Origin Symmetry - need educated help!?
820527
criterion is that changing sign of x changes the sign of y\n\nso a & b are symmetric, and other's aren't, since even powers and adding a constant both ruin the symmetry
746185
I can only answer 1 and 3 for you, cause i dont know the others, but here it goes:\n\n             1.) Think of this problem in terms of a right triangle. Draw a horizontal line. This line is the tidal force of 3* (10^3) N. Then draw a line on the end of the line perpendicular to the horizontal one. This line is your wind force of 6* (10^3) N. Connect the two ends of the lines with another line, to make a complete right angle triangle. This is where it gets a little tricky. The angle formed when the slanted line and the horizontal line meet is the angle you are trying to find. To find it, use the tan function, which is defined as opposite side's value (6*10^3 N) over the adjacent side's value (3*10^3 N). Lets use x as a variable for the unknown angle. So...\n\ntanx=(6*10^3) / (3*10^3)\nUse inverse of tan now (on calculators, you usually have to press shift or 2nd, then the tan key.) So...\n\nInverse tan ( (6*10^3) / (3*10^3) ) = x\n\ndo that in your calculator, and the value you get is the angle.\n\nFor the net force just subtract (3* 10^3) from (6*10^3) and the result will be your net force.\n\n3.) I dont know if u know this or not, but there is a formula that is part of Newton's second law. It is F=m*a.\n"F" being the Force, while "m" is the mass, and "a" is the acceleration. Just plug in the values you are given so that it comes out....\n\n390=270*a\n\n390/270=a ("a" is in meters per second squared)\n\nHope that helps
868256
a circle has infinite lines of symmetry\n\nso does a straight line\n\nso does a sin wave\n\nthere are many figures that have infinit lines of symmetry\na more interesting question might be whether there are an infinite\nnumber of figures that have infinite lines of symmetry. (there are)
496604
Try doing this with an asymmetric figure. The translation and reflection won't give the same result.\n\nHowever, you can get any translation by doing *two* reflections one after the other.
871837
Yes.\n\nSkew symmetry means that A^T=-A where A^T is the transpose of A. If (a_ij ) are the entries of A, this will say that\na_ji=-a_ij.\nWith i=j,\na_ii=-a_ii,\nso \na_ii=0.
279906
Some of the answers in this are ridiculous.  In the "G" lecture of the second degree, the candidate is told that it represents "geometry" which is the Liberal Art/Science most revered by Masons, and that, more importantly it is the initial letter of God.  both of these things are discussed in the same 5 minute lecture.  Sorry to dissapoint the answerer who had a whole theory worked up.
798763
echinoderms are secondarily radially symmetric whise ancestors were bilaterally symmetric.\nhence, secondary radial symmetry implies that members of a radially symmetrical group came from ancestors with bilateral symmetry.
243852
Hope this helps\n\n\nhttp://argyll.epsb.ca/jreed/math9/strand3/triangle_congruent.htm\n\nhttp://mathforum.org/geopow/fullsolutions/102596.solutions.html or www.mathforum.org
410020
Xcuse me dude, plz be clear abt wt u wanna ask abt the PRINCILES OF GEOMETeRY
293291
Bob, I used to enjoy your show before you started doing all those really expensive houses.  Then, of course, you left.\n\nOh, well.\n\nDon't you know, Bob, that isosceles triangles don't have "right sides?"  An isosceles triangle has two "legs" of equal length.  If it is a right isosceles triangle, the side opposite the right angle is called the hypotenuse.\n\nSay hi to Norm for me!
701505
As others have said, your current equation has x and y being 0, which would be the origin.\n\nIf you meant to type:\n2x - y = 0\n\nThen to solve this, isolate y:\ny = 2x\n\nThis is just a line going through the origin at a slope of 2 through points (1,2), (2,4), (-1, -2), (-2, -4), etc.
954443
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
570600
Ok, first just so you get the figure right, draw dots n name them like i have n at the places like m showing them below\n      A                B\n      \n    E                   F     \n\n   D                      C             G\n\nNow, join AB, D-C-G, EF , A-E-D and B-F-C  also.. join\n A-F-G .. GET IT?? and you'll have AE equal to ED as well as BF equal to CF as the median of the trapezium joins the mid=points of the non-parallel sides...\n\nNOW IN triangle BFA and triangle CFG ,you have\nBF congruent to CF ----   Proved tht F midpoint of BC\nangle BFA congruent to angle CFG--vertically opposite angles\n and as AB is parallel to CD (opposite sides of a trapezoid) it is also parallel to CG ( construction was to join D,C and G , remember??\n\ntherefore, abgle ABF congruent to angle FCG --- alternate angles for parallel lines AB and CG ..\n\nSO YOU NOW HAVE triangle ABF congruent to triangle CFG --- Angle-Side-Angle Test of congruency...\n\ntherefore, AF congruent to FG --- corresponding sides for congruent triangles\n\nNOW, IN THE BIG TRIANGLE ADG,\n  E is the midpoint of side AD - by construction\n  F is the midpoint of AG  --- Proved AF congruent to FG\n \ntherefore... EF parallel to DG --- By the Midpoint Theorem which states that-- A line passing through the midpoints of any two sides of the triangle is parallel to the third side (and has half the length of the third side )\n\nSO WE'VE NOW PROVED EF // DG\n .\n. . , EF IS ALSO // TO DC --- D,C AND F LIE ON THE SAME LINE.. I.E. THEY ARE COLLINEAR\n\n SO, EF//DC\nBUT DC IS PARALLEL TO AB ---- OPPOSITE SIDES OF THE TRAPEZIUM ABCD\n \n .\n. . , EF IS ALSO PARALLEL TO AB\n\n .\n. ., WE CAN PROVE THAT ----> AB // EF // DC  (REASON- both AB and DC are parallel to the same linE EF)  \n                     \n\n .\n. . WE HAVE PROVED THAT THE MEDIAN OF THE TRAPEZIUM (in this case GE)  IS PARALLEL TO BOTH THE BASES (i.e AB and CD )
983185
Take the first few derivatives of sin -- it is cyclical and you will see a pattern.  After you see the pattern, you will be able to say exactly what the 53rd derivative is, without having to actually take the derivative 53 times!\n\n~ ♥ ~
69753
I'm a bit rusty on this but I believe what you have is  the group of transformations on a pentagon. x is a reflection across an axis of symmetry, y is a rotation where a vertex moves to the next adjacent vertex.\n\nI think that each G/<x> is going to be a refection across a different access of symmetry so it's a group, and each G/<y> is going to be another cyclic group of order 5 generated by a rotation of 1,2,3,4 sides.\n\nHope this helps its been decades since I did heavy group theory.
902708
Try \n1. http://en.wikipedia.org/wiki/Eccentricity_%28mathematics%29\n\n2. http://mathworld.wolfram.com/Eccentricity.html
628627
Rotational symmetry: if something is rotated a fixed set of degrees about an axis, and it remains the same, its called rotational symmetry. eg, circle rotated about its center.\n\nLine Symmetry: If an object is symmetrical about any given axis. eg, diagonal of a square.\n\nCongruence: Implies exact match. Two toyota cars can be thought to be congruent.\n\nSimilarity: Implies the angles at the vertices match, but dimensions don't match. eg an equilateral triangle of area 1cm^2 and 100cm^2 are similar, not congruent.