Quadrilateral JKLM is a rhombus. If m<LMK=36, find m<KJM.?
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1.  Opposite angles in a rhombus are congruent.\n2.  Adjacent angles in a rhombus are supplmentary.\n\n<LMK is half of one of the smaller angles, and <KJM is one of the larger angles.  So <LMJ will be 72, and since <LMJ and <KJM are supplementary, <KJM is 108.
341464
A quadrilateral is a four-sided polygon with four angles. There are many kinds of quadrilaterals. The five most common types are the parallelogram, the rectangle, the square, the trapezoid, and the rhombus. \n The sum of the angles of a quadrilateral is 360 degrees. \n Angle CB  would equal 2 because it is congruent to angle AD.
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Doesn't a rhombus have equal diag.?\n\nIf you mean a parallelogram, the formula is:\n\n(diag. 1 x diag. 2) / 2.  \n\nSo:\n\n(8x10)/2\n\n80/2\n\n40 u^2
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No picture to be found.  \n\nAre M, N, O, P all on the same line?  \n\nIf so, NOQ is isoceles, which makes MNQ and QOP congruent angles.  Since two sides and the included angle uniquely define the third side, so MQ and QP are congruent. and MQP is isoceles.
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l = 5+w\n\nl*w = 36\n\np = 2l + 2w
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If M and N are midpoints of RT and ST, then MN is parallel to RS.  There is a property of triangles that says that if you have a base and a segment inside the triangle whose endpoints are midpoints of the other two sides, the segment is half the length of the base.\n\nSo in your case, we need to find the length of the base RS.  Since the points are on the x-axis, the length is simply |2a-2b|.  Thus, the length of the segment MN is (|2a-2b|)/2, or |a-b|.
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(  4 * (2)^(1/2)    ,    45 degrees)\n\n\n(   - 4 * (2)^(1/2)   , 225 degrees)\n\n\n\nand that big jumble on the start of each is 4 times the square root of 2, or the square root of 32, AKA (32)^(1/2). Math is hard on keyboards.
929836
v = b - u/k\nv - b = u/k\nk(v-b) = u
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you sure you have this problem right ???????\n\nhere's what i came up with\n\nm + ( m + 3 ) = ( m - 2 ) - ( m - 1 ) \nthis is wrong \n\nm + ( m + 3 ) = 2m + 3\n\n( m - 2 ) - ( m - 1 ) \n\n= ( m - 2 ) ( m + 12 ) \n\n= m^2  + 1m - 2m  + 2 \n\n= m^2 - 1m + 2
838340
2m - 3(9m + 1) = 7m + 4(7 + 5m)\n2m - (27m + 3) = 7m + (28 + 20m)\n2m -27m -3 =7m +28 +20m\n- 25m - 3 = 27m + 28\n- 25m - 27m = 28 + 3\n- 52m = 31\n52m = -31\nm = - 31/52
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To avoid any confusion I have called the point of intersection of MO and NP, Q\n\nΔMNP is isosceles (MN = MP)\n\nIn Δ's MNQ, MPQ\nMN = MP (Data)\n<NMQ = <PMQ (Data)\nMQ is common\nTherefore ΔMNQ ≡ ΔMPQ (SAS Property)\n\nTherefore NQ = PQ (Corresponding sides of congruent triangles are equal)\n\nAnd <MQN = < MQP (Corresponding angles of congruent triangles are equal)\nBUT <MQP is a straight angle\nTherefore <MQN = < MQP = 90°\n\nSo MO is the perpendicular bisector of PN
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Yes.\nAny square is a rhombus.\nSome rhombuses are squares (the ones that have all angles of 90 degrees).
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6 + 4(m+13) -9m\n6 + (4m + 52) - 9m\n58 - 5m
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decagon: S x 10\nrhombus: S x 4\nparallelogram: (L x2)+(W x2)\n\nragular fugures are figures that have the same lengths in all the sides ( which is the case with the decagon) so you just add up all the sides or multiply the side by the number of sides the figure has.\n\na rhmbus has all the sides of equal length so again you just add all the sides or multiply the side by 4.\n\nthe opposite sides of the parallelograms are of the same length so you multiply the lenght and width of the parallelogram by 2 and add the results and thats it.\n\ngood luck.
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(3m/n)(n^2/9) \n=(3mn^2)/(9n)\n=(3mn)/(9)\n=(mn)/3
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d.) 2m^2(6m^6+3m^4+7)
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Assuming that MQ is perpendicular to LN,\nthen angle LOM = 90 degrees.\n\nMO = (1/2)MQ = (1/2)18 = 9\nLO = 12\n\nUsing Pythagoras Theorem,\nLM^2 = 9^2 + 12^2\nLM^2 = 225\nLM = SQRT 225\nLM = 15