The average rate of change (AROC) of a function f(x) on an interval [a, b] is equal to the slope of the secant line to the graph of f(x) that passes through (a, f(a)) and (b, f(b)), a.k.a. the difference quotient given by
![f_{\mathrm{AROC}[a,b]} = \dfrac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=f_%7B%5Cmathrm%7BAROC%7D%5Ba%2Cb%5D%7D%20%3D%20%5Cdfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
So for f(x) = x² on [1, 5], the AROC of f is
![f_{\mathrm{AROC}[1,5]} = \dfrac{5^2-1^2}{5-1} = \dfrac{24}4 = \boxed{6}](https://tex.z-dn.net/?f=f_%7B%5Cmathrm%7BAROC%7D%5B1%2C5%5D%7D%20%3D%20%5Cdfrac%7B5%5E2-1%5E2%7D%7B5-1%7D%20%3D%20%5Cdfrac%7B24%7D4%20%3D%20%5Cboxed%7B6%7D)
(p)=(333+114)•3
because 114 pages are already read and if he needs 333 pages to reach 1/3 of the book you add 114 and 333 so you can find out how much pages is 1/3 of the book. then you multiply that number by 3 because there are 3 sections of 333+114.
Answer: CONFUSED LIKE REALLY CONFUSED BUT THANKS
Step-by-step explanation:
The formula to find<span> a </span>circle's area<span> (radius)</span>2<span> usually expressed as π ⋅ r 2 where r is the radius of a </span>circle<span>. </span>Area<span> of </span>Circle<span> Concept. The </span>area of a circle<span> is all the space inside a </span>circle's<span> circumference.</span>
Well dont you have to bring down the zeros than you will have the answer
