Points A and C represent vertices of rectangle ACKL. How many different coordinates can K and L have?
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considering both y values are -2 for A and C....\n\nwe know that a straight line exists between these two points, which means that it HAS to be 1 of the 4 sides.\n\ngiven that, the coordinates or K and L HAVE to be (3, infinity (EXCEPT -2)) and (-4, infinity (EXCEPT -2))
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i believe this is a continuation of the previous prob.\n\nit can be solved graphically. \ndraw the lines x + y = 2,  y - x = 2 and y = -2 on the cartesian plane. the feasible reason will be the area enclosed by the triangle formed by these 3 lines. the vertices being (0,2) , (4, -2) and (-4, -2). it can be proved that the max value of 2x + 3y will be one of these vertices. \n\nsubstituting, u can see the maximum occurs at (0,2) whence 2x + 3y = 6
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All squares are rectangles,\nall parallelograms are quadrilaterls\nand all squares are parallelograms
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To determine if it is a parallelogram:\nIf a quadrilateral has one pair of opposite sides which are parallel and equal in length, then the figure is a parallelogram.\n\nTo determine if it is a rectangle:\nIf a parallelogram has two pairs of opposite sides which are parallel and equal in length, then the figure is a rectangle.\n\nGraph the points and determine the distance between the points, or, in your case, two points have an X of 0 and two have X of 3\n\nBetween the (0, 2) and (0, 6), there is a distance of 4\nBetween the (3, -2) and (3,2) there is a distance of 4\n\nBetween the (0,6) and (3, 2) there is a distance of 3 horizontally.\nBetween the (0,2) and (3, -2) there is a distance of 3 horizontally.\n\nYour figure has two pairs of opposite sides that are equal in length and parallel.  You have a rectangle.
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All rectangles have equal diagonals.\nAnd a square is one example of a rectangle.
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Karnaugh map:\n\nCheck out the following link which will help.
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ZERO (O)  NO RECTANGLES..\n\n\nCHESS BOARD HAVE ONLY SQUARES..
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I suppose that it means all the lines that you can draw from every vertex to every other vertex.  For a rectangle it would be 2.  For a 5 sided figure.( I got up and sketched it) it would b 5.  You only count a line once, not both directions.\n\nAnother sketch suggests 9 for the hexagon.  Why don't you sketch a hexagon, draw and count the lines between vertices.
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Let the trianglas be ABC & BCD\nAD forms 2 triangles, ABD & ACD\nAB=AC since ABC is isisceles\nBC+BD since BCD is isosceles.\nangle ABC=angle ACB\nangle DBC=angls DCB so angls ABD=angle ACD so ABD is congruent to ACD by SAS\n\ncall the point where AD crosses BC E\n\nso both vertex angles are bisected.\nlook at ABC.\nAE bisects BAC & AE is a commin side for BAE & CAE.\nsince AB=AC, the triangles are congruent by SAS\ntherefore BE=CE\n& angle AEB= angle AEC which are supplements so they must be right angles.\nQED
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ANSWER : 25 square units.\n\n---------------------------------\n\nFirst off, draw a picture. \nSo, you have one point on the (0,0), one on the x-axis, one on the y-axis, and one on the diagonal line.\n\nLet's call the length of the horizontal line segment of the rectangle w.\n\nThe other side will then be (10-w).\n\nSo, the area of the rectangle is A = w*(10-w).\n\nIf you graph this, there is a maximum at the point (5,25), so 25 is the max area.\n\nTo verify this, take the derivative of A with respect to w, and set equal to zero.\nA' = 10 - 2w = 0\n2w = 10\nw = 5\nA = 5(5) = 25
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-E-N-G-L-A-N-D- 2 POINTS
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Not all Parallelograms are rectangles, But all rectangles are parallelograms.
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Not enough information. If the rectangle is also a square, then the length of each side would be 32/4 = 8 cm. Letting x be the length of one of an arbitrary pair of sides and y represent the length of a side from the other pair, you do know of course that x and y must satisfy the equation 2x + 2y = 32 cm --> x + y = 16 cm. Since you have two unknowns but just one equation, there are uncountably infinite many possibilities for (x, y).
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Square and Rectangle= lenght*height\ntriangle=base*height divided by 2
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y= -4x^2+8x-1\ny = -(4x^2-8x+1)\ny = -[(2x-2)^2 - 4 + 1]\ny = -[(2x-2)^2 - 3]\ny = 3 - (2x-2)^2\n\nSo the vertex is (1,3) and it is a maximum point.
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Construct meaning with a compass and straightedge? And by "adjacent" they're already at right angles? Just measure each side with the compass, move it so one leg is at the other corner, and draw a small arc approximately where the other leg extends. Do this for both sides, then the point of intersection is your missing point, and you can draw out the rectangle with the straightedge.\n\nIf the sides are not at 90 degrees, just turn one of them (measure one with the compass, draw a circle based on it, then take the two points of intersection and construct the perpendicular bisector, then take the point where that intersects....)
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(1)\nh = 3\nk = 4\n5 = a(0-3)² + 4\n5 = 9a + 4\n1 = 9a\n(1/9) = a\n\nY = (1/9)(x - 3)² + 4\n\n(3)\nY = (¼)x² + 4x -  7\nY = (¼)(x²+16x) - 7\n\n x² +16x = x² + 16x + (16/2)² - (16/2)²\n     = [x + (16/2)]² - (16/2)\n     = (x + 8)² - 8²\n     = (x + 8)² - 64\n\nY = (¼)[(x + 8)² - 64] - 7\nY = (¼)(x + 8)² - 16 - 7\nY = (¼)(x + 8)² - 23