Answer:
e. 
Step-by-step explanation:
It is given that,
The density of intergalactic space material is
kg per cubic meter.
And the volume of intergalactic space material is 
So the mass of that much intergalactic space material is,
, where 'm' is the mass, and 'v' is the volume.
Putting the values we get,


It is also given that the mass of lead is the same as the mass of the intergalactic space material. Therefore, the mass of lead is 
So the volume of lead is,


So the volume of lead is
.
386 divided by 20 is 19.3 so Greg averaged 19.3 miles per gallon.
Log w (x^2-6)^4
Using log a b = log a + log b, with a=w and b=(x^-6)^4:
log w (x^2-6)^4 = log w + log (x^2-6)^4
Using in the second term log a^b = b log a, with a=x^2-6 and b=4
log w (x^2-6)^4 = log w + log (x^2-6)^4 = log w + 4 log (x^2-6)
Then, the answer is:
log w (x^2-6)^4 = log w + 4 log (x^2-6)
The second one because domain refers to x, and since at (-3,-4) the point is hollow, the sign should be < without the line under it