+1982 What is the domain of the function f(x) = log (1/x + 12)?
A) {x ∈ R}
B) {x ∈ R | x < –12}
C) {x ∈ R | x > –12}
D) {x ∈ R | x ≠ 12}
E) {x ∈ R | x ≠ –12}
Which is equivalent to (-1 + 2i)(5i)?
A) -11
B) -8
C) -1 + 7i
D) 10 - 5i
E) -10 - 5i
+128407 I think this is supposed to be
f(x) = log ( 1 / ( x + 12) )
We cannot take the log of 0 or a negative.....thus......the domain is all real nubers > -12 ⇒ "C"
Second one
(-1 + 2i ) * 5i = -5i + 10i^2 = -5i - 10 = -10 - 5i
+1982 1. Since it is only possible to find the log of a positive number, 1/x + 12 > 0. Therefore, x + 12 > 0, and if 12 is subtracted from both sides, the inequality becomes x > –12, so the domain of the function is {x ∈R| x > –12}.
