Question

I need help studying for the Algebra 1 EOC please.

Answer 1

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Answer 2

Things I would go over is, inequalities, equations

Answer 3

I'm taking the practice test now, but I took Algebra 1 three years ago and I don't remember any of this stuff

Answer 4

Well, whatever you fail on the practice test study. That is what its for.

Answer 5

I'm not sure how to study for it, I tried to look up one problem but I can't find anything close to it. Could you help me with the problem I'm stuck on?

Answer 6

Maybe.

Answer 7

Yup

Answer 8

Thanks. I'll post the question. I just need guided through it, or instructions on what to do for it.

Answer 9

The table below shows the values for the function y=f(x).
x: -4, -1, 0, 3, 6
y: 7, -2, 3, -4, 5

Complete the table for the function y=f(1/5x).

Then it gives me a blank table.

Answer 10

Sorry it it's messy, I wasn't able to copy it from the practice test so I just had to type everything

Answer 11

@Arcadiouse

Answer 12

Are you ok with function notation?
Like if I write this,\[\large\rm f\left(\color{orangered}{-4}\right)=7\]does it make sense to you?
I got that information from the first chart.
`-4 in the input` and `7 is the output`.

Answer 13

I wouldn't say I'm an expert on function notation, but that doesn't look too unfamiliar to me

Answer 14

Notice with this new input,\[\large\rm f\left(\color{orangered}{\frac15x}\right)=?\]If we chose to put x=-4 like before,\[\large\rm f\left(\color{orangered}{-\frac{4}{5}}\right)=?\]It gives us a problem.
We don't know enough about the function to determine what the output would be when the input is -4/5.

Answer 15

But what we do know...
is that the output is 7 when the input is -4.

So what if we try something sneaky.
What if we try to change this entire input \(\large\rm \color{orangered}{\frac{1}{5}x}\) into \(\large\rm \color{orangered}{-4}\)?

Then we would be able to determine that we have 7 again, right?\[\large\rm f\left(\color{orangered}{\frac15x}\right)=f(\color{orangered}{-4})=7\]

Answer 16

Maybe I can put this in simpler words.. umm
We're going to take the old x values,
set them equal to this new input,
and solve for the new x values.\[\large\rm \frac15x=-4\]Multiplying by 5 gives us our first `new x value`, \(\large\rm x=-20\)
But our y values will all be the same.