Answer:
75
Step-by-step explanation:
100% --- 100
100% - 25% --- ?
75% --- × 100 = 75
dy/dt = - (1/5) ft/s = - 0.2 ft/s
Given
L = 5 ft
Qin = 25 ft³/s
Qout = 30 ft³/s
h = 10 ft
dy/dt = ?
We can apply the relation
ΔQ = Qint - Qout = 25 ft³/s - 30 ft³/s
⇒ ΔQ = - 5 ft³/s
Then we use the formula
Q = v*A
where Q = ΔQ, A = L² is the area of square base and v = dy/dt is the rate of change in the depth of the solution in the tank
⇒ ΔQ = (dy/dt)*L²
⇒ dy/dt = ΔQ/L²
⇒ dy/dt = (- 5 ft³/s)/(5 ft)²
⇒ dy/dt = - (1/5) ft/s = - 0.2 ft/s
13. 1/3
14. 1/20
15. 9/20
16. 9/200
17. 253/25
18. 39/10
Final answer is .
Given problem is .
Now we need to simplify this problem.
Apply formula
so we get:
Hence final answer is .
y=x-1
y=-2x-4
although I cant summon a graph for this one, I can give cords
for first graph (-2,-3),(-1,-2),(0,-1), (1,0),(2,1)
For second graph the slope is down 2 over 1, and begins at (0,-4).
(-2,0)(-1,-2),(0,-4),(1,-6),(2,-8)