Answer:
762 days
Step-by-step explanation:
Given

Let the rate be R.
So the rate change with time is represented as:

So:

To get the number of insects between day 0 and day 3, we need to integrate dR and set the bounds to 0 and 3
i.e.
becomes


Integrate
![R = 200t + \frac{10t^2}{2} + \frac{13t^3}{3} [3,0]](https://tex.z-dn.net/?f=R%20%3D%20200t%20%2B%20%5Cfrac%7B10t%5E2%7D%7B2%7D%20%2B%20%5Cfrac%7B13t%5E3%7D%7B3%7D%20%5B3%2C0%5D)
Solve for R by substituting 0 and 3 for t





<em>The population of insect between the required interval is 762</em>
what i would do is implant some number under 5 for the X's.
Then work it out as if 3 was x so for instance -4(8-3x3)
Answer:
<h2>- 417</h2>
Step-by-step explanation:
n - integer
The sum of n and 181 is equal to -236:
n + 181 = -236 <em>subtract 181 from both sides</em>
n + 181 - 181 = -236 - 181
n = -417
She needs 45 more sticks of butter. This is because she already has two cakes, and 47-2=45. She can bake one more cake with each stick of butter, which means the answer is indeed 45.
Https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-data-statistics/cc-6th/v/calculating-interquartile-range-iqr
This is a link that I gave to my little sister that helped her understand what interquartile range is. It has an example as well! I hope this helps!