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.
Volume is 1/3 bh.
11 times 12 is 132. 132 times 1/3 is 44
ANSWER : 44
Answer: f-(x)=6x-6
Step-by-step explanation: Negative times a positive is negative. Negative times a negative is a positive. Works the same with f(x) representation.
Answer
22
Hope this helped
Answer:
Try an app called connects. I cant help you with this problem but that app can. it has tutors and other things that can help.
Step-by-step explanation: