The correct answer is B: Humid air has higher pressure because of the heaviness of water. Hope this helps!!
<h3>
1. Algebraically determine how long will it take the ball to reach its maximum height? What is the ball’s maximum height?</h3>
We can use the concept of derivative to find this result, but since the problem states we must use algebraic procedures, then we solve this as follows:
- Step 1: Write the original equation:
![h(t)=-16t^2+25t+5](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2B25t%2B5)
- Step 2: Common factor -16:
![h(t)=-16(t^2-\frac{25}{16}t-\frac{5}{16})](https://tex.z-dn.net/?f=h%28t%29%3D-16%28t%5E2-%5Cfrac%7B25%7D%7B16%7Dt-%5Cfrac%7B5%7D%7B16%7D%29)
- Step 3: Take half of the x-term coefficient and square it. Add and subtract this value:
X-term: ![-\frac{25}{16}](https://tex.z-dn.net/?f=-%5Cfrac%7B25%7D%7B16%7D)
Half of the x term: ![-\frac{25}{32}](https://tex.z-dn.net/?f=-%5Cfrac%7B25%7D%7B32%7D)
After squaring: ![(-\frac{25}{32})^2=\frac{625}{1024}](https://tex.z-dn.net/?f=%28-%5Cfrac%7B25%7D%7B32%7D%29%5E2%3D%5Cfrac%7B625%7D%7B1024%7D)
![h(t)=-16(t^2-\frac{25}{16}t-\frac{5}{16}+\frac{625}{1024}-\frac{625}{1024}) \\ \\ h(t)=-16(t^2-\frac{25}{16}t+\frac{625}{1024}-\frac{5}{16}-\frac{625}{1024}) \\ \\ h(t)=-16(t^2-\frac{25}{16}t+\frac{625}{1024}-\frac{945}{1024}) \\ \\](https://tex.z-dn.net/?f=h%28t%29%3D-16%28t%5E2-%5Cfrac%7B25%7D%7B16%7Dt-%5Cfrac%7B5%7D%7B16%7D%2B%5Cfrac%7B625%7D%7B1024%7D-%5Cfrac%7B625%7D%7B1024%7D%29%20%5C%5C%20%5C%5C%20h%28t%29%3D-16%28t%5E2-%5Cfrac%7B25%7D%7B16%7Dt%2B%5Cfrac%7B625%7D%7B1024%7D-%5Cfrac%7B5%7D%7B16%7D-%5Cfrac%7B625%7D%7B1024%7D%29%20%5C%5C%20%5C%5C%20h%28t%29%3D-16%28t%5E2-%5Cfrac%7B25%7D%7B16%7Dt%2B%5Cfrac%7B625%7D%7B1024%7D-%5Cfrac%7B945%7D%7B1024%7D%29%20%5C%5C%20%5C%5C)
- Step 4: Write the perfect square:
![h(t)=-16[(t-\frac{25}{32})^2-\frac{945}{1024}] \\ \\ \boxed{h(t)=-16(t-\frac{25}{32})^2-\frac{945}{64}}](https://tex.z-dn.net/?f=h%28t%29%3D-16%5B%28t-%5Cfrac%7B25%7D%7B32%7D%29%5E2-%5Cfrac%7B945%7D%7B1024%7D%5D%20%5C%5C%20%5C%5C%20%5Cboxed%7Bh%28t%29%3D-16%28t-%5Cfrac%7B25%7D%7B32%7D%29%5E2-%5Cfrac%7B945%7D%7B64%7D%7D)
Finally, the vertex of this function is:
![(\frac{25}{32},\frac{945}{64})](https://tex.z-dn.net/?f=%28%5Cfrac%7B25%7D%7B32%7D%2C%5Cfrac%7B945%7D%7B64%7D%29)
So in this vertex we can find the answer to this problem:
The ball will reach its maximum height at ![t=\frac{25}{32}s=0.78s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B25%7D%7B32%7Ds%3D0.78s)
The ball maximum height is ![H=\frac{945}{64}=14.76ft](https://tex.z-dn.net/?f=H%3D%5Cfrac%7B945%7D%7B64%7D%3D14.76ft)
<h3>2. Algebraically determine what price will maximize the revenue? What is the maximum revenue?</h3><h3 />
Also we will use completing squares. We can use the concept of derivative to find this result, but since the problem states we must use algebraic procedures, then we solve this as follows:
- Step 1: Write the original equation:
![R(p)=-5p^2+1230p](https://tex.z-dn.net/?f=R%28p%29%3D-5p%5E2%2B1230p)
- Step 2: Common factor -5:
![R(p)=-5(p^2-246p)](https://tex.z-dn.net/?f=R%28p%29%3D-5%28p%5E2-246p%29)
- Step 3: Take half of the x-term coefficient and square it. Add and subtract this value:
X-term: ![-246](https://tex.z-dn.net/?f=-246)
Half of the x term: ![-123](https://tex.z-dn.net/?f=-123)
After squaring: ![(-123)^2=15129](https://tex.z-dn.net/?f=%28-123%29%5E2%3D15129)
![R(p)=-5(p^2-246p+15129-15129)](https://tex.z-dn.net/?f=R%28p%29%3D-5%28p%5E2-246p%2B15129-15129%29)
- Step 4: Write the perfect square:
![R(p)=-5[(x-123p)^2-15129] \\ \\ R(p)=-5(x-123p)^2+75645](https://tex.z-dn.net/?f=R%28p%29%3D-5%5B%28x-123p%29%5E2-15129%5D%20%5C%5C%20%5C%5C%20R%28p%29%3D-5%28x-123p%29%5E2%2B75645)
Finally, the vertex of this function is:
![(123,75645)](https://tex.z-dn.net/?f=%28123%2C75645%29)
So in this vertex we can find the answer to this problem:
The price will maximize the revenue is ![p=123 \ dollars](https://tex.z-dn.net/?f=p%3D123%20%5C%20dollars)
The maximum revenue is ![R=75645](https://tex.z-dn.net/?f=R%3D75645)
Answer:
I DONT KNOW DO YOU WANT TO BE MY FRIEND I DONT KNOW DO YOU WANT TO BE MY FRIEND I DONT KNOW DO YOU WANT TO BE MY FRIEND
Step-by-step explanation:
Answer:
The distance covered is 113.75 m
Step-by-step explanation:
As per the question:
The initial velocity of the train, v = 20 m/s
The final velocity of the train, v' = 6 m/s
Uniform deceleration, a = 1.6![m/s^{2}](https://tex.z-dn.net/?f=m%2Fs%5E%7B2%7D)
Or uniform acceleration, a = - 1.6![m/s^{2}](https://tex.z-dn.net/?f=m%2Fs%5E%7B2%7D)
<em>Here, the body decelerates, i.e., slows down at a uniform rate thus we take acceleration with negative sign.</em>
Now, to find the distance covered, s:
Using the eqn of Kinemetics:
![v'^{2} = v^{2} + 2as](https://tex.z-dn.net/?f=v%27%5E%7B2%7D%20%3D%20v%5E%7B2%7D%20%2B%202as)
![6^{2} = 20^{2} + 2(- 1.6)s](https://tex.z-dn.net/?f=6%5E%7B2%7D%20%3D%2020%5E%7B2%7D%20%2B%202%28-%201.6%29s)
![36 - 400 = - 3.6s](https://tex.z-dn.net/?f=36%20-%20400%20%3D%20-%203.6s)
![s = \frac{- 364}{- 3.6} = 113.75\ m](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B-%20364%7D%7B-%203.6%7D%20%3D%20113.75%5C%20m)