Answer:
Step-by-step explanation:
<h3>A.</h3>
The equation for the model of the geyser is found by substituting the given upward velocity into the vertical motion model. The problem statement tells us v=69. We assume the height is measured from ground level, so c=0. Putting these values into the model gives ...
h(t) = -16t² +69t
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<h3>B.</h3>
The maximum height is at a time that is halfway between the zeros of the function.
h(t) = -16t(t -4.3125) . . . . . has zeros at t=0 and t=4.3125
The maximum height will occur at t=4.3125/2 = 2.15625 seconds. The height at that time is ...
h(t) = -16(2.15625)(2.15625 -4.3125) = 16(2.15625²) ≈ 74.39 . . . feet
The maximum height of the geyser is about 74.4 feet.
So you have to double the numerator of the fraction which is 4 times 2 equals 8 so it’s 8/5 miles per hour or 1.6 mph
Answer:
I need a picture
Step-by-step explanation:
Answer:
Side of 22 and height of 11
Step-by-step explanation:
Let s be the side of the square base and h be the height of the tank. Since the tank volume is restricted to 5324 ft cubed we have the following equation:
As the thickness is already defined, we can minimize the weight by minimizing the surface area of the tank
Base area with open top 
Side area 4sh
Total surface area 
We can substitute
To find the minimum of this function, we can take the first derivative, and set it to 0
![s = \sqrt[3]{10648} = 22](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%5B3%5D%7B10648%7D%20%3D%2022)
What do you mean what are the solutions to?
I don’t understand the question?
