number 5 :)
@Seratul
@DanJS
What do you need help with??
#5? a,b,c?
@DanJS yes
do you know the difference in linear and exponential equation?
y=mx+b y=A*b^x
yes
linear model has a constant slope, or change in values, you see option2 is linear since it increases by 100 each time
a exponential will change by some multiple value each time, you can see from the ratios 1210/1100 = 1.1 1331/1210 = 1.1 this is exponential , the multiplier is 1.1
oh yes i do
can be used to describe the value of the investment after a fixed number of years using option 1 and option 2? Explain your answer.
i copy and paste from question
part A
well, option A wont be able to be constant but option B should be constant right
yes, option B is linear, you can use two points from the table to get the equation for the line
how?
(1 , 1100) and (2 , 1200) m=100 y - 1100 = 100*(x - 1) y - 1100 = 100x - 100 y = 100x + 1000
oooooh
okay so i think we answered part A , right >\?
did we answer everything it was asking tho?
okay so we got 1 fnction for B ,so now for A
a) (copied from what i typed above) --- linear model has a constant slope, or change in values, you see option2 is linear since it increases by 100 each time, ---a exponential will change by some multiple value each time, you can see from the ratios 1210/1100 = 1.1 1331/1210 = 1.1 this is exponential , the multiplier is 1.1
same way right?
Option A is not linear, it increases by a multiple of 1.1 each time, not a constant value
Part A is everything you just copied ?
yes i know but it says write a function at least once for each option
general exponential equation y = A*b^x initial value A, multiplier base b
i've seen that but idk what to do with it atm
wait so A=1100 and B= 1.1 (like you said)
you need to figure out the values for A and b y = A*b^x you have 3 points (x,y) given on the table, you can use them in the equation
right b=1.1, that is the common multiplier y = A*(1.1)^x
you can use one of the points (x,y) to get the value for A, using the first point (1,1100) y = A*1.1^x 1100 = A*(1.1)^1
so how do we get A?
divid 1100 by 1.1?
\[\large y=A*(1.1)^x\] yeah just put in a point value for x and y, and solve for A
so we just finished part B correct?
yeah y=1000*1.1^x
okay part C:
they want it as a function y=f(n) f(n) = 1000*1.1^n f(n) = 100n + 100
do i use both or one of them? or the one which produces more profit
they want n=20 years, put that in for both options..
The exponential will at some point in time become the better option. The exponential will give larger and larger increases in investment as the years get larger. Each year the current amount is multiplied by 1.1, and the amount keeps growing each year and so the increases will also grow each year. as opposed to the linear option, this one just increases by the same 100 each year.
So is olve for option B with the function and explain thats it better than Option A, correct @DanJS
This is what you got for the equations. Option A -- > f(n) = 1000*1.1^n Option B -- > f(n) = 100n + 100
The last part asks about the values at 20 years, n=20 Option A -- > f(20) = 1000*1.1^20 = 6727.5 Option B -- > f(20) = 100*20 + 1000 = 3000
and then i explain which one is better and why right
@DanJS
im not sure what to write for part B..
you just write the equations, and explain one is exponential and one is linear, like above
f(n) = 100n + 100 how do we have a intial value of 100? do you mean 1100?
@DanJS
wait whats the original function for option 1 and option 2 .. something doesnt make sense
@sunnnystrong can you help me explain part B?
that was mistyped once, sorry, f(n)=100n + 1000
where did you get 1000?
go back through the whole thread again
and why isnt it like 100^n
Hmmmm... well it looks like you already did the work with @DanJS but \[f(n)=100n + 1000 \] \[f(n)=1000(1.1)^n \]
i dont understand why 100n and not 100^n
Ohh ... okay well let me see if I can explain haha As you can see.. Option two is increasing at a constant rate. As @DanJS said... Therefor it isn't 100^n
oh
Option 2 is an arithmetic sequence. Each new term is 100 more than the previous term. What about Option 1? Note that the 2nd increase is greater than the 1st increase. This is typical of exponential growth functions.
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