Given AB=8, AM=7, AC=14, AM is a median, what equation should be use to get the length of BC?
919171
To answer this question, you need to know the shape that these points make.  Are they on a line? or are they some two-dimensional shape?
1057756
lets say length= L\nwidth=W\nperimeter=P=22=2L+2W\nL=W+7\n\n22=2(W+7)+2W\n22=4W+14\n8=4W\n2=W\n9=L
337108
6b + 24a + 45\n= 3*2*b + 3*8*a + 3*15\n= 3(2b + 8a + 15)
827919
a = 0.5bc^2\n\n2a = bc^2\n\n2a/c^2 = b
949749
m=cab/a-b\n\nMultiply both sides with (a-b) you would get\nma-mb=cab\ni.e. cab+mb=ma\n      b(ca+m)=ma\ngiving you b=ma/ca+m
1075528
(7.3-.5-2.5)/-.9 = b\n\nb = -4.8
937830
T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
869858
P(B|Ac) = P(B and Ac)/P(Ac)\n\nP(B and Ac) = P(B) - P(A and B)  (It's easiest to see that if you look at a Venn diagram.  It's the stuff that both in B but not in A.)\n= 0.25-0.1\n= 0.15\n\nP(Ac) = 1-P(A) = 1-0.3 = 0.7\n\nSo P(B|Ac) = P(B and Ac)/P(Ac) = 0.15/0.7 = 15/70 = 3/14\n\nP(Bc|A) = 1 - P(B|A)\n= 1 - P(B and A)/P(A)\n= 1 - 0.1/0.3\n= 2/3\n\nAnswers would be similar for the next part.\n\nP(B|Ac) = 2/5\nP(Bc|A) = 17/20
705715
ab(a + b - 2c) + bc(b + c - 2a) + ca(c + a - 2b) >=? 0\n\na²b + ab² + b²c + bc² + ac² + a²c - 6abc >=? 0\n\nabc(a/c + b/c + b/a + c/a + c/b + a/b) - 6abc >=? 0\n\nabc(a/c + b/c + b/a + c/a + c/b + a/b - 6) >=? 0\n\nNow look at your terms inside.  You have:\na/c + c/a\nb/c + c/b\na/b + b/a\n\nThese are reciprocals.  Regardless of what numbers you pick for a, b, c (which must be 1 or higher), the sum will be no smaller than 2.  (1/1 + 1/1 = 2).\n\nAt a minimum you have 6, but it can never result in something smaller than 6, therefore the amount in the parentheses will be 0 or greater.  The product of this with a positive amount (abc) will also be 0 or greater.\n\nAlternatively,\na²b + ab² + b²c + bc² + ac² + a²c - 6abc >=? 0\na(b² + c² - 2bc) + b(a² + c² -2ac) + c(a² + b² - 2ab) >=? 0\na(b-c)² + b(a-c)² + c(a-b)² >=? 0\n\nHere you have positive integers and squares (which are always 0 or greater), so the sum will be greater or equal to zero.
853494
x = (3 + 1)/2 = 2\ny = (2 - 1)/2 = 1/2\n\ntherefore the median is\n(2, 0.5)
493498
all the above answers are correct, but here's an example:\n\ngiven: 15,18,12,21,11,17\n\nin order: 11,12,15,17,18,21\n\nnow originally, the median would be the two numbers, 15 and 17.  but the median is only one number, or in this case, the average between the two middle numbers, 15 and 17.  so:\n\n(15+17)/2 = 16\n\nso the median's 16 - yay.
472370
-11a + 7 < -12a + 18 \n a + 7 < 18            .( Added 12a to both sides)\n a < 11                .(Subtracted 7 from both sides)\n\na < 11 (FINAL answer- a is less than 11)
921810
2(4x+7)=2x-4\ndistribute\n8x+14=2x-4\n-2x from feach side\n6x+14=-4\n-14 from each side\n6x=-18\ndivide by 6\nx=-3\nand the next one\n-2(18-3y)=7y+2y\ndistribue and add light terms\n-36+6y=9y\nsubtract 6y from each side\n-36=3y\ndivide by 3\n-12=y\n:) welcome
799288
must be perpindicular to BC\nslope of BC is (0-(-1)/(6-1)=1/5\nslope of altityde is -5\ny=-5x+b & it imcludes point 3,2\n2=-5*3+b\n2=-15+b\nb=17\nthe equation is y=-5c+17
745771
first find the midpoint of AB which is C(9,-2).\nNow obviously the one-fourth distance will be the midpoint of AC which is ...well I let you do this. This is simple knowing the formula for the midpoint .\nMathguy did well too but you are supposed to apply, I think, midpoint formula.
139064
i think i know how to do it, one minute...\nhey is it 7k-14/(2k-6)(k-4) -6K+18/(2K-6)(K-4) = K+4/(2K-6)(K-4)\nor (7k-14)/(2k-6)(k-4) -(6K+18)/(2K-6)(K-4) = (K+4)/(2K-6)(K-4)\ni added parenthesses\n\n>>>ok so that's what you need to do:\nlet's say (2k-6)(k+4) = X\nso you get:\n7k -14/X -6k +18/X = k +4/X\n7k -6k -k = 4/X +14/X +18/X\n0k = 0X\n0 = 0*(2k-6)(k-4)\n=> every k will give us a legal equation!\n\ni really hope it's true, have u got the answer?
815598
a+b+ c+ab+cb+ac = abc + 1\nor a+b+c - abc = 1- ab-cb-ac\nor (a+b+c - abc)/(1-ab-bc-ac) = 1\n\nputting a = TAN A , b= tan B, c = tan C\n\nwe get LHS = tan(A+B+C) = 1\nor A+B+C = tan ^-1 = pi/4+npi\n\n chose any a and b then c can be chosen as\n\ntan^1(pi/4 - tan A - tan B)