Pythagorean theorm help plz?
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Draw a picture of a square with length 90, representing the bases.  The diagonal represents first to third base; call it x.  By Pythagorean Theorem,\n\nx^2 = 90^2 + 90^2\n\nx^2 = 16200\n\nx = 127.3
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use synthetic  divsion method \ni think this link would help\nhttp://www.purplemath.com/modules/synthdiv.htm
520168
Try this:\n\nhttp://www.algebra.com/
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(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
665741
http://en.wikipedia.org/wiki/Pythagorean_theorem#Proofs\n\nCheck below for the "proof of converse".  It's the most straightforward there is.\n\nBasically, you try to construct a triangle with x,y, and z, only to discover that it is congruent to a right triangle!
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D\n\nSeriously, what class are you in? Algebra? Geometry? You chould know this!! It's the pythagorean theorm- [a2 +b2 =c2] (2's = squared).  You have a and b, the two adjacent sides of the triangle.  Therefore, it is the square root of 50, or approx. 7.07.  Thats SUPER important, make sure you remember it!
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Pythagoras:\nAC squared = AB squared + BC squared\nTherefore\nAC = square root of quantity (AB^2 + BC^2)\nYou can do the math.
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You **did** say that the Pythagorean Theorem was a given, right? I think that your INTENT is to prove the converse of the Pythagorean Theorem, right?\n\nLet x, y, and z be positive integers such that\nx^2 + y^2 = z^2.\nClearly z > x and z > y (otherwise either x^2 or y^2 must be less than zero; prove this for yourself if you must).\n\nLet x and y be the sides of a right triangle that form the right angle.\nThen, by the Pythagorean Theorem the length of the hypotenuse h is\nh = √(x^2 + y^2)\nh = √(z^2) . . . . . . . . (given equation above)\nh = z . . . . . . . . . . . (given z is positive) \n        \nTherefore, by side-side-side congruence, any triangle whose sides are of lengths, x, y, and z, where x^2 + y^2 = z^2, must be a right triangle with hypotenuse z.\n\n\nASIDE TO HOOD...: You are using the Pythagorean Theorem to prove the Pythagorean Theorem. Although the questioner's intent is not perfectly clear, I don't think the intent was to prove "if x, then x". If so there are easier ways to do so.
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I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
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1 - A is enough because with the lengths of 2 sides you can use pythagoras' theorem to work out the length of the other
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the answer is pi 3.14. use the pythagoran therom with x= pi(r2) then compensate for the remainder and it'll work itself out. If that doesn't work, try FOIL. first, outside, inside, last.
947787
In this equation, you have a right triangle.  After 3 hours, the first car has traveled 150 km south and the 2nd car ahs traveled 210 km west.  You can find the between these two cars at 3 hours by using the Pythagorean Theorem since you have a right triangle.\n\nDistance = sqrt(150^2+210^2) = 258.07 km\n\nThe general form of the Pythagorean Theorem is:\n\na^2 + b^2 = c^2, where:\n\na = distance of car 1\nb = distance of car 2\nc = distance between cars.\n\nTake the derivative to find how the rates are related to one another.  Take the derivative with respect to time:\n\n2a(da/dt) + 2b(db/dt) = 2c(dc/dt)\n\nIn this problem, we are looking for dc/dt.  These are our known values:\n\na = 150 km\nb = 210 km\nc = 258.07 km\nda/dt = 50 km/hr\ndb/dt = 70km/hr\n\nPlugging these values for the equation above, solve for dc/dt:\n\ndc/dt = [2a(da/dt) + 2b(db/dt)]/[2c] ------->\n\n[(2*150*50) + (2*210*70)]/(2*258.07)\n\nTherefore, dc/dt = 86.02 km/hr.  Now change to km/sec:\n\n(86.02 km / 1 hr)(1 hr / 60 min)(1 min / 60 sec) = 0.024 km/sec\n\n-----------\n\nHope this helps
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Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
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http://www.gomath.com/Questions/question.php?question=27271\nthere are your answers
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What the Pythagoras theorem states is that the square of the hypotenuse is equal to the sum of the squares of the two other sides (lets call them "a" and "b").\nSo you have to satisfy this condition:\n \nhypotenuse^2 = a^2 + b^2\n \nIn the case hypotenuse = sqrt(3),\n \n3 = a^2 + b^2\n\nSince there are an infinite amount of pairs (a, b) that satisfy this equation, there are infinite triangles.\n \nIf you´re thinking of the special cases (the ones you memorize for geometry and trig) I can only remember the case of sides 1, sqrt(2), sqrt(3). But I´m pretty sure there were 2 other cases (with different hypotenuses).
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Pythagorean Theorem: To find the hypotenuse:\nSquare each leg. Add the results. Take the square root of the result. \n\n7^2 + 9^2 = c^2
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ok i got the answer to this\ni just solved it now and here's the solution.if u have time and patience then u can go through it ........\n\ntan a{tan (a+60)} + tan a {tan (a -60)} + tan (a+60)tan(a-60)\n\n= tan a(tan a + tan60) / (1-tan a tan60) + tan a (tan a - tan 60) / (1+tan a tan60) + {(tan a + tan 60) / (1- tan a tan 60) } {(tan a -tan 60)/1 + tan a tan 60)}\n\nnote ^ is to the power of,in case u didn't know.and one more thing,make sure u know all the formulae used in this problem.this will help u understand my steps better! and sqrt = sqaure root of .\n\n=( tan^2 a+ tan a sqrt3) / (1 - sqrt3tan a) + (tan ^2a - sqrt 3 tan a )/(1 + sqrt 3 tan a ) + (tan ^2a - 3)/(1 - 3 tan^2 a )--- (this particular thing (tan ^2a - 3)/(1 - 3 tan^2 a ) i got it using the formula (a+b)(a-b)=a^2-b^2 after mulipying {(tan a + tan 60) / (1- tan a tan 60) } {(tan a -tan 60)/1 + tan a tan 60)}\n\nand now that becomes\n= {(1 + sqrt3 tan a )(tan ^2 a+ sqrt 3tana) + (1 - sqrt 3 tan a)(tan^2 a - sqrt3 tan a) + tan^2 a -3} / {1 - 3tan^2a}\n=( tan^2a + sqrt3tan^3a + sqrt 3 tana + 3tan^2a + tan^2a - sqrt 3tan ^3 a - sqrt3tan a + 3tan a  + tan^2 a -3) / ((1 - 3 tan^2 a )\n\nnow u can cancel out + sqrt3tan^3a  and - sqrt3tan^3a ,+sqrt 3 tan a and -sqrt 3 tana .\n\nwhich becomes\n= (9tan^2a - 3) / (1 - 3tan^2a)\n= -3(-3tan^2a + 1) / (1 - 3tan^2 a) \n= -3\n\n\nall the best for your studies dear friend.