Log₄20-log₄ 45+log₄144=
log₄(20/45)+log₄144= (log_a b- log_a c=log_a (b/c) )
log₄[(20*144)/45]= (log_a b +log_a c=log_a (b*c) )
log₄(2880/45)=
log₄(64)=n ⇔ 4^n=64 (log_a x=n ⇔ a^n=x)
4^n=4³ ⇒n=3 (64=4*4*4=4³)
Answer: log₄20-log₄ 45+log₄144=3
The variable 'T' equals -20.
I believe you cant fix one rather you need a new one. It runs about 300 plus labor
Answer/Step-by-step explanation:
a. Using two points on the line (0, 2) and (4, 12), find the gradient:
Gradient = ∆y/∆x = (12 - 2) / (4 - 0) = 10/4 = 2.5
Gradient = 2.5
b. The gradient, 2.5, represents the unit rate. That is, the volume of the oil in the tank per second.
i.e. 2.5 litres per second
c. The value, L = 2 represents the y-intercept, which is the starting value or initial volume of oil in the tank at 0 secs.
