How many trapezoids does it take to complete a design?
814748
The obtuse angles of the trapezoids will be (in degrees) 90+invtan((57.35-29.27)/(2*27.54)), which is approximately 117 degrees.\n360 degrees minus the sum of two of these angles (126 degrees) is the interior angle of whatever regular polygon you are trying to make.\nThe interior angle of a regular hexagon is 120 degrees.\nThe interior angle of a regular heptagon (7-sided polygon) is approximately 128.57 degrees.\n\nTherefore, you could put 6 of these tables into a circular arrangement and have some extra space between the front edges of the tables, or you could put 7 of these tables into a circular arrangement (a better fit) and have a little bit of extra space between the back edges of the tables.\n\nIt would also look more interesting to have heptagonal arrangements of tables in the room.
936075
The answer comes from the definitions of the terms used.  Let's go through them:\n\nA trapezoid is a quadrilateral with one pair of opposite sides parallel.  The parallel sides are called the BASES.\n\nThe other two sides are called the LEGS.\n\nThe MEDIAN of a trapezoid is a LINE SEGMENT connecting the mid points of the legs of a trapezoid.  Hence the median is // to the bases.\n\nIt is the same for a parallelogram but since both opposites sides are // it doesn't matter which ones form the base and which ones form the legs.  The median will be // to the base you chose.\n\nHope that helps.
910632
lines, polylines, blocks, circles etc... look in the help files if you have the software\n\nhttp://www.we-r-here.com/cad/tutorials/index.htm
837569
I'm afraid the previous answers got this one wrong.  \n\nYou DO need 10  posts per side.  That's right.  But the corner posts do double duty.  Each corner post counts as one of the ten in both sides.\n\nThe correct answer is 36 posts are needed.
79526
You could shortcut creating a design by creating a SlideMaster... which is a template design of sorts. There's one for the regular slide, and one for the Title slide. Hope it helps.
755601
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!
1134906
Here's a link to a PDF file I wrote consisting of all the steps involved (though not quick enough to beat the other two answerers...)\n\nhttp://prash4.esmartdesign.com/ans2.pdf
326809
The area of a trapezoid is the height times the average of the bases, or h*(b1+b2)/2.  In your case, your bases are the top and bottom of your figure (parallel to the x-axis), and your height is along the y-axis.  So,\n\nh = a\nb1 = c\nb2 = b\n\nA = a(b+c)/2
644195
Computer Aided Design
600899
The following site not only explains to you the use of Multiple Design Templates but also has a presentation on the process.  Good luck! \n\nhttp://www.webmayfair.com/easilyexpandyourdesignoption.htm
185652
since your ddoing coordinate geometry you should at least tell what shape you're trying to get.\nI'm assuming it's a parallelogram so u should make it (1,8)
58756
It depends on a few things.  How long do you want the strand to create your arch?  How many balloons will you use in each cluster to create your strand?  The typical clusters are made of either three or four balloons.  It also depends on the size of balloon you use.  Most arches are made with 11" balloons.  \n12 x 3' \n\nI prefer to use 11" balloons in clusters of 3.  The typical length of the strands I do is about 24'.  I can usually plan on about 12 balloons for every 3' of the strand.  So a 24' strand takes around 96 balloons.  I would round that up to about 120 to account for poppage and damaged balloons.  \n\nIf you're looking to have one done I would rcommend you call a party supply store that sells balloons.  If they don't do it, they will have the best recommendations and references for you.  I've also included a site below that sells the instructions on how to make one.  Good luck and have a happy balloon day!
1070589
Assuming you mean as 4 shapes equivalent to the original L shape, the answer is given in the attached image.
601270
I can do it... but let me think a little on it.  Can I take the full 4 months?  :-)\n\nOkay, I think I have a solution... took me about 5 minutes I think:\n\nIf I move the dot from B1 -> B2 and C1 -> C2, then B and C will still have 4 dots, and row four will still have 4 dots, but now row two will have 4 dots.\n\nYou didn't say they couldn't share the same square!!!!\n------------\nI'm back\n\nYou didn't exactly state what a "row" was.  I would have called that a "line" but maybe that's what you meant.\n\nOkay, I have a couple revised solutions.  A "row" is now defined as 4 dots that are colinear (on the same line) but each point must be distinct.  This invalidates my previous answer.\n\nOne solution matches with what has already been stated, which is:\nA4 --> B5  and  D4 --> C5.  This makes two pairs of overlapping lines each with 4 dots.\n\nHowever, my new solution is to take C1 and place it halfway between columns B and C and halfway between rows 2 and 3.\n\nThen you have a line on row 4, a line on column B, and two diagonal lines!!!!\n\nOne dot moved, 4 lines created!!!!\n\nAnd if you *must* move two dots, then just move B1 --> B5 and you still have a valid answer.
1104691
Analysis\nDesign\nCode\nImplementation\nTests
595914
You find the area of the trapezoid (the base of the figure) and then multiply by the height of the figure.
978173
You need to start off with a six-pointed star (constructed with two equilateral triangles).\n\ntriangle AEC and triangle FBD\n\nNow connect A to D, then connect B to E, and finally C to F. If you \ndrew everything correctly, those three segments should meet at a point \nin the center of the star.  \n\nThe 3 line segments you just drew plus the six sides of the star make \n9 lines. There are 5 points of intersection on each line, and the \ntotal number of intersections is 19.