Answer:
9.0 ft
Step-by-step explanation:
Let the distance from the bottom of the board to the edge of the wall be represented as "x"
Angle measure = θ = 53.13°
Hypotenuse = 15 ft
Adjacent side = x ft
The trigonometric ratio we would apply would be CAH:
Thus,
Cosine θ = Adjacent/Hypotenuse
Plug in the values
Cos 53.13° = x/15
Multiply both sides by 15
15 * Cos 53.13 = x
9.00002143 = x
x ≈ 9.0 ft (nearest tenth)
Therefore, distance from the bottom of the board to the edge of the wall = 9.0 ft
Answer:
x(t) = 581.66 lb
Step-by-step explanation:
From the given information:
Consider the salt quantity in the tank at time t be x(t) lb.
∴ the rate of change of salt content in the tank is 
where:
the rate of inflow = salt conc. × flow rate = 3 lb/gallon × 7 gallons
=21 lb/s
rate of outflow = salt conc. in the tank × flow rate
= x/250 × 9
= 9x/250 lb/s
∴


This is a 1st order linear differentiation,
The integrating factor if 
∴


at t = (0) and x(0) = 100 lb
Hence;



∴
after time t = 1 minute i.e 60 s


x(t) = 581.66 lb
Start here: A = 18 ft^2 = L * W. Next, L = 2W.
Merging these 2 formulas lets us eliminate L:
18 ft^2 = (2W) * (W) = 2W^2.
So W^2 = 9 ft^2, and W=3 ft. Please use L = 2W (from above) to calculate L.
Answer:
2
Step-by-step explanation: