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.
Answer:
Well i believe the answer is obtuse
Step-by-step explanation:
Answer:
x^5+2x^4-7x^3+11x^2+28x
Step-by-step explanation:
1. 2x + 50 = 100
subtract 50 on both sides and then divide by 2.
x = 25
180 - 100 = 80
m
2. 7x - 4 = 87
add 4 on both sides and divide by 7.
x = 13
180 - 87 = 93
m
3. 5x - 2 = 38
add both sides by 2 and divide by 5.
x = 8
180 - 38 = 142
m
4. 14 + 6x = 134
subtract both sides by 14 and divide by 6.
x = 20
180 - 134 = 46
m
5. x = 53
m
6. x = 14
m
Answer:
4.5 cm
Step-by-step explanation:
The ruler says it all..... (why do you need help with this? What grade????)
Hope this helps, have a good day :)