No. Writing it as a biconditional would mean that "If a number is odd, then it is a prime number" is also true. The original statement is true, but the converse is false.
When you multiply 6 by 600 it equals 3,600
6x600=3,600
a+b+c=0
[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]
[a^2+b^2+c^2+2ab+2ac+2bc=0]
[a^2+b^2+c^2=-(2ab+2ac+2bc)]
[a^2+b^2+c^2=-2(ab+ac+bc)] (i)
also
[a=-b-c]
[a^2=-ab-ac] (ii)
[-c=a+b]
[-bc=ab+b^2] (iii)
adding (ii) and (iii) ,we have
[a^2-bc=b^2-ac] (iv)
devide (i) by (iv)
[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
Answer:
deez nus
Step-by-step explanation:
Answer:
You should be thinking about a negative number. In fact, you should be thinking about a number that is the opposite of 5! -5! So, the additive inverse of 5 is -5
Step-by-step explanation:
