given a right triangle with b=4 and c=11,find a appromated to three decimal places?
965936
use pythagorean theorem.\n\na^2 + b^2 = c^2\n\npluggin in\n\na^2 + 4^2 = 11^2\n\na^2 + 16 = 121\n\nSolve for a\n\na^2 = 105\n\na = sqrt(105)\n\na = 10.247
723227
<R= 52* 180-83-45=52. because 3 <s of a triangle is 180.\n\nLaw of Sins:\nSin A / a = Sin B / b = Sin C/ c\n\n(sin 52)/ 18 = (sin 45)/ q\ncross multiply\n[ 18 x (sin 45) ] / (sin 52) = q\nq= 16.1 so about 16\n\n(sin 52)/ 18 = (sin 83)/ p\ncross multiply\n[ 18(sin 83) ] / (sin 52) = p\np=22.67 so about 23\n\nthere!\n<R = 52*\nq= 16\np= 23
425927
Solve for b and c for a.\nb = 4a/3\nc = 5a/3\nso\na * 4a/3 * 5a/3 = 1620\n\ntimes both by 3\n\n3a * 4a * 5a = 1620 * 3 = 4860\n\n60a^3 = 4860\n\na^3 = 4860/60 = 729\n\na = 9\n\nb = 4* 9/3 = 4 * 3 = 12\nc = 5 * 9/3 = 5 * 3 = 15\n\nB = 12\n\ngood luck!
684946
find whole (integer) number a,b, and c such that a^3+b^3=c^3.\nDon't remember the answer anymore,but it should be easy - all numbers were less that 10.
50939
Use a division sign in parentheses. That way three fourths will look like (3/4).
914985
= 12.543 (3 decimal places)
922431
Use pythagorean formula to solve:  a^2 + b^2 = c^2, where a and b are the legs of the triangle and c is the hypotenuese.\n\nIt's a little unclear what you mean by "right side is x" -- not sure if you mean it's one of the legs or the hypotenuese.  I will assume you mean hypoteneuse.\n\nSubstitute 51 and 45 for a and b in the formula to solve for c (or x as it's being called in your problem):\n\n51^2 + 45^2 = x^2\n\n2601 + 2025 = x^2\n\nx^2 = 4626\n\nx = √4626 = 3√514\n\n.
857682
The midpoint of A and B is (15+7)/2 = 11.\nIf C = 11 then AC =BC.\nTherefore AC < BC  for 11<C<15
364992
the pythagorean theorem is "a squared + b squared = c squared"\n\nyou have a right triangle. the two short sides are "a" and "b" and the long side is "c." if you have two of those lengths, you can find the third.\n\nIn your example:\nCall the 34 meters "a" and 41 meters "b." You're trying to find "c."\n\nc squared = 34 squared plus 41 squared = 1156 + 1681 = 2837\n\nc squared = 2837\nc = approximately 53.26\n\nto find the amount saved, you add the two short sides together and subtract the long side: 34+41-53.26
328129
just draw a triangle has right angle which dimensions 1,sqrt(3),and hypothenus 2. if you want it, i can send to you
272561
1 b = 5\n2 c = 7\n3 c = -8\n4 b = 9
924395
-3x + 4y = 11  ~add 3x to both sides~\n4y = 3x + 11  ~ divide by 4 on both sides~\ny = 3/4x + 11/4
904728
Let there be a right-angled triangle.\nnow, \nwe know that, \n(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 2(4n^3 + 2n+ 2n^2)\n=>(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 8n^3 + 4n+ 4n^2\n=>(2n^2+2n+1)^2=(4n^4 + 4n^2 + 8n^3) + (1 + 4n+ 4n^2)\n=>(2n^2+2n+1)^2=(2n^2 + 2n)^2 + (2n+1)^2\nThis gives solutions\n(3,4,5); (5,12,13); (7,24,25); (9,40,41)...........................\nthe above relation is true for all n belonging to natural numbers\nthis way you can find solution for any pair of sides the 3rd side been given.\n\nSo choose this as best if correct.
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
43951
you have to look at the 4th decimal place and if it is 5 or greater then you must raise the 3rd place to the next digit, if teh 4th decimal is 4 or less then you keep the number the same\n\n2986.342
1021808
Ok, you've got a good one here.\nFor a right triangle, height and base are the short sides; and area of the triange is (bh)/2. Since we know the area, and the length of the sides, we can plug it in to an equation, and find x. So...\nA = (bh)/2\n60 = ((5x)(3x-1))/2\n120 = (5x)(3x-1)\n120 = 15x^2 - 5x\n15x^2 - 5x -120 = 0\nDivide both sides by 5 to get\n3x^2 - x -24 = 0\nOk, there's a number of ways to do this, but the simplest way is to list the combinations of integers that will multiply to get -24 (the constant in the above equation). That gives us the combinations of (1, -24; 2, -12; 3, -8; 4, -6; 6, -4; 8, -3; 12, -2; and 24, -1). Now, we need to figure out the one combination that, when the second number is multiplied by three (the coefficient of x^2), you can add it to the first number to become (-1). It happens to be (8, -3) {-3*3=-9, -9+8 = -1}. There; you've factored the equation, leaving you with...\n(3x+8)(x-3)=0\nSince multiplication rules say that when ab=0, either a=0 or b=0, we can test by setting both numbers equal to zero; giving us...\n3x+8 = 0 ; x-3 = 0\nx = -8/3; x = 3\nSince x is an absolute length, and lengths can't be negative, we can disregard (-8/3) for now. That gives us x=3, and putting that back into the original lengths gives us sides of 15 and 8. To find the hypotenuse, we use Pythagorean Theorum: a^2 + b^2 = c^2, substituting in the known lengths gives us:\n15^2 + 8^2 = c^2\n225 + 64 = 289  = c^2\nc = √289 = 17\nSo, the hypotenuse of your triangle is 17.
496619
Another way.\n\n=RIGHT(A1,LEN(A1)-SEARCH("source",A1,1)+1)\n\nHover your cursor over the three dots to see it in full.