  not the size....or if at one time its filled.......A + B ......"density is less"........."density is more".is a wide scope.....especially If you want liters for an answer.
293745
width = 5 \nlength = 12
262716
You have not provide suficiant information. The hieght is of no cosequence to the diameter since you failed to provide a base dimension of lenght or hieght of which could be the same. The water level is not realitive because no parementers have been established. Is the vessel full? Further more: a diameter of a spout is meaningless. Absolute stupidity.
676788
depends how ur gonna make it\nif life a book, then 8" - 5.75"\nif three fold, then 8" - 3.83"
435033
Think of the bottom of box B. It's 3ft by 3ft. So you could fit 4 of box A in the bottom layer because 2 box As are three feet wide. Then you could fit 4 more in the next layer, because box B is 3 feet high, and two box As are 3 feet high. So its a total of 8.
336711
Think of it as a small bowl with high sides.  It's a convenient container for mixing and heating small amounts of materials.
500452
There are 12/5 litres which must be multiplied  by 7/4. That gives us 84/20. Divide both numbers by 4 and you get 21/5, or 4 and 1/5.\n\nAt least I hope so :)\n\nAnd for Sazi, I hope I've explained it clearly enough that you can talk your teacher through it.
599704
Since one litre = 1000cm^3,   We can construct a cylindrical container with surface area S = 2pir^2 +2pirh.(I am using pi since I don' t have the greek symbol.)  We wish to minimize S subject to the constraint that V = pir^2h = 1.  Now, rewrite the equation for S as S = 2pir^2 + 2pir(1/pir^2) = 2pir^2 + 2/r).  Now, differentiate S wrt r and you get ds/dr = 4pir - 2/r^2 = 0 => r = 1/pir^2.  Now , use h = 1/2pir^2 to solve for h.\n\nFor the rectangular boxes, the problem is impossible without some relation between the three sides..\n\nLet me know if you don't understand.