Question

no idea how to start

Answer 1

just for 7.5

Answer 2

like the directions said start with plotting the graph

Answer 3

I meant after the graph sorry

Answer 4

what would be the line of best fit then what would be the slope of that line of best fit ?

Answer 5

at the start would be week 0

Answer 6

hmm wait let me draw a super accurate graph brb

Answer 7

exponential growth yes because not by the same amount each week. as the number of weeks increase the amount of growth increases more than the week before

Answer 8

oh wait so I wouldn't just have one answer? I would have a different answer per week?

Answer 9

could I not just use the formula y=Ae^(kt) (once I figure out A from the line of best fit)

Answer 10

week 4 - week 3 more than week 3 - week 2 so each week the growth is more

Answer 11

Thank you! My friend said the answer was 2.5, and I thought there must be a different amount each time so... I was confused

Answer 12

\[y = ab^t\]
use endpoints to find a and b
\[ab = 14\]
\[ab^6 = 115\]
\[\rightarrow 14b^5 = 115\]
\[b =( \frac{115}{14})^{1/5} = 1.52\]
This is the multiplication rate
\[\rightarrow a = \frac{14}{b} = \frac{14}{1.52} = 9.19\]
This is the initial amount of weeds

Answer 13

@3mar

Answer 14

Thank you both, tremendous help :) <3

Answer 15

yes @triciaal

Answer 16

@3mar nvm the question is now closed but if I tag you it means you may help on that question at my special request thanks

Answer 17

Thank you for your confidence @triciaal

For all of you, some said \(~~y=a*b^{kx}~~\) , other said \(~~y=h*e^{qx}~~\), but I think that the best fit and most accurate curve of that growth is "quartic equation"

Here is my approach!
https://www.desmos.com/calculator/4b1tc2kg2k

Take that and see!

Answer 18

That is an interesting approach 3mar, thank you for sharing

Answer 19

No problem.
You are welcome!

I hope I helped you.

Any Help... Any Time...

Answer 20

Are you satisfied, @amorfide?