Upon a slight rearrangement this problem gets a lot simpler to see.
x^3-x+2x^2-2=0 now factor 1st and 2nd pair of terms...
x(x^2-1)+2(x^2-1)=0
(x+2)(x^2-1)=0 now the second factor is a "difference of square" of the form:
(a^2-b^2) which always factors to (a+b)(a-b), in this case:
(x+2)(x+1)(x-1)=0
So g(x) has three real zero when x={-2, -1, 1}
Answer:
73.5
Step-by-step explanation:
circumference = 2πr , where r = radius
given radius = 11.7 so r = 11.7
C = 2πr
==> plug in r = 11.7
C = 2π(11.7)
==> multiply 2 and 11.7
C = 23.4π
==> multipl 23.4 and π
C = 73.5 ( rounded to the nearest tenth )
Step-by-step explanation:
To solve a system of equations, we can add the two equations and solve for one of the remaining variables -- let's try to eliminate the 
variable when we add the two equations together.
Right now, there's a 
term in the first equation, and a 
term in the second equation, so if we add those together, we'll be able to eliminate the variable altogether and solve for 
.
However, when we also have a 
term in the first equation and 
term in the second equation, so adding these together will also eliminate the term, leaving a 
on the left-hand side of the equation.
If we add the two numbers on the right side of the equation, we get 
, which does not equal , meaning there are no solutions to this system of equations.
Answer:
40ft
Step-by-step explanation:
Please kindly see the attached file for more explanation
