Each curve completes one loop over the interval
![0\le t\le2\pi](https://tex.z-dn.net/?f=0%5Cle%20t%5Cle2%5Cpi)
. Find the intersections of the curves within this interval.
![6-6\sin\theta=6\implies 1-\sin\theta=1\implies \sin\theta=0\implies \theta=0,\theta=\pi](https://tex.z-dn.net/?f=6-6%5Csin%5Ctheta%3D6%5Cimplies%201-%5Csin%5Ctheta%3D1%5Cimplies%20%5Csin%5Ctheta%3D0%5Cimplies%20%5Ctheta%3D0%2C%5Ctheta%3D%5Cpi)
The region of interest has an area given by the double integral
![\displaystyle\int_\pi^{2\pi}\int_6^{6-6\sin\theta}r\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%5Cpi%5E%7B2%5Cpi%7D%5Cint_6%5E%7B6-6%5Csin%5Ctheta%7Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
equivalent to the single integral
![\displaystyle\frac12\int_\pi^{2\pi}\bigg((6-6\sin\theta)^2-6^2\bigg)\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac12%5Cint_%5Cpi%5E%7B2%5Cpi%7D%5Cbigg%28%286-6%5Csin%5Ctheta%29%5E2-6%5E2%5Cbigg%29%5C%2C%5Cmathrm%20d%5Ctheta)
which evaluates to
![9\pi+72](https://tex.z-dn.net/?f=9%5Cpi%2B72)
.
Answer:
TU=36
Step-by-step explanation:
use the side lengths of PQR to determine the proportionality of the two triangles. 9/2=4.5 so, multiply 8 by 4.5 to find TU.
Find range ,median mode and mean of the following numbers -19,-11,-19,-18,-13,-19,-18,-11.0, 11,12
IrinaK [193]
Range= 31 because 12- -19 = 31 you subtract the lowest with the highest.
Mean= 8.3
Mode = -19 because you mostly see -19 there’s 3 -19 . :)))
I think we’re talking about the 15, the smallest number for 15 would be 3. Because 3 x 5 = 15