Just wondering, since the maths questions don't seem to get any further than "solve 3x - 16= 2" or "how do i find the other angle of a triangle". Pathetic questions barely even hard enough for secondry school education. \nAny realy mathematicians out there?!
130621
I teach statistics to Actuarial postgraduate - so I think I am a real mathematician.\n\nSo don't give up hope yet - but I do believe the bias has changed massively since it went from the alpha to beta stage.  We can make a difference!
276301
Your statement is incorrect.  It reads:\n\n"In the image below, there are right angles. How would I prove that they're right angles without a protractor in any triangle?"\n\nActually, it should read:\n\n"In the image below, there are some angles."\n\nAnything else is pure conjecture.  (That means a guess.)\n\nIf you would like us to say something about those angles, you need to tell us how the figure was constructed.  After you tell us that, and ask your question more clearly, your answer will be obtained in short order.\n\nedit...\n\nYou have added some text in your extra text file that doesn't help at all:\n\n"I know that these green points are right angles..."\n\nSo you think a point can also be an angle.  It will be a long time before you get someone (who knows and understands geometry) to agree with that.\n\nIf you start over and explain what you want to know, in correct terms, you may have a chance at getting an answer.
428245
It looks as if there's one big triangle, but in fact the smallest angle of the green triangle isn't the same as the smallest angle of the red triangle, and so although it looks as if there's one big triangle cut up, it isn't so.  And rearranging it makes the hole appear.  I can do a diagram showing just where the extra area unit is, but not on here.\n\nDo you know trigonometry? If so, notice the tangents of the two angles I mentioned are different.  One is 2/5 and the other is 3/7.\n\nIf you don't know trigonometry, just notice that the green triangle has the sides about the rightangle in the ratio 2:5, the red triangle has them in the ratio 3:7.  It's the swapping of the positions of those two triangles that takes up the extra area.  The green triangle has the smaller angle, so when it's the lower triangle the "hypotenuse" of the big triangle has an inward kink.  When the red triangle is the lower one it has an outward kink.  The space between these two "hypotenuses" is one square unit of area.
347375
Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
306075
PART (a)\nUsing Pythagoras’ Theorem in triangle ABC,\nAB^2 + BC^2 = AC^2 = 85 [since AC = sqrt(85)]\nBoth sides AB and BC has to be positive integers.\nBy trial we take AB = 6, BC = 7 [since 6^2 + 7^2 = 36 + 49 = 85]\n\nAgain, using Pythagoras’ Theorem in triangle CDE,\nCD^2 + DE^2 = CE^2 = 85 [since CE = AC = sqrt(85)]\nBoth sides CD and DE has to be positive integers.\nBy trial we take CD = 2, DE = 9 [since 2^2 + 9^2 = 4 + 81 = 85]\n\nThus we have the length of four sides of the hexagon ABCDEF,\nAB = 6, BC = 7, CD = 2, DE = 9.\n\nNow, let us concentrate on triangle AFE.\nIn any triangle, the sum of any two sides is always greater than the third side.\nSo, in triangle AFE, AF + FE > AE\ni.e. AF + FE > sqrt(85) = 9.22 [since AE = AC sqrt(85)]\n\nAs length of 4 sides of the hexagon are fixed so for minimum perimeter the sides of triangle AFE has got to be minimum. As AE = sqrt(85) is also fixed so the other two sides has to be minimum and hence their sum has to be minimum i.e. (AF + FE) has to be minimum. But AF + FE > 9.22 and also both AF and FE has to be positive integers, \nso (AF + FE)min = 10.\n\nNow, 10 = 1+9 = 2+8 = 3+7 = 4+6 = 5+5 \n\nSides AF and FE cannot attain any pair of these values (1,9; 2,8; 3,7; 4,6; 5,5) as length of each side of the hexagon has got to be different positive integers and we already have the nos. 2, 6, 7, 9 as length of 4 sides.\n \nSo, AF + FE = 11 [next positive integer to 10]\nAgain 11 = 1 + 10 = 2 + 9 = 3 + 8 = 4 + 7 = 5 + 6\n\nThus, AF and FE can attain 2 pairs of values, either 1 and 10 or 3 and 8. \nSo the minimum value of (AF + FE) of possible values of AF and FE is 11.\nHence the minimum value of perimeter of the hexagon is \nAB + BC + CD + DE + EF + (FE + AF)\n= 6 + 7 + 2 + 9 +11\n= 35 \n\nPART (b)\nTo find a minimum value of AC, AC has to be less than present value sqrt(85). Further there must be two pairs of positive integers whose sum equals AC^2 (which is less than 85). So those positive integers has got to be in the range 1 through 9. The squares of these positive integers are presented below. One cannot find two pairs from these numbers such that the sum of each pair is equal to any number less than 85. \n1 4 9 16 25 36 49 64 81\nSo the minimum value of AC is sqrt(85) and the corresponding perimeter of the hexagon is 35.\n\nTo get the figure of this hexagon give me your email id so that I can send it as attachment.
492416
I'm assuming you are referring to the disguised triangles in the attached link... (link 1).  I'm sure there are some rigorous mathematical proofs you could supply, or automated ways to count the triangles, but I basically decided to just use the "brute force" method of counting.  The hard part is recognizing all the different triangles while not double counting...\n\nWhat I finally did was to create several copies of the diagram and tried to color the triangles.  You can color the individual triangles (24), right off the bat.  Now if you try to pair up triangles, you can pair them one way (12) or another using the flipped image (12).  Grouping them further in groups of three or four triangles, I counted another set of (7) and the flip for another (7).  Then I noticed four other ways to group four triangles (3), (3), (3), (3).  Lastly, there are two large triangles composed of the inner sixteen triangles (1) and (1).\n\nAltogether I got seventy (76) triangles.  You can double check my work using the attached image (link 2).  I think I got them all and didn't double count...
188270
A triangle is 180 degrees... so 28 + 58 = 86. so now solve this: 180-86=....94!! Tah Dah!
382050
Looks as though this is some kind of assignment. No? Maybe you ought to think it all through and provide your own answers. You can do it!!\n\nChow!!
437660
I have problems with math questions here.  Are people just trying to get somebody else to do their homework?
312805
Trigonomery is much harder than geometry. You do more than angles, and solving for sides there, but not impossible. If you know your graphs and algebra(NEEDED)really well. Go ahead, but be warned, its a bit harder.
474700
It helps if you know something about Pythagorean triples (integer solutions to the Pythagorean equation for right-angled triangles).\n\nIf you do, you will know that there are no primitive triples in which the difference between the legs is 3, The smallest such differences are 1, 7 and 17 (always prime, but not all primes will serve).\n\nTherefore this must be a non-primitive triple which is a multiple by three of a primitive triple with the legs differing by 1. There is a series of such triples, starting with (3, 4, 5) which can be generated from one another.\n\nSo we divide 87 by 3 and get 29 and recognise that this must be the triple (20, 21, 29), the second in the series.\n\nSo plainly the answer is length = 63 cms and width = 60 cms (i.e. 3 x 21 and 3 x 20)\n\nTHE ALGEBRA \n\nL = B + 3\nL^2 + B^2 = 87^2\n2 B^2 + 6B + 9 = 7569\nB^2 + 3B - 3780 = 0\n(B - 60) (B + 63) = 0\n\nAs rectangles in the real world have sides of positive value, ignore B = -63 and take B = 60, yielding L = 63,\n\nBut none of that is necessary if you know your triples!
554546
They taught us SOHCAHTOA back in OL trig, but I never really felt the need for it, I mean just picture a right angled triangle and the longest side is the hypotenuse,h. Take an angle and the side opposite it is h.sin x , where x is the angle. The other side is h.cos x
388618
Not clear on what conditions there are on the third triangle -- e.g. not sure what's parallel to what, because it's lines, not triangles, which are parallel to each other.  On the other hand, certainly it should be possible to draw a triangle by extending some sides!  We need a facility for drawing diagrams here.
405073
uhh,,i will have to ask my maths teacher in college,,\nthanx for the superb question,,
472382
Very understandable....I just posted a question about my pre-calculus homework and i didnt ask for the answer, i just need help finding how to get it....i'd like to know how to do it myself, i dont just want someone to tell me the answer....does me no good @ all!
278634
Memorize that for any triangle having sides of length a, b, and c and angle C opposite side c,\n\nc^2 = a^2 + b^2 - 2abcos C .\n\nNote that for C = 90 degrees where cosine of 90 degrees is zero, the rule reduces to the pythagorean theorem.
358979
the angles are just unknown if i get you correctly.\nthere are sin, cos, and tan proportions you can work out to find the rest. i think you only need three of the measurements to work out the rest of the triangle, like 2 sides and an angle or 2 angles and a side.  not positive.