f(x) = 2/9 - 9/2 x=9
I think I have the answer, but I'm not sure that it's right.
it doesn't depend on x. for every x value, answer is 2/9 - 9/2 what is your answer?
I forgot to put the x after the 2/9
So f(x) = 2/9x - 9/2.
so you should put 9 instead of x. 2/9*9-9/2=2-9/2 can you continue?
What I did was this: f(x) = 2/9x - 9/2 2/9 times 9/1 and got 18/9. I subtracted 9/2 from 18/9 and got 9/7. Is that right? I'm supposed to make it an ordered pair. I put (9, 9/7). Is that right?
nope. 2/9*9/1=18/9=2*9/1*9 9's are canceled. and it's 2/1=2 if you want to subtract, you must make denominators equal. 2/1=2*2/1*2=4/2 4/2-9/2=4-9/2=-5/2 do you understand?
Yes. Thank you. So the ordered pair would be (9,2)?
nope. (9, -5/2) which i find out (:
How did you get the -5?
4/2-9/2 means that you should substract the numerator values and you should write the same denominator. 4-9/2=-5/2 ok?
\[f(x) = \frac{2}{9}x - \frac{9}{2}\] \[f(9) = \frac{2}{9}(9) - \frac{9}{2}=2 - \frac{9}{2}=\frac{4}{2} - \frac{9}{2}\rightarrow -\frac{5}{2}\]
Can you maybe explain a little more. I'm still not understanding.
You have your initial equation \[f(x) = \frac{2}{9}x - \frac{9}{2}\] You are given x = 9 You plug that in for x in your equation. \[f(9) = \frac{2}{9}(9) - \frac{9}{2}\] In the first part, the two nines cancel and you are left with '2' So now you have \[= 2 - \frac{9}{2}\] To be able to do the subtraction you need to get a common denominator of 2 \[2 = \frac{4}{2}\] So we have, \[=\frac{4}{2} - \frac{9}{2}\] To subtract you just take \[=\frac{4-9}{2}=\frac{-5}{2}\]
Thank you. I understand now.
No problem!
Well, there is one thing that I still don't understand. How did you get 4?
When you are adding or subtracting fractions, you need a common denominator. We have \[2 - \frac{9}{2}\] We need to have something over 2 on the left side. So you do this. \[\frac{2}{1}\times(\frac{2}{2})=\frac{2\times 2}{2}=\frac{4}{2}\]
Okay. Thank you. I didn't understand where the 4 came from. But now I do.
No problem! Do you have any more questions or concerns?
No thank you. I am finished with my homework. I just wasn't sure that the answer was.
Well, if you ever have any problems that you need help with, feel free to tag me in a question. (@AshleyNicoleBanks )
( @AshleyNicoleBanks )
There we go, now it decides to work.
I'll be sure to. Thanks.
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