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.
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)
31 C for 3 ABC A for 1 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.
In this case you want to group the terms so they have at least two terms in common. First step group and rewrite it: abc + a'bc + a'b'c' + a'b'c + ab'c' + abc' = Use the rule Identities x(y+z)=xy+xz: bc(a+a') + a'b'(c'+c) + ac'(b'+b) = Use the rule Identities x+x'=1: bc (1) + a'b'(1) + ac'(1) = Use the rule Identities x(1) = x: bc+a'b'+ac'