Question

Here are my steps, could you tell me where I'm wrong, I know it is in my method of cross cancelling I just cant think of correct way:

f(x)=−(3 5 )x+4 if f(3)
−(3 5 )(3)+4
−(3(3) 5 )+4
−(9 5 )+4
−(9 5 )(5)+4(5)
−9+20
11

Answer 1

f(x)=−(3/ 5 )x+4 if f(3)
−(3/ 5 )(3)+4
−(3(3)/ 5 )+4
−(9 /5 )+4
−(9 [5]? )(5)+4(5)/[5] <- right in here
−(9+20)/5
11/5

Answer 2

and keeping the "-" outside is bad karma

Answer 3

\[f(x)=−\frac{3}{5}x+4 \ \ \ \ at\ f(3)\]
\[=−\frac{3}{5}(3)+4 \]
\[=\frac{−3(3)}{5}+4 \]
\[=\frac{−9}{5}+4 \]
\[=\frac{−9}{5}+4\frac{5}{5} \]
\[=\frac{−9}{5}+\frac{20}{5} \]
\[=\frac{−9+20}{5} \]
\[=\frac{11}{5} \]

Answer 4

Amazing, Thank you very much hopefully I can work through it with her now. Off to go see, wish me luck, lol

Answer 5

good luck