Let's say "p" people were going to the expedition initially, and the cost for each was "c", now, we know the total cost is 1800, so for "p", folks that'd be 1800/p how much each one cost, namely, how many times "p" goes into 1800.
well, prior to leaving, 15 dropped out, so that leaves us with " p - 15 ", and the cost "c" bumped up to " c + 27 " for each.

![\bf 1800p=1800(p-15)+27[p(p-15)] \\\\\\ 1800p=1800p-27000+27(p^2-15p) \\\\\\ 0=-27000+27(p^2-15p)\implies 0=-27000+27p^2-405p \\\\\\ \textit{now, let's take a common factor of }27 \\\\\\ 0=p^2-15p-1000\implies 0=(p-40)(p+25)\implies p= \begin{cases} \boxed{40}\\ -25 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%201800p%3D1800%28p-15%29%2B27%5Bp%28p-15%29%5D%0A%5C%5C%5C%5C%5C%5C%0A1800p%3D1800p-27000%2B27%28p%5E2-15p%29%0A%5C%5C%5C%5C%5C%5C%0A0%3D-27000%2B27%28p%5E2-15p%29%5Cimplies%200%3D-27000%2B27p%5E2-405p%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20let%27s%20take%20a%20common%20factor%20of%20%7D27%0A%5C%5C%5C%5C%5C%5C%0A0%3Dp%5E2-15p-1000%5Cimplies%200%3D%28p-40%29%28p%2B25%29%5Cimplies%20p%3D%0A%5Cbegin%7Bcases%7D%0A%5Cboxed%7B40%7D%5C%5C%0A-25%0A%5Cend%7Bcases%7D)
well, you can't have a negative value of people... so it has to be 40.
so, 40 folks were initially going, then 15 dropped out, how many went on the expedition? 40 - 15.






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And we're done.
Thanks for watching buddy good luck.
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(a) With a unit rate of 8, repeated (iterated) 72 times, we can simply multiply 8 by 72 to get the net change: 8 × 72 = 576 fish.
(b) For the unit rate (i.e. rate of change per time - minutes in this case) we can simply divide the total water pumped out by the total time it took to perform this. So: 48 ÷ 8 = 6 gallons/min.
Hope I could help you! :)
Step-by-step explanation:
here
-11r ≥ 220
-11r/-11 ≥ 220/-11 (dividing both side with -11 to eliminate -11 )
r ≥-20