<span>x in (-oo:+oo)
((5-9*x)^1)/2 = ((4*x+3)^1)/2 // - ((4*x+3)^1)/2
((5-9*x)^1)/2-(((4*x+3)^1)/2) = 0
(5-9*x)/2-((4*x+3)/2) = 0
(5-9*x)/2+(-1*(4*x+3))/2 = 0
5-1*(4*x+3)-9*x = 0
2-13*x = 0
(2-13*x)/2 = 0
(2-13*x)/2 = 0 // * 2
2-13*x = 0
2-13*x = 0 // - 2
-13*x = -2 // : -13
x = -2/(-13)
x = 2/13
x = 2/13</span>
Any log function/equation could be transformed in an exponential one:
log₂(x) = 4
x= 2⁴ and x = 16
I believe the answer is C.
Answer:
The formula d=1/2n+26 relates the nozzle pressure n (in pounds per square inch of a fire hose and the maximum horizontal distance of the water reaches d (in feet).
Step-by-step explanation:
4!•3! = 144
you’re welcome :)
