Melody

Here is the pic I used

I started with vanilla and let the number be x and then I worked up from there.

We are told that 3x=x+120

so solve that and you can work it out from there.

One of them does work.

I didn't use the graph.

The graph was just checking that the algebra was correct.

Slope is just rise over run

Rise is just the difference in y values and run is the difference in x values.  Easy peazy.

(just be careful about the negatives.)

You do not know what the slop is but you do know that the slope between any two points will be the same as between any other 2 points.

\(\frac{3--5}{-a+2-3}=\frac{2--5}{2a+3-3}\\ \frac{8}{-a-1}=\frac{7}{2a}\\~\\ 16a=-7a-7\\ 23a=-7\\ a=\frac{-7}{23} \)

check

\((3,-5),\quad (2\frac{7}{23},3),\quad (2\frac{9}{23},2)\)

LaTex:

\frac{3--5}{-a+2-3}=\frac{2--5}{2a+3-3}\\
\frac{8}{-a-1}=\frac{7}{2a}\\~\\
16a=-7a-7\\
23a=-7\\
a=\frac{-7}{23}

Well I am glad you are entertaining yourself Ginger :)

I have not done this question but this is how I would start.

Let the cost of a pin be p cents

Let the cost of a keychain by k cents

Let the entire amount of her money be  M cents

Now start setting up true equations.

See if you can do something with that.

You are welcome :)

Let's see

\(f(x)=\frac{x^5-1}3, Find f^{-1}(-31/96) \)

\(f(x)=\frac{x^5-1}{3}\\ y=\frac{x^5-1}{3}\\ 3y=x^5-1\\ 3y+1=x^5\\ x=\sqrt[5]{3y+1}\\~\\ f^{-1}(x)=\sqrt[5]{3x+1} \)

\(f^{-1}(-31/96)\\ =\sqrt[5]{3*\frac{-31}{96}+1}\\ =\sqrt[5]{\frac{-31}{32}+1}\\ =\sqrt[5]{\frac{1}{32}}\\ =0.5\)

LaTex:

f(x)=\frac{x^5-1}{3}\\
y=\frac{x^5-1}{3}\\
3y=x^5-1\\
3y+1=x^5\\
x=\sqrt[5]{3y+1}\\~\\
f^{-1}(x)=\sqrt[5]{3x+1}

Which ones have you already discounted? Why?

As shown??  where is the pic?