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Mathematics 13 Online
OpenStudy (anonymous):

Someone please help with y=2^x?

OpenStudy (anonymous):

Someone really really sweet helped me earlier today, but I dont think they are online.

Mehek (mehek14):

the table is linear because the equation for it is \(y=2x\)

Mehek (mehek14):

if you have the same ordered pairs as from the table

Mehek (mehek14):

so then yes the table matches the graph

Mehek (mehek14):

and to find the 10th, just look at the y when x = 10

OpenStudy (anonymous):

so wait this would be linear or exponential? I'm all confused again :/

OpenStudy (anonymous):

okay wait the table when graphed is linear and then the equation I get using exponential regression, has to be - exponential?

OpenStudy (anonymous):

Okay I found a website that does exponential regression using a table. & I think I have to do y=ab^x my numbers are a= 2.550657097 and b=1.259112119 are these right?

OpenStudy (anonymous):

you don't need to do exponential regression for this because it's linear. The line of best fit is exactly \[y=2x\]

OpenStudy (anonymous):

I think I just thought it was expo because that's what the whole lessons about but yeah I mean it looks linear, but I still have to use expo regression (or constant ratio) to get an equation for this table and my rate of changes.

OpenStudy (anonymous):

oh wait did you fill out that table or did they give it to you?

OpenStudy (anonymous):

no I had to fill it in.

OpenStudy (anonymous):

they gave me the first two and i did the rest

OpenStudy (anonymous):

ok so each round DOUBLES # of rounds | ppl receiving forward 1 2 2 4 3 8 4 16

OpenStudy (anonymous):

That would make it exponential

OpenStudy (anonymous):

oh dang really? I knew my table was wrong, but I didn't know exactly how.

OpenStudy (anonymous):

okay well at least that is cleared up.

OpenStudy (anonymous):

i'm going to redo this graph in google to see how it looks

OpenStudy (anonymous):

ok. I think it should go smoothly from there

OpenStudy (anonymous):

okay yeah it looks exponential now

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

those rates of changes aren't going to be correct because the points changed

OpenStudy (anonymous):

oh yeah sorry sorry I realized that as I posted them.

OpenStudy (anonymous):

We did exponential earlier. Once you have the graph you can add the trendline

OpenStudy (anonymous):

How did we get exponential? I remember just the equation being on the side there.

OpenStudy (anonymous):

yeah you have to add the equation from the advanced options where you go to edit the chart

OpenStudy (anonymous):

something like advanced edit I think

OpenStudy (anonymous):

yeah that's it

OpenStudy (anonymous):

okay so I have y=1e^(0.693x) how do I graph this over my original graph?

OpenStudy (anonymous):

I think it graphs automatically

OpenStudy (anonymous):

or I guess the line is just the same because I'm not seeing a new line just the one I had

OpenStudy (anonymous):

yeah they'll be very close together for this one because you made up the points to match the function. y = e^(0.693x) is basically the same as y = 2^x

OpenStudy (anonymous):

okay yeah when I did advanced edit and all that stuff, the dot graph I had just changed to the plots being connected with a line, so I guess that they overlay eachother?

OpenStudy (anonymous):

yeah that sounds right

OpenStudy (anonymous):

how do I figure that out? Something with the 10th round?

OpenStudy (anonymous):

@peachpi

OpenStudy (anonymous):

my connection is so bad rn I don't know if I'm seeing replies

OpenStudy (anonymous):

your equation is y = 2^x 50000 = 2^x Solve for x

OpenStudy (anonymous):

15?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

awesome! Again I appreciate your help so much. Can't thank you enough <3

OpenStudy (anonymous):

you're welcome

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