Find all real value of x such that f (x) = 0. $f ( x ) = \frac { - 9 x + 8 } { 5 }$
A)x = $\frac { 8 } { 9 }$
B)x = $\frac { 8 } { 45 }$
C)x = ± $\frac { 8 } { 9 }$
D)x = $- \frac { 8 } { 9 }$
E)x = ± $\frac { 8 } { 45 }$

Question

Quizplus · Accepted Answer

To find the value of $x$ for which $f(x) = 0$, set the function equal to zero and solve for $x$: $0 = \frac{-9x + 8}{5}$. Multiplying both sides by 5 to eliminate the denominator gives $0 = -9x + 8$. Adding $9x$ to both sides and then dividing by 9 gives $x = \frac{8}{9}$.

Find All Real Value of X Such That F (X) f(x)=−9x+85f ( x ) = \frac { - 9 x + 8 } { 5 }f(x)=5−9x+8​

Find All Real Value of X Such That F (X) $f ( x ) = \frac { - 9 x + 8 } { 5 }$