We have the given I = 10^-1 and I₀ = 10^-12.
Simply plugging in the values to the equation, we have:
L = 10log(10^-1/10^-12)
L = 110 Db
The answer is D.
(I used a scientific calculator in solving for L).
I assume that the given equation above is 0.9(x+1.4)−2.3+0.1x=1.6 and not 0.9(x+1.4)−2.3+0.1x=1.60.9(x+1.4)−2.3+0.1x=1.6, I think there is a typo error on this. Based on equation I assumed the answer is 2.64.Thank you for posting your question here, I hope my answer helps.
3×2y-(1/3×3)=5×3
6y-1=15
6y-1+1=15+1
6y=16
y=8/3
Y cannot be 8

this formula is used to determine gpm (gallons per minute) of flow with a known hose diameter (d) and nozzle pressure (np). The constant (29.7) is consistent when figuring flow with in gallons with a know pressure expressed in psi (pounds per square inch).
Scenario example: a firefighter is using a handline to fight fire with an 1 3/4 inch line with a nozzle pressure of 100 psi. How much water is the firefighter flowing when fighting fire?
Answer:
D. Minimum at (3, 7)
Step-by-step explanation:
We can add and subtract the square of half the x-coefficient:
y = x^2 -6x +(-6/2)^2 +16 -(-6/2)^2
y = (x -3)^2 +7 . . . . . simplify to vertex form
Comparing this to the vertex for for vertex (h, k) ...
y = (x -h)^2 +k
We find the vertex to be ...
(3, 7) . . . . vertex
The coefficient of x^2 is positive (+1), so the parabola opens upward and the vertex is a minimum.