Will give medal ......Can someone check my work to see if im doing it right. .A scientist is studying the growth of a particular species of plant. He writes the following equation to show the height of the plant f(n), in cm, after n days: f(n) = 10(1.02)n When the scientist concluded his study, the height of the plant was approximately 11.04 cm. What is a reasonable domain to plot the growth function?
Part a i think the domain to plot is either (0-5) or (1-5)
@johnweldon1993 @SithsAndGiggles or anyone else that might be able to check and help me :)
Well first, think about what the function is It is the relation of the height of the plant at any given times 'n' Where the graph crosses the y-axis, is when 'n' = 0...or the beginning height of the plant
As far as the domain goes, from the equation \[\large 10(1.02)^n\] What is the minimum time we can use? well we cant use last week, the data is for now until whenever the plant decides to die So I would use your (0 to 5) choice
so is my part A and part B is correct @johnweldon1993
Part A looked good Part B, specify that it is the height of the plant at time 0 *the starting height*
then part c i would use the formula to find the average rate of change between n=1 and n=5
Right \[\large \frac{f(5) - f(1)}{5 - 1} = ?\]
y^2-y^1 over X^2-X^1
I dont know how you wrote it like that on here lol
Which is 4 over 4 which is= to 1
Lol learn the latex...use the equation button below the textbox But wait its 4/4? I know the bottom is 4 but the top doesnt sound right
oh thanks ill try it
would it be f(4)
No, the way I write f(5) and f(1) simply means what does the function equal when you plug in 5 for n? and what does the function equal when you plug in 1 for n?
so i have to figure out the formula 10(1.02) n which is 10(1.02)5 =which is 11.041
and 10(1.02)1 =10.2
right!!
\[\frac{ 11.041-10.2 }{ 4}\]
exactly!
I'm actually headed out now....but you have it now
which is \[\frac{ 0.841 }{ 4}\]
is that all i need to do :)
Thank you so much for helping me understand this problem @johnweldon1993
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