If 30 playing cards of length12 cmand 6 cm are used to make a pyramid  with  4 cards in the base,find the area covered by the front side of the pyramid.\n\nAns:576sqrt(3)\n\n\nhow this problem can ve solved ?\n\nhttp://www.fotosearch.com/comp/BDX/BDX380/bxp66034.jpg  here is an exampl of playcard pyramid.\n\nnow, problem is  how do i calculate front side area ?\n\nwe have 30-4=26 cards to maniplate.\n\ncan you please tell  how tol slove this
209087
there are 4 cards in the base.So the length of the base is 48. We need to calculate the height. The front side of the pyramid forms a triangle having sides of equal length, so all angles of the triangle are 60 degrees. Using the Pythagoras theorem, the height is found to be 24sqrt3. The area of the front side is 48*24sqrt(3)/2 = 576sqrt(3)
511444
I wish I had Adobe Acrobat so I could e-mail you the lengths of the squares because explaining this without the diagram will be confusing, but I will try.\n\nThe important thing is to remember that the problem stated that this entire figure is a QUADRILATERAL THAT IS NOT DRAWN TO SCALE. So you can't assume that this is a square.\n\nLet's start with the side marked 9cm and 8cm (the bottom side). The small square in between the two is 1cm around (the difference between the 9 and 8 cm squares). Because of that 1 cm square, the larger square on top of the 9cm square is 10cm around (the side of the 9cm square + the side of the 1cm square). So on the left hand side of the quadrilateral, you now have a side of 9cm and 10cm.\n\nWe established the 1 cm square. The square on top of the 8 cm square is 7cm in length (8-1). So the last square on the bottom of the quadrilateral is 15cm (8cm+7cm). So you now have the total length of the bottom length of the quadrilateral (9+8+15)== 32cm. You also now have a partial length of the right hand side of the quadrilateral.\n\nThe square on top of the 7cm square is 4cm around. Subtract the 1cm side from the 7cm side (6cm), then subtract it from the 10cm side...(4cm). Now that you know that the square is 4cm around, then the remaining length on the 7cm side of the square beneath the 4cm square is 3cm.  (7-4). We already established the 15cm square. Add the 3cm length to 15cm to get 18cm and the remaining side of the right part of the quadrilateral, 33cm (18 +15).\n\nNow that you know 18cm square, subtract the 4cm square from it to leave you a length of 14cm. That leaves you with the remaining side of the left hand of the quadrilateral (10+9+14) ==33cm. It also leaves you with the remaining upper side of the quadrilateral, 14cm to give you (18+14)cm == 32cm\n\nSo you have 32+32+33+33==130cm, the perimeter of the quadrilateral, which is a rectangle.\n\nThis is just one way to solve this problem. There are others. For example, once you deduced the 4cm square, you could have added that side to the 10cm side to give you 14cm.
414399
12,000,198 microscopical creatures\n\nit's 7 x 7 x 7 x 7 x 17 x 7 x 6 x 7 = 12,000,198\n\nBy the way, why is this in Words and Wordplay not Math?\nAm I missing something?
331738
I can't see your image, but if it is like any of these other pictures (see link), the problem is with the slopes of the triangles.  They may appear the same, but they aren't and that is where you lose a fraction of a square.  These small differences eventually add up to a full square if you rearrange things differently.\n\nThe key is in the slopes.\n\nI wish I could see the picture because I like area paradox puzzles like these.\n\n-----\n\nI'm back and can see your picture now... it is just like the first paradox in my link.\n\nThe answer, as I said is the slopes of the triangles.  One triangle has a rise of 2 for a run of 5 (slope = 0.40) and the other has a rise of 3 for a run of 8 (slope = 0.375).  So despite how it looks in the picture, that is not a straight line along the top.\n\nIn the top case, the slope causes a slight bend inward on that line.  In the bottom case, the slope causes a slight bend outward on that line.  If you take the difference between these two you get 1 sq. unit which accounts for the "missing" square.  The second link shows this visually.
540250
I was able to draw 16 different forms (8 with their symmetrical reversed forms).  There may be more but I don't have the time to work on it.\n\nI took another look at the problem and realized that in addition to the left-right symmetry there is also a top-bottom symmetry that adds at least another 14 answers to the 16 that I came up with earlier.  So there are at least 30.\n\nCould be more, but it is my bed time.
83501
ok so for A, the first set has 2 seats. so between the 2 seats, theres 2.5feet of steel. and then from each seat to the end, theres 2 more feet (on each side) so thatz 4 feet + 2.5feet = 6.5 feet for the first set.\nfor the 2nd set, 3 seats so 2*2.5 between the seats (2.5 between seats 1 and 2 and then 2.5 more between seats 2 and 3) + 4 =9feet of steel for the 2nd set.\nfor the 3rd set, theres 4 seats so that means (3*2.5) + 4 = 11.5feet of steel.\n\nso you should get the pattern hopefully. if theres 5 seats, then 4*2.5 + 4. 6 seats is 5*2.5 + 4.\nso then the expression would be [(n-1)*2.5] + 4. i dont really know how to explain why its correct. i guess you could just say  it works and it makez sense. for n seats, theres n-1 gaps between the seats and each gap is 2.5feet long. and then theres alwayz 2 feet on both ends added to it which is the +4. \n\nanywayz hope this helps. let me know if u need mroe help or whatever. laterz
152103
Make right triangles with the height as one leg, a radius (segment from center of base to one of its vertices) as the other leg, and lateral edge as the hypotenuse.  The radius will equal the length of the base edge as it's a side of an equilateral triangle.  So 10^2 + 15^2 will equal lateral edge squared.\n\nSlant height would be the hypotenuse of a right triangle with legs of the pyramid's height, and the apothem (distance from center of base to midpoint of a side.)  You can get the apothem from Pythagorean theorem from the right triangle formed by half a base edge, a radius, and the apothem.  Then use the apothem in another Pythagorean theorem to get slant height.\n\nLateral area = 6 times the area of any one triangle.  Each triangle's area is 1/2 times 10 times slant height.\n\nVolume = 1/3 (area of base)(height).  The area of the base is 1/2 (apothem)(perimeter).\n\nThose are the steps.  Hope you can figure them out from this typing!
458485
1)  Notice that 6^2=6*6=36; 6^3=36*6=216; everytime you multiple a power of 6 by 6, the units digit remains 6.  Now what about powers of 5?\n\n2)  If r is total minutes and the first 3 minutes is $1.00, and the rest of r costs $0.20 per minute, then the total cost would be $1.00 + $0.20(r-3) = \n1 +0.2r + 0.6 =\n1.6 + 0.2r\n\n3)  3*10^4 = 3*10,000 = 30,000, and 4*103 = 412.\nNow divide 30,000 by 412 to get their ratio, and then multply that number by 100 to convert to percent difference.\n\n4)  Not enough info - you need the height too.  If the same height, obviously the can with the larger radius\n\n5.1)  65*20% =65*0.2 = 13\n\n5.2)  Again, not enough info. From the info given, all you can say about one-half the number of people who preferred green was that it was between 0 and (65-13)/2, aka 26
304439
Sorry for repeating some of the previous answers.\n\n(1) The probability of getting no aces (~X) is:\n\nP(~X) = (48/52)*(47/51)*(46/50)*(45/49)*(44/48) = 35673/54145\n\nso:\n\nP(X) = 1-P(~X) = 18472/54145\n\n(2) I agree with mathgirl826. The last reply is almost correct except that there are 4*3 = 12 possibilities for the selection of the two suits (not 6). Once the suits are selected there are C(13,3)*C(13,2) hands, so:\n\nP(Y) = 12*C(13,3)*C(13,2)/C(52,5) = 429/4165\n\n(3) blahb31's analysis looks good, but maybe a slightly easier way to do it is to first calculate P(Y|~X), the probability of Y when no aces are drawn. It's just the same as P(Y) would be if the deck contained only 48 cards with 12 cards (no aces) in each suit:\n\nP(Y|(~X)) = 12*C(12,3)*C(12,2)/C(48,5) = 110/1081\n\nThen we can use (this follows from P(X and Y) = P(Y) - P((~X) and Y)):\n\nP(Y|X)\n= P(X and Y)/P(X)\n= (P(Y) - P(~X)*P(Y|(~X)))/P(X)\n\nAll of the values in the above expression have already been calculated:\n\n= (429/4165 - (35673/54145)*(110/1081))/(18472/54145)\n= 1947/18472\n\nThis is about 2% higher than P(Y). The event Y which indicates a relative lack of diversity in suits, is somewhat correlated with a decreased probability of duplication in rank values, and therefore a slightly higher probability of finding any given rank such as an ace.
245759
let us take each lobe as 3kgs(assumption) then they will have 24kgs of lobe. that is each get 8 kgs...so the person with  3 lobes will have 9kgs and other 15kgs. so 3 lobe will give 1 kg to the hunter and 5 lobe fellow will give 7kg to the hunter. so if each kg costs 1 bronze coin.they will divide 1:7. Are you clear?\n\nBut i missed all your other questions gal!!!!!!!! waht you have done...explaining this i missed all other questions.\n\nis the previous one right?\n\nBut i wont be the first to answer them.....
543635
OOOOOOOOOh i LUV LUVLUVLUV challenge 24!!!! ok her are some ideas:\n7,2,2,1=(7-1)(2+2)\n6,8,2,1=(8/4)(6*1)              ***All taken from Challenge 24 cards\n3,3,6,2=(3+3)(6-2)                     (single digit)\n9,2,5,1=(9-1)(5-2)\n~* ah what the heck i dont have time to finish this. I'llget back to you later, soon. and for future reference, DO buy challenge 24 cardsand so you can use them often. the only annoying thing is, it doesnt come w/anser key u have to figure it out yourself:(
269850
1) Let x:the length , y:the width of the picture.Note:x,y>0\nThe first equation we have is : x=y+4 (1)\nThe combined area has the length is x+4,the width is y+4       \nso the second equation we have is: (x+4)*(y+4)=140 (2)\nReplace (1) in (2)-->(x+4)x=140--> x^2+4x-140=0 -->x=10 or x=-14 .since x>0 then x=10(cm)-->y=6(cm).\n2)  Let x: the length,y:the width.Note:x,y>0\n    The first equation we have is x=y+3 (1)\n    The second is: the area=x*y=550 (2)\n    Replace (1) in (2)--> (y+3)y=550--> y^2+3Y-550=0 --> y=22 or y=-25.Since y>0 then y=22(cm)-->x=25(cm)\n3)  Let x,y:two positive integers.Supposed that x>y>0 \n    we have two equation:x*y=240(1) and x=y+8(2)\n    Replace (2) in (1)-->y(y+8)=240--> y^2+8y-240=0 -->y=12 or y=-20.Since y>0 then y=12-->x=20.
505701
Each day she is spending  $ 3.25 on the card  ($1.00 daytime and $2.25 in night)  So it will last for a week.n  Since 7 x $3.25 = $22.75 < $30.\n\nEach week she is spending $3.25 x 7 days = $22.75 per week\n\nfor 13 week, 13 weeks x $22.75 = $295.75 for 13 weeks\n\ndivide $295.75 by 30 for $30 cards   295.75/30 = 9.858  round it off to 10.\n\nshe will need minimum 10 cards.
383443
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!
139489
ON the answer card are the answers to all the possible multiplication combinations using the numbers 0-12.  The child should pick up a question card and then point to the appropriate answer on the answer card.  The answers are not in order, so the adult has to know the answers on his/her own.  good luck :D
206530
Surface area:\nCylinder(lateral):2 times pie times radius times height\ncylinder(total): 2 times pie times radiuse times height + 2 times pie radius squared\nTriangular prism(total): find the area of every side and add all together to get surface area(hint:area of a triangle=1/2bh or bh/2)\nCone(lateral):pie times radius times length\nCone(total):pie times radius times length + pie squared\npyramid(total):find area of ever side and add all together \narea of a triangle=1/2bh bottom of the pyramid is length times width
20918
are you guys still doing homework on Friday night?\nFind the common demoninator for the three numbers.\n2x3=6\n6x3=18\n2x9=18\nso 18 is the common denominator\n2x9=18\n9x2=18\n3x6=18\nWhat ever you do to the bottom, you also do to the top.\nNeed more help....\npinkstealth@yahoo.com