find the coordinates of all of the corner points?
916482
Sorry, but I can't graph and shade here, but if you've already done that, then you'll see that inequality 1 meets with inequalities 2 & 4 and that inequality 3 also meets with inequalities 2 & 4.\n\nSo now you can set up 4 simultaneous equations to find the corner points.\n\nLet Sim 1 be : -x + 3y = 20 and 13x + 3y = -50 (1 & 2)\nLet Sim 2 be : -x + 3y = 20 and 5x + 2y = 36 (1 & 4)\nLet Sim 3 be : 13x + 3y = -50 and -3x + 5y = -34 (2 & 3)\nLet Sim 4 be : -3x + 5y = -34 and 5x + 2y = 36 (3 & 4)\n\nSim 1 gives the point (x, y) = (-5, 5)\nSim 2 gives the point (x, y) = (4, 8)\nSim 3 gives the point (x, y) = (-2, -8)\nSim 4 gives the point (x, y) = (8, -2)
411393
Yes 10, think of vertices like the corners.
89202
Hi again, just graph the three boundary lines and shade the correct sides.  This will give you a shaded triangle with vertices (2,3), (4,5), and (4,0)  [I think; I don't have graph paper handy].  The maximum and minimum values must occur at one of these 3 sets of coordinates.\n\nfor (2,3), 3x + y = 9\nfor (4,5), 3x + y = 17\nfor (4,0), 3x + y = 12\n\nSo 9 is the minimum value and 17 is the maximum.\n\nCheck it out yourself to be sure the graphs really intersected where I thought they did!
502159
?\n\n(x-coordinate, y-coordinate)\n(0,0)\n(1,3)\n(2,6)\n(3,9)\n\nCos y-coordinate is 3 times of x-coordinate?
943665
this is impossible to answer, there are many shapes that could be made this way.  do you have any other information like right angles or angles at corners?
333747
Scroll down some on the website I give you and you will find it. The title is "Plot Coordinates on a Map"
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
873152
lol you need a graph for the first part!!\nand for the rest use:\ndistance formula: d= [squared root of (x1-x2)squared + (y1-y2)squared]\nmidpoint formula:  [(x1+x2)/2 , (y1+y2)/2]\n\nAnswers:\nd= squared root of 68\n----------------------------\nm= [(-4+4)/2, (-1+ (-3)/2)]\nm= (0,-2)
880255
I can just give you a link to help you solve it.\n\nhttp://www.ticalc.org/pub/83plus/basic/math/geometry/
598966
1- line segment\n2- angle\n3 -triangle\n4 - square, rectangle, rhombus, diamond, parallelogram, or quadrilateral\n5-pentagon\n6-hexagon\n7-heptagon
838358
i can't give you the answer but i used my ti-83 and i can tell you the points of the graph...\n(1,0) (2,4) (3,8) (4,0)\n\nHope this helps
736970
Here are some real tough ones.(Looked tough to me)\n\nProve that cot(90 - theta)sin(90 - theta) = sin theta\n\nFind Cos(3x)\n\nFind Sin(3x)
86738
The angles all add up to 180 degrees.\n\nIf you know the lengths of all the sides you can use the "Law of Tangents" to find all the angles.  This is advanced geometry and Trigonometry, so rather than type, I am including a link to a pretty good explanation.\n\nhttp://answers.google.com/answers/threadview?id=38131
577034
Since the radius is 44, you could go either 44 units left, right, up or down from the center and find a point on the circle itself.  So (44,0) would be one such point.  However, these points are not in a quadrant; they are on an axis.  All the other points would fit in the circle's equation, x^2 + y^2 = 44^2.  So pick a number for the x-coordinate, square it, subtract this from 44 squared, then do the square root.  That gives you the y-coordinate.
967408
Don't you have a book with a diagram of all those points?
1016603
If you use degrees, there are 360 points.\n\nIf in mills, there are 6400 points.
1106173
The direction (angle) of |u| and |v|?