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|dw:1437111756117:dw|

No one brave enough to tackle this one? lol

Did you want to go through all the parts?

This actually doesn't look too bad. I'm assuming that your class wants you to do this by hand?

Yes.
I know it should be easy which makes me feel stupid for asking...

No, no worries. It's only easy because I've had to pull my hair out on these in the past.

Have you already gone ahead and graphed the data for part a)?

Yes. I dont see the relation though

OK, I'm going to go ahead and use technology on my side to see the graph. Hold on one moment.

Alright, thank you!

OK, here's what I see:

Your graph is much more proportional than mine,haha, and I see the curve now

Well, I cheated with technology. :) Alright, so we see the curve now.

It's actually very straight, so a line should do a good approximation.

So its linear? is that the relationship?

Yep, one of the few words in math that actually makes some sense. :)

So how do we make this into a linear equation? Take a random pair of coordinates?

Yeah I don't know. Our teacher had as watch a video about it before answering this question.

So, they've chosen the random points for us.

OK, looks good

So, m=807.692

I think you mean 80.769, I believe it was 42,000 vs. 420,000, but I might be mistaken.

Actually, wait... The points were (0,0), and (520,40,000) typo on my part. Sorry.

Oops it was 40,000. Guess we dragged the 2 from 520 by mistake

Yep, that happened. :) So, I'm getting 76.923

yeah haha. So 40000=m(520)
So m=76.923

Yay!!

So our equation is y = 76.923 x, where x is the distance and y is the velocity.

I'll go ahead and plot this over my previous graph, one second.

Here we go

So for part C, I plug in the velocity and y=(76.923)(60,000) ?

Careful... the velocity goes into the y, not the x.

Almost linear, do I needto change answer A?

Actually it would be a good idea to write your equation as: v = 76.923 d to keep that clear.

Okay, then linear it is. And galaxy Hydra is 780 (megaparsec?) away?

Let me check.

Yep, I agree.

Now to estimate the age of the universe with our slope.

Does the m stand for megaparsec?

yes

OK, I think all they want is to just plug in our m = 76..... into the equation.

Do I have to simplify it at all?

I'm getting a really big number...

Me too :(

I'm using mathway, what are you using?

Just searching directly in Google the following
9.5*10^11 / 76.923
gives me this
12350012350

I got 12348888600.026 so... 1.23x10^11 ? Does that sound reasonable?

Sure, our model isn't perfect anyway, so that's a good estimate.

So I have no lead for what part E wants me to do

Just real quick I think you want 1.23x10^10

Oh okay. Sounds good. Good to double check, ty

Ohhh okay!! Thank you so much!!!

Your welcome! Not so bad hopefully! :)

Not as bad as I thought it would be at all XD guess I'm just afraid of big numbers

m = 100 -> 9,500,000,000 = 9.5 x 10^9
m = 50 -> 19,000,000,000 = 1.9 x 10^10

Yes! B) Yay team! Have a good *enter time of the day here*

lol Sounds good, you too!