the simplified expression is:
6s + 9
Answer:
Follows are the solution to this question:
Step-by-step explanation:
is invertible lines transformation
![S[T(x)]=x \ and \ V[T(x)]=x \\\\t'x \ \varepsilon\ 1 R^n\\\\](https://tex.z-dn.net/?f=S%5BT%28x%29%5D%3Dx%20%5C%20and%20%5C%20V%5BT%28x%29%5D%3Dx%20%5C%5C%5C%5Ct%27x%20%5C%20%20%5Cvarepsilon%5C%201%20R%5En%5C%5C%5C%5C)
T is invertiable linear transformation means that is
and 
Let
so,
![s[T(u)]=v[T(u)]\\\\s(v)=v(v) \ \ \forall \ \ v \ \ \varepsilon \ \ 1 R^n](https://tex.z-dn.net/?f=s%5BT%28u%29%5D%3Dv%5BT%28u%29%5D%5C%5C%5C%5Cs%28v%29%3Dv%28v%29%20%5C%20%5C%20%20%5Cforall%20%5C%20%5C%20v%20%5C%20%5C%20%5Cvarepsilon%20%5C%20%5C%201%20R%5En)
M < 7 / 12
(1) < 7 / 12
(1) < 0.583333333 (false statement)
(-1) < 7 / 12
(-1) < 0.583333333 (true statement)
(-9) < 7 / 12
(-9) < 0.583333333 (true statement)
(-5) < 7 / 12
(-5) < 0.583333333 (true statement)
B C D maybe?
Answer: x = -3/4 can not be a rational zero of the polynomial.
Step-by-step explanation:
We have the polynomial:
6x^5 + ax^3 -bx -12 = 0.
The theorem says that:
If P(x) is a polynomial with integer coefficients, and p/q is a zero of P(x) then p is a factor of the constant term (in this case the constant term is -12) and q is a factor of the leading coefficient (in this case the leading coefficient is 6.).
The factors of -12 (different than itself) are (independent of the sign).
1, 2, 3, 4 and 6.
So p can be: 1, -1, 2, -2, 3, -3, 4, -4, 6, -6.
The factors of 6 are:
1, 2 and 3, so q can be 1, -1, 2, -2, 3, -3.
Then the option that can not be a zero of the polynomial is
x = -3/4
because the number in the denominator must be a factor of the leading coefficient, and 4 is not a factor of six.
