helo,i want to ask what are the following perfect triangle in each solutions of this?please help me!!!?
92345
Can you complete the question ? \n\nFYI : Perfect triangles are triangles with sides of integer length and having numerically equal integer area and perimeter.\n\nEg. Triangle with sides of length 6,8, and 10 units.
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
716347
Of three points, A,B,C\n Check the slope of A-B,  B-C, A-C.  If any of the slopes are the same, you have parallel lines and thus no triangle.  So, to do (2), do (4) first.\n  If you know the distance between A-B, B-C, A-C, you can calculate the area with the formula, so (5) also depends on (4).\n\n You write your own code.  yA isn't a site to do your homework, since if you don't do it yourself, you don't pass the class!
859897
***************\nQUESTION 1\n***************\nOn the first question, we first have to find the equation that we will be solving for.  \n\nAssume that x is one integer.\nWe know that x+1 is the other integer because they are consecutive.\n\nSo we now put the wording into an equation\n1/3(x + (x+1)) = x - 5\n\nNow we solve the equation\n1/3(2x + 1) = x - 5\n2/3x + 1/3 = x - 5\n16/3 = 1/3x\n16 = x\n17 = x + 1\n\nOur two integers are 16 and 17.\n\nNow we check our answer.\n1/3 of 33 = 11 & 16 - 5 = 11 We are good! :)\n\n\n***************\nQUESTION 2\n***************\nEvery hour, Amy and Randy together walked 3.6 + 4.2 km apart from each other.\n\n3.6 + 4.2 = 7.8\n\nSo, every hour they will be 7.8 more km apart from each other.  We only want 10 minutes though, and there are 60 minutes in an hour. To get how far apart they will be in 10 minutes, we take:\n7.8 * 10/60 = 1.3\n\nRandy and Amy will be 1.3 km apart after 10 minutes.\n\nWe are done! :)
914532
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
776798
Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
1167473
Here is the answer to your question (using the Pythagorean Theorem):  a^2 + b^2 = c^2, a and b are legs of the triangle, c is the hypotenuse.  So c = 26.\n\nso you have:  a^2 + b^2 = c^2\n                       a^2 + b^2 = 26^2\n                       a^2 + b^2 = 676\n                       \nNow, you need two squares that add up to 676.  The only squares that will work are 10 and 24.  So, here is what it looks like:\n\n                     10^2 + 24^2 = 26^2\n                     100 + 576 = 676\n                          676 = 676\nSince you get the same thing on both sides, you have the right answer.  This means that 10 and 24 are the lenglths of the other two sides of the triangle.\n\nHope this answers your question.\n\nChris (Math Teacher)
787980
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
502657
Try this link and read it all the way through.  It's short.\n\nhttp://mathcentral.uregina.ca/qq/database/QQ.09.00/tom3.html\n\nThis link is better:\n\nhttp://mathforum.org/library/drmath/view/56760.html\n\nJust scroll down.
943161
The faster way to solve this questions is using graphical method. Draw the respective triangles (choosing appropriate scale) and you will get answer.
502102
well chinaren got it started but I believe his explianation is weird.\nLets assume x=3^t\nthen log3(x)+log2(x+1)=3\n=>log3(3^t)+log2(3^t+1)=3\n=>tlog3(3)+log2(3^t+1)=3\n=>log2(3^t+1)=3-t\n=>3^t+1=2^(3-t)\n=>3^t+1=2^3/2^t\n=>(3^t+1)*2^t=2^3\n=>3^t*2^t+2^t=8\n=>6^t+2^t=8\nfrom here there is one one solution possible as t=1\nso x=3^1=3
607226
1.What are the applications for proving 2 Triangles  similar?\n2.What is the formula that is used to calculate the area of a i}Trapezium\n  ii}Hexagon\n3.How can you prove 2 circles are similar.\n\n\nHope this helps..I hope this is what u were asking for?
551883
The answer is C\nwell, one side's length of a triangle should be shorter than the other 2's plus (2 sides plus should be longer than one side)\nthen: 3x+4+8x+2>6x-1\n      8x+2+6x-1>3x-4\n      6x-1+3x+4>8x+2\n--> x>-7/5\n    x>3/11\n    x>-1\nThus, the answer should be x>3/11\n:)
176413
Great puzzle, took me awhile to make it to f=1, g=4. The letters are from a-i: 9,8,7,2,6,1,4,5,3.\n\nPlugged in this is: (9 * 87 + 2/6) / (14 + 5/3) or (783 + 1/3) / (15 + 2/3) which equals 50.\n\nAnswer logged Sunday, March 12th, 3:20 pm with no other valid answers yet. Sorry to have to write that, but I've seen people change their answers too many times in the past.
897859
I find this an interesting question, and I suspect there are several possible methods for solving it.  However, I recommend we deal with it straightforwardly by solving for, say, the base length, height ... any 'variables' found within these polygons' area formulas.\n\nThe understanding here is that both polygons are regular polygons (the triangle is therefore equilateral, the hexagon has 6 equal-length sides, etc.).\n\nThe most creative solutions (and least mathematical) would probably involve cutting up one of the shapes and "re-gluing" it to form the other, thereby gaining some hint at the relationship between their perimeters.\n\nI normally prefer straight math, and would recall:\n\n A(triangle) = (1/2) base * height\n A(hexagon) = ([3sqrt(3)]/2) (side)^2\n\nBut let's "logic" our way through this one first.  Think of the hexagon as being composed of 6 equilateral triangles, each containing 1/6 of the overall area.  Because the original hexagon had an area in 1:1 ratio with the given triangle, we now can state:\n\n Area (SmallTriangle) = (1/6) (BigTriangle)\n\nRecalling a particular theorem concerning similar polygons ("The ratio of their areas will be the square of the ratio of their sides"), we can take this further:\n\n Ratio of BigTri:SmallTri Areas = 6:1\n\nSo,\n\n Ratio of BigTri:SmallTri Sides = sqrt(6):1\n\nPicture now how the small triangles would be "glued together" to form their parent hexagon.  Only the "outside side" would contribute to the hexagon's perimeter ... SIX such sides would be needed.  In short:\n\n Ratio of hexagon perimeter:BigTri perimeter = 6sqrt(6):1\n\nHope this (is correct and) helps!
487186
A hexagon is made up of six equilateral triangles -- triangles with equal sides.  You need to split the triangles in half, in which case you'll get 12 right-triangles.\n\nDraw a diagram!\n\nIf the hypotenuse of the right triangle is x, one of the other sides is x/2 (because you split an equilateral triangle in half) and the other side is half 1.875.  You can calculate x using Pythagoras's Theorem ("the square of the hypotenuse ...).  The width of the screw at the widest point is 2*x.\n\nTo solve (B), it should be enough to draw another diagram.  You just need to allow for the screw head and the walls of the socket.