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Does a by b plus b by c plus c by a equals 3 imply that abc is the cube of an integer when a b c are integers?
No. Here is a proof by counterexample that it does not.Given ab + bc + ca = 3:Assume toward a contradiction that abc is a cube. Then a = b = c.Without loss of generality, let a = 2, b = 2, and c = 2.Then ab = 4, bc = 4, and ca = 4.ab + bc + ca = 4 + 4 + 4 = 12.Therefore, 12 = 3, which is false, and so the original statement is false.
Factor 3a plus ab plus 3c plus bc?
a(3+b)+c(3+b) * * * * * This is easy to finish: . . . = (a + c)(3 + b).
Ab equals 2x plus 1 metersbc equals x plus 2 meters angle abc equals 30 the area of the triangle abc is 3 m2 calculate the value of x give your answer correct to 3 significant figures?
Area = 0.5*AB*BC*sin(ABC) = 0.5*(2x+1)*(x+2)*0.5 = 3 So, (2x+1)*(x+2) = 12 2x2 + 5x + 2 = 12 2x2 + 5x - 10 = 0 x = 1.31 (to 3 sf)
In triangle ABC side AB is 9 cm shorter than side AC while bc is 3cm longer than side AC if the perimeter is 48 cm find the lenghts of the three sides?
AB + AC + BC = 48 AB + (AB +9) + (AB + 9 + 3) = 48 Solve and AB = 9 So AB = 9, AC = 18 and BC = 21
Why does b times ab equal ab squared and not equal ab plus b squared?
b*ab = ab2 Suppose b*ab = ab + b2. Assume a and b are non-zero integers. Then ab2 = ab + b2 b = 1 + b/a would have to be true for all b. Counter-example: b = 2; a = 3 b(ab) = 2(3)(2) = 12 = ab2 = (4)(3) ab + b2 = (2)(3) + (2) = 10 but 10 does not = 12. Contradiction. So it cannot be the case that b = 1 + b/a is true for all b and, therefore, b*ab does not = ab + b2
