<h2>The test is worth 15 points.</h2><h3 /><h3>Double-check:</h3><h3>15 x 1.2 = 18</h3>
Answer:
The radius would be 4.8
Step-by-step explanation:
The circumerfence of a circle can be found use the equation C=2
r. So since we are given the circumerce we need to rearrange the eqaution to it looks like
=r. when we do this out we find that the radius equals 4.8.
Answer:
15.07 Euro
Step-by-step explanation:
The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.
Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.
By the fundamental theorem of calculus,
![\displaystyle f(5) = f(0) + \int_0^5 f'(x) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%285%29%20%3D%20f%280%29%20%2B%20%5Cint_0%5E5%20f%27%28x%29%20%5C%2C%20dx)
The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so
![\displaystyle \int_0^5 f'(x) \, dx = \frac{5+2}2 \times 2 = 7](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E5%20f%27%28x%29%20%5C%2C%20dx%20%3D%20%5Cfrac%7B5%2B2%7D2%20%5Ctimes%202%20%3D%207)
![\implies \max\{f(x) \mid 0\le x \le5\} = f(5) = f(0) + 7 = \boxed{13}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cmax%5C%7Bf%28x%29%20%5Cmid%200%5Cle%20x%20%5Cle5%5C%7D%20%3D%20f%285%29%20%3D%20f%280%29%20%2B%207%20%3D%20%5Cboxed%7B13%7D)