In mathematics I know I am going to see quadrant grids soon in fifth grde math.But how do you use one?
132061
You are usually given a point to plot.  The point is written as (-4, 2).  The first number represents x and the second represents y such as (x,y)\n\nTo plot the point, you start at the origin (middle) which is (0,0)\n\nThe x-axis is the horizontal line that goes left and right.  You put your pencil on the origin *but don't make any marks*\n\nLook at the first number in the ordered pair (-4,2)\nIt's -4.   You move across to -4 (on the x-axis...the horizontal line) and stop your pencil there but don't make a mark.\n\nThen look at the second number (-4,2) and you see it's 2.  Since 2 is positive you move up to spaces and then plot your point.  If it was negative, you would move down that number of spaces on the y-axis (vertical line).\n\nHope this helps.  Plotting points is fun.
90027
Add them together and divide by 5.\n\nWow. I could seriously have you people do my homework for me. Every night. I'll type out the problems and see how you all do. :P JP
217227
Properties of mathematics makes it work. I'm not sure exactly what it has you do but essentially it does this.\n\nIf your number is X. You add 5 to it then subtract 5 from it and you get your original number. Of course the nice ones hide the fact that your doing exactly this. \n\nDavid Copperfield did a TV bit once where he had a 3X3 square. You start out at any square you want and move according to his directions. After each move he took away a square. In the end you always end up on one square. That was cool. Anyone heard of it?
286577
Short Answer:\nThe diagonal line or hypotenuse is not a straight line.  It bends either way thus both figures are NOT triangles.\n\nLong answer:\n\nThe two diagrams are not perfect triangles; they are quadrilaterals. Why is that significant? Look at the "diagonal line" on each of the diagrams. You may be able to see, if you look closely, that the supposed diagonal line is not really a straight line. Rather, it's two connected segments (two straight line segments) with a slight angle to it where the two segments meet.\n\nThe top diagram bends in a convex manner, so that there is slightly more area contained below the diagonal. In the bottom diagram, the bend is in the opposite direction.\n\nTo help you see this better, look at the green traingle in the top diagram. It's dimensions are 2 units high, and 5 units long. The Red triangle is 3 units high. In order for the two line segments that make up the combined diagonal to be a straight line, the Red triangle would therefore need to be 7.5 units long (2/5 = 3/x; therefore x=7.5). Instead, it's 8 units long. If that Red Triangle were only a half a unit shorter along the base, then the diagram would all line up, and we wouldn't be trying to solve this problem!\n\nThe four pieces that make up the upper diagram "appear to fit exactly" on a triangle that is half of a rectangle 13 units long and 5 units high. (Count the squares: 13 across, 5 down.) If "appearances are correct", then that would imply that the total area of the four pieces is half of 13x5, or half of 65. So you would think that the total area of these four pieces is 32.5 square units. Let's total them up and see:\n\nIn the top diagram, you can see that the two irregular pieces combine to make a 3x5 rectangle. So these two pieces are 15 square units (you can count the squares, as well! One is 7 squares, the other is 8.) That's part of our total.\n\nThe Red Triangle is half of a 3x8 rectangle. So it's 12 square units. So add 12 to our total.\n\nThe Green Triangle is half of a 2x5 rectangle. So it's 5 square units. So add 5 to our total.\n\nTotal Area of the four pieces is 15+12+5 square units, or 32 square units. So the top quadrilateral is covering only 32 squares out of the 32.5 square units that a triangle would cover. It LOOKS LIKE it covers 32.5 squares, but really it only covers 32!\n\nLet's check the bottom diagram. Obviously the four pieces only cover 32 square units, using the same math as above. There's a 33rd square which is empty.\n\nIf you take the top diagram, and rotate it 180 degrees, so that it fits perfectly on the bottom diagram, you'll see that the red triangles are butted against each other, as well as the green traingles. Therefore, if the top diagram has a slight angle caused by the line segments, the bottom diagram has a slight angle exactly complimentary to the top one. Therefore, if the top diagram undershoots the 32.5 square units by just a half of a square, then the bottom one OVERshoots it by half a square.\n\nAdd up the two half squares, and you have the whole hole.\n\nFour pieces that cover 32 squares, plus four pieces that cover 32 squares, plus one empty square equals 65 squares, which is the area of rectangle 13 units long and 5 units high.\n\n(7+8+12+5) + (7+8+12+5) + 1 = 13 x 5\n\nClear as mud, eh?
139562
now, if you add that it has 4 sides and all right angles, you probably can pass that off as the definition of a rectangle.\n\notherwise you just have a parallelogram.
527662
first divide [0,2pi] into 6 intervals [0,2pi/6],[2pi/6,4pi/6],[4pi/6,6pi/6] amd so on\n\nNow let's look to [0,2pi/6]. Since cos is periodic the graph on other intervals will be the same as on [0,2pi/6]\nEven on [0,2pi/6] we have to restrict more and exclude the case when cos(6 theta) is negative since R is always non-negative.\n Now if we take R=2cos(theta) you will obtain a circle, since it's R=2cos6(theta) this circle would be compressed, you obtain a "petal"\nNow since this repeats 6 times you get 6 "petals"( but on the graph you see only one)
479157
I think you are referring to magic numbers making up a magic square wherein you arrange such numbers in a grid and get the same sum for all the numbers in the same row, same column and diagonal.\n\nFor example, take the numbers 1 -9 and arrange them in a 3 x 3 grid (3 rows and 3 columns).  Set the 5 in the middle and arrange the rest around it so as to get a total of 15.  You should get something like this arrangement.\n\n     6     1     8     =     15\n     7     5     3     =     15\n     2     9     4     =     15\n\n     =     =     =      two diagonals each add up to 15 as well\n    15    15   15
478166
http://math.arizona.edu/~lega/UG/Colonies/Colonies.htm\n\nhttp://www.magma.ca/~scimat/B_subtil.htm\n\nhttp://www.pnas.org/cgi/content/full/0600829103/DC1
164006
All squares are rectangles,\nall parallelograms are quadrilaterls\nand all squares are parallelograms
40936
explain sample:\n(12,5)\ngraph (12,5) first thing you have to know, x=12, y=5. so on the graph, you have to move 12 space to the right, then move 5 space up. (if the number is negative, then move oppositional direction), after that place a dot on the graph. that is it. sample and easy, what grade are you in?
489471
Unfortunately, the people above have misused geometry.\n\nThe way that you'd solve this in a geometry class would involved quite a bit of trigonometry. Here's a way to solve it without using as much. (I worked it out both ways and both methods give the same result. I also checked the result.)\n\nI'm assuming that your octagon is regular.\n\nDraw a regular octagon. Call the length of each side a.\n\nNow use vertical and horizontal lines to draw gridlines on the octogon. This divides the octagon into four triangles, four rectangles, and a square in the middle.\n\nThe square has area a^2..\n\nThe triangles are 45-45-90 triangles having hypotenuse a. This means that both sides are a / square root of 2. This means that each triangle has area:\n\n1/2 * (a / sqr(2)) * (a / sqr(2)) = a^2 / 4.\n\nThe rectangles have one side that is a side of the octagon (a) and one side that is a leg of the triangle (a / sqr(2)).\n\nThis means that the rectangles have area:\n\na * (a / sqr(2)) = a^2 / sqr(2).\n\nThe octagon's area is equal to the sum of the areas of its regions:\n\n4 triangles + 4 rectangles + 1 square\n\n4*(a^2 / 4) + 4*(a^2 / sqr(2)) + a^2 = \n\na^2 + 2sqr(2)*a^2 + a^2 = \n\na^2 (2 + 2sqr(2)) = 64.\n\nSolve for a, and you get approximately 3.6407 feet.\n\nThat is to say, an octagon having a side length of 3.6407 feet has the same area as a square having a side length of 8 feet.\n\nThis seems contradictory until you realize that an regular octagon isn't made up of just squares and triangles. If they were, then the "diagonal" sides would be longer than the "horizontal" and "vertical" sides.
377442
Simply draw a chart that has each of the 6 variables at right angles to each other, draw lines between the linked variables such coke-greenbike (#3), then use process of elimination.\n\nYeah, right... draw a 6-dimensional hypercube....\n\n Jeez, I've spent about 3 hours on this, and so far it looks like Carla, who was on the yellow bike which was 1st in line, was Bigbird and brought rootbeer and jelly beans.  Either Emily, who was dressed as either dracula or the clown, or Daniel dressed as the other, brought the candycorn.  But at the moment I have a contradiction, which means that somewhere I made a logical error.  I'll try to get back to this sometime later - good luck
588506
I got  the first part. just get the cosx by itself. then you wind up with the cosx = -(the square root of 2/2)\nwhich is 225+360n and 315 +360n(in degrees)\nSorry i can't help you on the other one. Good luck, just keep trying new things with it, would be my advice.
195541
I like to add conceptual dimensions like color, (red, green, and blue each get one) temperature, pitch, volume, etc. Then the nine dimensional origin (0,0,0,0,0,0,0,0,0) would be a point located at the center of the 3-D grid colored 50% grey at zero degrees centigrade exhibiting middle C at zero decibels.\n\nI could also imagine extra dimensions represented in the form of trajectories, but more often than not I am thinking in the terms above.
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get some graph paper. plug in numbers for x and calculate its equivalent y value. do that 5 times, and you should have your curved line. If not, use higher or lower numbers than you used the first time ( I like starting with 0) Then  draw a line perpendicular to the most bent part (aka axis of symmetry, aka cut the line in two so both sides of the new line look like mirror images), find 2 points on that cutting line, and there you go.
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Since you have to divide the area into five rectangles, you will need rectangles from:\n\n0 to 1, 1 to 2, 2 to 3, 3 to 4, 4 to 5\n\nSince you are using the right endpoints, you need to be using the second value of each division; that is, in the "x to y" examples, you'll be using the y endpoints 1,2,3,4, and 5.\n\nThe area of each rectangle will then be 1*f(x), since the width of each rectangle is 1.  Therefore, the total sum formula should be as follows (with * representing multiplication).\n\n1 * f(1) + 1*f(2) + 1*f(3) + 1*f(4) + 1*f(5)\n\nOr, quite simply, f(1) + f(2) + f(3) + f(4) + f(5) to solve using the right endpoints.  Calculating this, we get\n\n(25 - 1^2) + (25 - 2^2) + (25 - 3^2) + (25 - 4^2) + (25 - 5^2)\n(24) + (21) + (16) + (9) + 0 = 70\n\nTo calculate using the left endpoints, recall that our intervals for the rectangles were:  0 to 1, 1 to 2, 2 to 3, 3 to 4, and 4 to 5.  In this case, we use the LEFT endpoints and take the FIRST value of each interval; that is, 0,1,2,3, and 4.  The width of the rectangle doesn't change at all (the width is 1), so our left endpoint summation formula because\n\n1 * f(0) + 1 * f(1) + 1 * f(2) + 1 * f(3) + 1 * f(4)\n\nor, quite simply\n\nf(0) + f(1) + f(2) + f(3) + f(4).  We calculated f(1) through f(4) already above, so we'll copy those values.  But f(0) = (25 - 0^2) = 25, so our total now becomes:\n\n25 + 24 + 21 + 16 + 9 = 95.\n\nArea using right endpoints = 70\nArea using left endpoints = 95
282203
You asked for the nth term, so here it is\n\nFigure: n\n\nSquares: n(n+1)/2\n\nDots:  (n^2 + 5n + 2)/2\n\neg. n = 5\n\nsquares = 5 x 6/2 = 15\n\ndots: (25 + 25 + 2)/2 = 52/2 = 26