No solution but the equation is true
Answer:
The possible rational roots are: +1, -1 ,+3, -3, +9, -9
Step-by-step explanation:
The Rational Root Theorem tells us that the possible rational roots of the polynomial are given by all possible quotients formed by factors of the constant term of the polynomial (usually listed as last when written in standard form), divided by possible factors of the polynomial's leading coefficient. And also that we need to consider both the positive and negative forms of such quotients.
So we start noticing that since the leading term of this polynomial is
, the leading coefficient is "1", and therefore the list of factors for this is: +1, -1
On the other hand, the constant term of the polynomial is "9", and therefore its factors to consider are: +1, -1 ,+3, -3, +9, -9
Then the quotient of possible factors of the constant term, divided by possible factor of the leading coefficient gives us:
+1, -1 ,+3, -3, +9, -9
And therefore, this is the list of possible roots of the polynomial.
According to the identity if a+b+c=0
then a3+b3+c3=3abc
a3+b3+c3/abc=3
a2*a/bc*a+b2*b/ca*b+c2*c/ab*c=3
cancel a,b,c in all the fraction then you get
<span>a²/bc+b²/ca+c²/ab=3.
</span>hence proved
The answer to this question is C.2 I think
First, you'd find the mean of all the numbers by doing 10+9+16+5+10 = 50/5 = 10. Second, you'd subtract all values from the mean by doing
16 - 10 = 6
10 - 10 = 0
10 - 10 = 0
10 - 9 = 1
<span>10 - 5 = 5
Lastly, finding the mean of all 5 numbers would be, 2.4! Hope this helps!! :D</span>