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1. ### Solve the ODE dm/dt = am^(3/4)-bm?

... 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. ### What is the answer to bm+7r+rm+7b?

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. ### Find the median BM and tan of MBC?

...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. ### Medians AN & BM in ΔABC are lengths 6 & 9 correspondingly and intersect at pt K. Angle AKB = 30, find area of ΔABC?

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

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

5. ### can u factor ax-bm+an-bn?

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. ### There is a triangle, ABC, with points Am, Bm, Cm at the centre on each side. Show..?

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. ### M is equidistance from A and . AM-4x+10 and BM=5x-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. ### . M is the midpoint of segment AB and AM = 4x – 2 and BM = 5x – 10. Find the length of AB.?

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. ### FACTORING HELP: 2 questions: 1) abx^2 + (an + bm)x + mn and 2) 6xy +30x - 10y - 9x^2 - 25?

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. ### Prove AB^2 + AC^2 = 2(BM^2 + AM^2)?

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