Master theorem: t(n) = 3t(n/3) + o(log n)? hello, i've been trying to solve the bounds for the recurrence stated in the title: i know a = 3, b = 3, and f(n) = o(log n), n^log(a) of base b --> n^log(3) of base 3 --> n^1 = n. so i get θ(n). now i proceed to use case 1 of master theorem: if f(n) = o(n^log(a-e) of base b) for some constant e > 0, then t(n) = θ(n^log(a) of base b) since f(n) < n^log(a) of base b, asymptotically. i believe the solution should be θ(n) but where i am stuck is proving that f(n) is polynomially smaller, when using case 1, the ratio of f(n) / n^log(a) of base b: = log n / n, so that the solution θ(n) is true any with this is appreciated

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

2. angles qrt and qst both measures 156 degrees.

step-by-step explanation:  

we have been given a diagram of a circle and we are told that measure of arc qvt is 156 degrees.  

we can see from our given diagram that angle qrt and angle qst subtends to arc qvt.

since we know that angles subtended by the same arc at the circumference are always equal.

by angle subtended by same arc theorem, measure of angle qrt and angle qst will be equal to the measure of arc qvt.

therefore, angles qrt and qst both measures 156 degrees and 2nd statement is the correct choice.