Question

I am trying to show my daughter how to solve, I keep getting messed up on the fraction part. Could someone show me how to to complete?
f(x)=-(3/5)x+4 if f(3)

Answer 1

f(x) = \(\cfrac{-3x}{5}+4\)

f(3) = \(\cfrac{-3(3)}{5}+4\)

Answer 2

to add them you have to either get common denominators or do some algebraing

Answer 3

like say:

\[N=\frac{-9}{5}+4\]
\[(5)N=\frac{-9(5)}{5}+(5)4\]
\[5N=-9+20\]
\[5N=11\]
\[\frac{5N}{(5)}=\frac{11}{(5)}\]
\[N=\frac{11}{5}\]

Answer 4

I just found the Equation button to make problem appear exactly as it is on her paper, could you please review and make sure working it out as you showed is still correct when retyped correctly:


\[f(x)=-\left(\begin{matrix}3 \\ 5\end{matrix}\right)x+4\] When f(3)

Answer 5

its the same ...

\[-\frac{3}{5}x=\frac{-3x}{5}=-3\frac{x}{5}\]
and any other arrangement that is suitable :)

Answer 6

Thank you. She is transferring from a private christian school to a public school and is covering math that other students have already learned and we are trying to learn at home to keep her up with the class. Thanks Again

Answer 7

i would advise, for problems like this, to review the 5 basic properties of algebra and the "order of operations"

Answer 8

is the answer 11? Want to make sure I got it right before I show her how I solved.

Answer 9

is the answer 11? Want to make sure I got it right before I show her how I solved.

Answer 10

the answer is 11/5; now this can take on a few forms of its own, but as is is a good answer

Answer 11

11/5

2' 1/5

2.2

all are the same value

Answer 12

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:

\[f(x)=-\left(\begin{matrix}3 \\ 5\end{matrix}\right)x+4\] if f(3)
\[-\left(\begin{matrix}3 \\ 5\end{matrix}\right)(3)+4\]
\[-\left(\begin{matrix}3(3) \\ 5\end{matrix}\right)+4\]
\[-\left(\begin{matrix}9 \\ 5\end{matrix}\right)+4\]
\[-\left(\begin{matrix}9 \\ 5\end{matrix}\right)(5)+4(5)\]
\[-9+20\]
\[11\]