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  1. ... variables: dt = dm/(am^(3/4) - bm) Integrate both sides: t = ∫ dm/(am...^(1/4)) ==> bt/4 = ln [(a - bm₀^(1/4))/(a - bm^(1/4))] Take exponentials of both ...

    1 Answers · Science & Mathematics · 10/08/2013

  2. bm + 7r + rm + 7b bm + rm + 7b + 7r (bm + rm) + (7b + 7r) m(b + r) + 7(b + r) (m + 7)(b + r)

    3 Answers · Science & Mathematics · 06/12/2006

  3. ...sin(alpha) = a/c and cos(alpha) =b/c. The median BM is the hypotenuse of a right triangle and ...sqrt [3(sin(alpha) )^2 +(sin(alpha))^2+ ( cos(alpha) )^2] BM = c/2 sqrt [3(sin(alpha) )^2 +1] tan(MBC...

    1 Answers · Science & Mathematics · 19/08/2013

  4. i) Median AN & BM intersect at K; so K is the centroid of the triangle ABC...

    1 Answers · Science & Mathematics · 21/06/2016

  5. ax-bm+an-bn just take the common terms ax + an - bm - bn= a(x+n) - b(m+n)

    4 Answers · Science & Mathematics · 09/05/2009

  6. I'll assume Am is between A and B, Bm is between B and C, and Cm is between C and...

    7 Answers · Science & Mathematics · 04/11/2007

  7. ...and B, and that AM=4x-10. If that's true, then AM = BM (picture a line with M exactly between points A and B...

    1 Answers · Science & Mathematics · 22/10/2009

  8. AB = AM + BM AB = 4x - 2 + 5x - 10 AB = 9x - 12 AM = BM 4x - 2 = 5x - 10 4x - 5x = -10 + 2 -x = -8 x = 8 substitute x = 8 into AB; AB = 9(8) - 12 AB = 72 - 12 AB = 60

    6 Answers · Science & Mathematics · 27/01/2009

  9. abx^2 + (an + bm)x + mn abx^2 + anx + bmx + mn factor by grouping ax(bx + n) + m(bx + n) (ax + m)(bx + n) -9x^2 + 6(y + 5)x - 5(2y + 5) 3x[-3x + 2(y + 5)] - 5(2y + 5) 3x(-3x + 2y + 10) - 5(2y + 5)

    2 Answers · Science & Mathematics · 11/09/2011

  10. AB^2+AC^2=4+4a^2+4b^2, M is (a+1,b) so BM^2+AM^2=(a-1)^2+b^2+(a+1)^2+b^2=2+2a^2+2b^2 so 2(BM^2+...

    1 Answers · Science & Mathematics · 17/02/2012

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