Area of a circumference=π*r²for diameter=21.2 cm---------> r=10.6 cmA=π*10.6²--------> A=π*112.36 -------> A=352.81 cm²if 2π radians (full circumference) has an area of -----------------> 352.81 cm² 3π/5 radians-------------------------------------> XX=[(3π/5)*(352.81)]/2π---------> X=105.84 cm²
10(h)+5(h)+7, because 10 and 15 are charged per a hour, so they have a h(house). 7 dollars is a fee that you only pay once. I hope this helps :)
With the provided vertices and the focus point, the hyperbola will look like the one in the picture below.
notice the "c" distance from the center to the focus point.
since it's a vertical hyperbola, the positive fraction will be the one with the "y" in it, its center is clearly half-way between the vertices at -5, 5.
its major axis or traverse axis goes from -2 up to 12, so is 14 units long, therefore the "a" component is half that, or 7.
![\bf \textit{hyperbolas, vertical traverse axis } \\\\ \cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h, k\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bhyperbolas%2C%20vertical%20traverse%20axis%20%7D%0A%5C%5C%5C%5C%0A%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20b%5E2%7D%3D1%0A%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Acenter%5C%20%28%20h%2C%20k%29%5C%5C%0Avertices%5C%20%28%20h%2C%20%20k%5Cpm%20a%29%5C%5C%0Ac%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%0A%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%0A-------------------------------)