abc = 158
Do you mean F = abc + abc + ac + bc + abc' ? *x+x = x F = abc + ac + bc + abc' *Rearranging F = abc + abc' + ab + bc *Factoring out ab F = ab(c+c') + ab + bc *x+x' = 1 F = ab + ab + bc *x+x = x F = bc
Angle abc.
ab'c + abc' + abc = a(b'c + bc' + bc) = a(b'c + b(c' + c)) = a(b'c + b) = a(c + b) I'm not sure if there's a proper name for that last step, or multiple steps to get to it, but it is intuitively correct. b + b'c is equivalent to b + c. Here's a quick truth table to show it: bcb'b'cb'c+bb+c0010000111111000111100 11
The simplest answer is 1 and 2 and 3. 1+2+3=6 1x2x3=6 Other valid answers: 0, 0, 0 -1, 0, 1 -x, 0, x 1, 1.5, 5 The general formula here is: abc=a+b+c abc-a=b+c a(bc-1)=b+c a=(b+c)/(bc-1) If you choose two random numbers b and c you will ALWAYS have a number a that satisfies the conditions (unless bc=1)
The real answer is Bc . Hate these @
If we replace ( d ) with 6 in the equation ( abc + abc = cdd ), we can rewrite it as ( 2abc = c66 ). This implies that ( c66 ) represents the number formed by ( c ) followed by 66. To solve for ( abc ) and ( c ), you would need additional information about the values of ( a ), ( b ), and ( c ).
AC is congruent to DF.
31 C for 3 ABC A for 1 A
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.
Suppose ABC is a triangle. There is nothing in the question that requires the triangle to be right angled. Suppose AB is the side opposite to angle C and BC is a side adjacent to angle C. Then AB/BC = sin(C)/sin(A)
A is a fire caused by burning solids. B is a fire caused by burning liquids. C is a fire caused by electricity. BC fire extinguishers should not be used against type A fires, but ABC fire extinguishers can be used against all three fires.
Let A, B, C are digits. Number ABC is divisible by eleven if and only if A+C-B=0 or A+C-B=11