Given :
In the early spring, the trout at Big Blue Lake swim at a depth of 30 feet below sea level.
When the lake warms up in the summer, the trout swim 20 feet deeper than that , d = 20 feet .
To Find :
At what position relative to sea level do the trout swim in the summer.
Solution :
Depth in spring from surface ,
.
Let , depth in summer is D .
Now , depth in summer relative to sea-level is :
![D=D_s+d\\\\D=30+20\ feet\\\\D=50\ feet](https://tex.z-dn.net/?f=D%3DD_s%2Bd%5C%5C%5C%5CD%3D30%2B20%5C%20feet%5C%5C%5C%5CD%3D50%5C%20feet)
Therefore , trout swim 50 feet deep below the sea level in the summer .
Hence , this is the required solution .
Answer:
D) 0.80×0.40 so the answer is D
Answer:
7 Hours per day
Step-by-step explanation:
35/5=7
If you are working 35/hrs a week, and you only work 5 days, to find out how many hours you work each day, you need to divide the amount of hours per week and the days per week.
Hence, 35/5=7.
Answer:
2 is the answer
Step-by-step explanation:
Answer:
a) ![v(t) = 3t^{2} - 16t + 2](https://tex.z-dn.net/?f=v%28t%29%20%3D%203t%5E%7B2%7D%20-%2016t%20%2B%202)
b) The velocity after 3 seconds is -3m/s.
c)
and
.
Step-by-step explanation:
The position is given by the following equation.
![s(t) = t^{3} - 8t^{2} + 2t](https://tex.z-dn.net/?f=s%28t%29%20%3D%20t%5E%7B3%7D%20-%208t%5E%7B2%7D%20%2B%202t)
(a) Find the velocity at time t.
The velocity is the derivative of position. So:
.
(b) What is the velocity after 3 seconds?
This is v(3).
![v(t) = 3t^{2} - 16t + 2](https://tex.z-dn.net/?f=v%28t%29%20%3D%203t%5E%7B2%7D%20-%2016t%20%2B%202)
![v(3) = 3*(3)^{2} - 16*(3) + 2 = -19](https://tex.z-dn.net/?f=v%283%29%20%3D%203%2A%283%29%5E%7B2%7D%20-%2016%2A%283%29%20%2B%202%20%3D%20-19)
The velocity after 3 seconds is -3m/s.
(c) When is the particle at rest?
This is when
.
So:
![v(t) = 3t^{2} - 16t + 2](https://tex.z-dn.net/?f=v%28t%29%20%3D%203t%5E%7B2%7D%20-%2016t%20%2B%202)
![3t^{2} - 16t + 2 = 0](https://tex.z-dn.net/?f=3t%5E%7B2%7D%20-%2016t%20%2B%202%20%3D%200)
This is when
and
.