In parallelogram ABCD, if m<A=5x+30, and m<D=x, find m<C.?
483665
<c=<a, <a+<d=180 (laws of parrallelograms....?!) \ntherefore 180=(5x+30)+(x)\n180=6x+30\n150=6x\nx=25\ntherefore<c=5x+30\n<c=5(25)+30\n<c==125+30\n<c=155 assuming <c is not less than a right angle (look at you picture in the text book)
1093701
Parallelogram:\na. 4 sides\nb. 1 set of opposite sides is parallel\nc. 1 more set of opposite sides being parallel\n\nQuadrilateral:\na. 4 sides\nb. 1 set of opposite sides is parallel\n\nAs can be seen above, the conditions for the Parallelogram is a super set of the Quadrilateral. A Quadrilateral needs one more condition to become a Parallelogram (another set of opposite sides being parallel).
975217
Line AC = 6, line BC= 2x+2, line AB=x+1\nSo 2x + 2 = 6\n2x = 4\nx = 2 <-- option C
38332
(x-2)/(x^2-2x-15)+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(x+2)/(5-x)\n(x-2)/((x+3)(x-5))+(-2-x)(x+3)/(x-5)(x+3)\n\n[(x-2) + (x+3)(-2-x)] / [(x+3)(x-5)]\n\nANSWER :\n(x^2+4x+8)/((x+3)(5-x))
19885
(5) x (2x^2-3x+5) I think
1181427
ABC+ABD+C'D\n=ABC+ABD(C+C')+C'D\n=ABC+ABCD+ABC'D+C'D\n=ABC(1+D)+C'D(AB+1)\n=1ABC+1C'D\n=ABC+C'D
115023
Imagine a `speed` x `time` chart. Plotting the data, we will have parallelogram whose area will the length the car walked between the two speeds.\n\nWe'll have, then, 30*((5.56+13.89)/2) = 291.75m
627684
C=5/9(F-32)\nF=(9/5)C +32
375748
All squares are rectangles,\nall parallelograms are quadrilaterls\nand all squares are parallelograms
749939
Nobody can help you because it's false.\n\nLet a1 = 1, a2 = 4, b1 = 2, b2 = 3, and n = 7\nThen a1 + a2 = 5, b1 + b2 = 5\nBut a1*a2 = 4, while b1*b2 = 6.\n\nYou are talking about modular arithmetic, right?
590231
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
227193
∫[(2x^4) - (4/x^5)]dx=\n\n∫(2x^4)dx - ∫(4/x^5)dx=\n\n2∫x^4dx - 4∫[x^(-5)]dx=\n\n2*[x^5/5] - 4*[x^-4/-4]=\n\n(2/5)x^5 + x^-4 + C=\n\n(2/5)x^5 + 1/x^4 + C
743138
join AC\nnow youhavetwo triangles ABC and ADC\nin triangle ABC <CAB+<B+<ACB=180*.............(1)<s of a triangle\nin triangle ADC <CAD+<D+<ACD=180*............(2)--------do--------\nadding (1) and (2)\n<ABC+<ADC+<B+<D+ACB+<ACD=360*\n=><DAB+<B+<BCD+<D=360*\nhence proved
1154468
make a common denominator\n\na/b - c/d = a*d/b*d - c*b/d*b = (a*d-c*b)/b*d
1168391
c=1\na=2\nn=3\nCan = 1X2X3 = 6\nA=1 \nn=2 \ny=3\nb=4\no=5\nd=6\ny=7\n1X2X3X4X5X6X7= 5040\n\ng=1\no o = 4\nd=3\n\n1X4X3= 12\n\nA=1\nT=2 \n1X2=2\nM=1\na=2\nt=3\nh=4\n1X2X3X4= 24
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
747277
5a+6b-3a+b=?\nSimplified= (2a+7b=?)  or (2a = -7b) or (7b = -2a)\nThey are all the same\n\n15x-3y-7x-4y=?\nSimplified = (8x-7y=?)  or (8x = 7y) or (-8x = -7y)\n\n15m-17n-23m+25n=?\nSimplified -8m+8n=? or (8n=8m) or (n=m)