Answer:
15 gallons
Step-by-step explanation:
Given that:
Pumping rate is modeled by the equation :
Q(t)=45-t gallons per minute ; where t is in minutes
Number of gallons in tank after 30 minutes ;
Q(t)=45-t
Q(30) = 45 - 30
Q(30) = 15
Hence, Number of gallons in tank after 30 minutes is 15 gallons
Answer:
Angle F
Step-by-step explanation:
They have the same degrees
The answer is:
-20a+24b-40c+12
Answer:
Any shape with three angles
Step-by-step explanation: Tri means three
Answer:
A) Radius: 3.44 cm.
Height: 6.88 cm.
B) Radius: 2.73 cm.
Height: 10.92 cm.
Step-by-step explanation:
We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.
a) We can express the volume of the soda can as:
![V=\pi r^2h=256](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2h%3D256)
This is the constraint.
The function we want to minimize is the surface, and it can be expressed as:
![S=2\pi rh+2\pi r^2](https://tex.z-dn.net/?f=S%3D2%5Cpi%20rh%2B2%5Cpi%20r%5E2)
To solve this, we can express h in function of r:
![V=\pi r^2h=256\\\\h=\frac{256}{\pi r^2}](https://tex.z-dn.net/?f=V%3D%5Cpi%20r%5E2h%3D256%5C%5C%5C%5Ch%3D%5Cfrac%7B256%7D%7B%5Cpi%20r%5E2%7D)
And replace it in the surface equation
![S=2\pi rh+2\pi r^2=2\pi r(\frac{256}{\pi r^2})+2\pi r^2=\frac{512}{r} +2\pi r^2](https://tex.z-dn.net/?f=S%3D2%5Cpi%20rh%2B2%5Cpi%20r%5E2%3D2%5Cpi%20r%28%5Cfrac%7B256%7D%7B%5Cpi%20r%5E2%7D%29%2B2%5Cpi%20r%5E2%3D%5Cfrac%7B512%7D%7Br%7D%20%2B2%5Cpi%20r%5E2)
To optimize the function, we derive and equal to zero
![\frac{dS}{dr}=512*(-1)*r^{-2}+4\pi r=0\\\\\frac{-512}{r^2}+4\pi r=0\\\\r^3=\frac{512}{4\pi} \\\\r=\sqrt[3]{\frac{512}{4\pi} } =\sqrt[3]{40.74 }=3.44](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdr%7D%3D512%2A%28-1%29%2Ar%5E%7B-2%7D%2B4%5Cpi%20r%3D0%5C%5C%5C%5C%5Cfrac%7B-512%7D%7Br%5E2%7D%2B4%5Cpi%20r%3D0%5C%5C%5C%5Cr%5E3%3D%5Cfrac%7B512%7D%7B4%5Cpi%7D%20%5C%5C%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B512%7D%7B4%5Cpi%7D%20%7D%20%3D%5Csqrt%5B3%5D%7B40.74%20%7D%3D3.44)
The radius that minimizes the surface is r=3.44 cm.
The height is then
![h=\frac{256}{\pi r^2}=\frac{256}{\pi (3.44)^2}=6.88](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B256%7D%7B%5Cpi%20r%5E2%7D%3D%5Cfrac%7B256%7D%7B%5Cpi%20%283.44%29%5E2%7D%3D6.88)
The height that minimizes the surface is h=6.88 cm.
b) The new equation for the real surface is:
![S=2\pi rh+2*(2\pi r^2)=2\pi rh+4\pi r^2](https://tex.z-dn.net/?f=S%3D2%5Cpi%20rh%2B2%2A%282%5Cpi%20r%5E2%29%3D2%5Cpi%20rh%2B4%5Cpi%20r%5E2)
We derive and equal to zero
![\frac{dS}{dr}=512*(-1)*r^{-2}+8\pi r=0\\\\\frac{-512}{r^2}+8\pi r=0\\\\r^3=\frac{512}{8\pi} \\\\r=\sqrt[3]{\frac{512}{8\pi}}=\sqrt[3]{20.37}=2.73](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdr%7D%3D512%2A%28-1%29%2Ar%5E%7B-2%7D%2B8%5Cpi%20r%3D0%5C%5C%5C%5C%5Cfrac%7B-512%7D%7Br%5E2%7D%2B8%5Cpi%20r%3D0%5C%5C%5C%5Cr%5E3%3D%5Cfrac%7B512%7D%7B8%5Cpi%7D%20%5C%5C%5C%5Cr%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B512%7D%7B8%5Cpi%7D%7D%3D%5Csqrt%5B3%5D%7B20.37%7D%3D2.73)
The radius that minimizes the real surface is r=2.73 cm.
The height is then
![h=\frac{256}{\pi r^2}=\frac{256}{\pi (2.73)^2}=10.92](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B256%7D%7B%5Cpi%20r%5E2%7D%3D%5Cfrac%7B256%7D%7B%5Cpi%20%282.73%29%5E2%7D%3D10.92)
The height that minimizes the real surface is h=10.92 cm.