Algebra 2, If the measure of one angle is five less than two-thirds of the measure of its supplementary angle?
918511
Supplementary angles add up to 180 so here is your starting equation.\n\n(2/3x-5)+x=180\n\nadd 5 to both sides\n\n2/3x + x = 185\n\n2/3x + x is going to equal 5/3x\n\n5/3x = 185\n\nMultiply both sides by 3/5\n\nx = (185*3)/5\n\nx = 555/5\n\nx = 111\n\n2/3x-5 = (2*111)/3 - 5\n\n= 222/3 - 5\n\n= 74 - 5\n\n= 69\n\n69 + 111 = 180\n\nTada!!
261181
assuming that 2x+5=0, then I deny that solution. The Pair(3,-5) workes for 6y-8=-54 only.
808242
Yes. These are intriguing geometry problems. What is YOUR question? Be specific. Post ONE question and be specific about where you are stuck. Right now, this just looks like you want someone to DO your homework for you. But you wouldn't cheat on your homework like that, would you?\n\nHere's a hint -- supplemental angles add up to 180\n\nFor example, if one angle is 60 degrees, its supplement is 120
914527
This is easy.\nComplementary angles add up to 90.\nSupplementary angles add up to 180.\n\nSo if two angles are complementary and one is 40, the other is 50.\n\n40 + x = 90\nx = 90 - 40\nx = 50\n\nIf two angles are supplementary and one is 40, the other is 140.\n\n40 + x = 180\nx = 180 - 40\nx = 140\n\nIf an angle has a measure of 'x':\nThe complementary angle is 90 - x\nThe supplementary angle is 180 - x\n\nDoes that make sense?  If not I can try to explain it another way...
161223
law of cosines : a^2 = b^2 + c^2 -2bc * cos(A)\nwhere a, b,c are side lengths and A is the angle oppisite side a.
623818
We can find angle B by subtracting the others from 180:\n\nB = 180 - 32 - 44\n= 104\n\nThe length of the sides can be found using the law of sines:\n\n(sin A)/a = (sin B)/b = (sin C)/c\n\nSo,\n\n(sin B)/b = (sin C)/c\n(sin 104)/b = (sin 44)/20\nb = 20(sin 104)/(sin 44)\n~ 27.94\n\n(sin A)/a = (sin C)/c\n(sin 32)/a = (sin 44)/20\na = 20(sin 32)/(sin 44)\n~ 15.26
555107
MY guess is that your trying to find the Hypotenuse of a triangle.  Thus A^2+B^2=C^2 or in this case H^2+W^2 = C^2, 25 + 196 = C^2, 221 = C^2, C = 14.86.  Missing side is 14.86.
722736
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.
1151668
The x-coordinate decreased by 4 - (-2) = 6 and the y-coordinate decreased by 2 - 0 = 2\n-3 - 6 = -9\n5 - 2 = 3\nAnswer: Q'(-9,3)
21860
Let's call these angles A and B.  And let's call C the adjacent angle in the middle of these two angles.\n\nThen we will have:\n\nA + C = 180\nB + C = 180\n\nAs a result, A = B = 180 - C
716170
since 2 is a root you take (x-2)*(x-2) and -3 is one also so you multiply (x+3) now you know it is a fifth degree so you need two more so you can take (4x^2+1) and multiply\n(x-2)(x-2)(x+3)(4X^2+1) you do the algebra
102966
This is only possible if it is an improper fration such as 3/2.  Which to simplify would be 1 1/2.
1139471
Yeah where's angle 1 and 2.\nBut remember the angle opposite the longest side has the greastest measurement.
1078035
C-120... Two supplementary angles must equal 180.... 60+120=180\nB-32 feet... 8*4=32
968872
Let the shorter leg be 'x'\nLonger leg = 'x + 1'\nHypotenuse = 'x + 2'\nx^2 + (x + 1)^2 = (x + 2)^2\nx^2 + x^2 + 2x + 1 = x^2 + 4x + 4\n2x^2 + 2x + 1 = x^2 + 4x + 4\nx^2 + 2x + 1 = 4x + 4\nx^2 -2x - 3 = 0\nx^2 + x - 3x - 3 = 0\nx(x + 1) -3(x + 1) = 0\n(x - 3)(x + 1) = 0\nEither x - 3 or x + 1 = 0\nIf x + 1 = 0, x = -1 which cannot be true for the triangle\nIf x - 3 = 0, x = 3\nTherefore, the sides are:\nShorter leg = 3cm\nLonger leg = 3 + 1 = 4cm\nHypotenuse = 3 + 2 = 5cm
918358
theorm of Pythagorus states that if one of the angle of triangle is right angle{90degrees} than the hypotenuese square{the side opposite to right angle} is egual to the sum of squares of the sides forming the right angle \ni.e. (Hyp)square =one side square + second side square \n a2=b2+c2
27599
If you are talking about trisecting an angle using a compass and straight edge, then it is not and unsolved problem, it is problem for which there is no solution.\n\nThere is an area of abstract algebra called Galois theory. One of the things that you learn in Galois theory is that there exist certain angles that cannot be trisected using the kind of techniques that you would use to bisect an angle or line. Flat out can't be done. I recall, but don't quote me on this, that 60 degrees is one example.\n\nAnother thing that comes from Galois theory is that there is no formula, that will give you the roots of a 5th degree polynomial. You are probably familiar with the quadratic equation which gives you the roots of a second degree polynomial. There are formulas that you can use to get the roots of 3rd and 4th degree polynomials. But there can't be a formula that will work for a general 5th degree polynomial.\n\nI know more than one high school student who wasted a lot of time trying to devise a method for trisecting angles when he/she heard it was a great unsolved problem. Their teachers neglected to tell them that it was not only a great unsolved problem, but that it was a great unsolvable problem.