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:

Step-by-step explanation:
Given
Variation: Directly Proportional
Insects = 10 million when Rate = 2 million
i.e.
when 
Required
Determine the differential equation for the scenario
The variation can be represented as:

Which means that the rate at which the insects increase with time is directly proportional to the number of insects
Convert to equation


Substitute values for
and P

Make k the subject



Substitute
for k in 


Answer:
43
Step-by-step explanation:
You are going to want to substitute the x of the function f(x) with the function of g(x) and replace the x in g(x) with -4
(f+g)(-4) = 4+3((-4)^2-3)
=43
hope this helped!
(b
if you flip up the squares on the right you can see it creates the perfect right triangle.