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ac + cb = ab = 9 2x - 1 + 3x = 9 5x -1 = 9 So 5x = 10 Thereby x =2. Also ac = 3 and cb = 6
the midpoint of
The problem is meaningless without a diagram but I am guessing that ABC make a triangle and D is on the extension of AB beyond B. In that case we use the exterior angle theorem to get CBD = C + A, so 125 = 90 + A and A = 35.
Cb, Db, Eb, Fb, Gb, Ab, Bb, Cb
Let the sides be a,b,c,(opposite to BC,AC,AB): Let the angle be enclosed at vertex A.Let R be the length of the angle bisector.The formula to find R is:cb= R*R + ([{(c-b)(c-b)-2bc.(cos A-1)}^1/2]/c+b)^2.cb
C is the midpoint of Ab . then AC = BC. So AC= CB.
ac + cb = ab = 9 2x - 1 + 3x = 9 5x -1 = 9 So 5x = 10 Thereby x =2. Also ac = 3 and cb = 6
the midpoint of AB.
C is not on the line AB.
Use Pythagoras' theorem to find the length of the 3rd side
C is not on the line AB.
The real answer is Bc . Hate these @
between A and B
the midpoint of
the midpoint (apex) Between A and B (Apex)
The general form for a double-displacement reaction is AB + CD -> AD + CB, where two compounds swap anions or cations to form two new compounds.
Ad < dbad < cbad + dc = cbDC + CB = DBAD < CB