MEDALS!! PLEASE HELP ME

Calculate the average rate of change for the graphed sequence from n = 1 to n = 3.

Question

MEDALS!! PLEASE HELP ME


Calculate the average rate of change for the graphed sequence from n = 1 to n = 3.

anonymous · Answer

attachment

lexi724 · Answer

http://calculator.tutorvista.com/math/506/rate-of-change-calculator.html

anonymous · Answer

Site your source...

lexi724 · Answer

put in 2 values (x,y) values in that calculator and youll find the rate of change it helped me so much and who @rileychampion365

anonymous · Answer

You: "Free Rate of Change Calculator Online | Calculator@TutorVista.com." Free Rate of Change Calculator Online | Calculator@TutorVista.com. N.p., n.d. Web. 01 Apr. 2015.

lexi724 · Answer

We found URL: http://calculator.tutorvista.com/math/506/rate-of-change-calculator.html Article Title: Free Rate of Change Calculator Online | Calculator@TutorVista.com Website Title: Free Rate of Change Calculator Online | Calculator@TutorVista.com We could not find Publisher/sponsor Publication date

anonymous · Answer

sorry i just dont understand any of this

lexi724 · Answer

there ya go

anonymous · Answer

i dont get it!! idk how im supposed to get the answer if idk how to even solve it

anonymous · Answer

what do i put in the blanks in that link that you gave me?

lexi724 · Answer

the imputs (x,y)

anonymous · Answer

so n = 1 i put for x1? and n = 3 i put for y2?

anonymous · Answer

hello...?

lexi724 · Answer

GIVE me a moment

anonymous · Answer

alright

lexi724 · Answer

there thats an example

anonymous · Answer

my answer choices are 
A) 1
B) 2
C) 3
D) 6

anonymous · Answer

$\Huge\frac{f(a)-f(b)}{a-b}$

This is the average rate of changed equation. basically the same as y2 -y1 over x2 minus x1 (all exponents, of course)

anonymous · Answer

if you plug in everything correctly, then you would get 

$\Huge\frac{16-2}{4-1}$

anonymous · Answer

just simplify this expression. That is how you use the formula @flvsucks

anonymous · Answer

Hope this helped!