The sequence is an geometric progression;
Nth=a₁r^(n-1)
a₁ is the first term=-1
r=ratio=a₂/a₁=6/-1=-6
Nth=-1(-6^(n-1)
NTh=an
a₁=-1(-6⁰)=-1
a₂=-1(-6¹)=6
a₃=-1(-6²)=-36
a₄=-1(-6³)=216
therefore this serie is rising at an increasing fast speed, it is an geometric progression.
So, you had done everything right so far (other than squaring the 2), but that was only half of the question.
to find the least common multiple, you need to first figure out what the prime factors have in common.

each have two twos. both have one 5, so we know our answer will look something like

now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.
one has one 3 and one has two, so we need two threes. now our equation is

what's the only number we have to deal with? 7...
how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes
Answer:
C
Step-by-step explanation:
because it doesn't have an x number written twice like for example
(2,0) (2,8) (3,0)
Answer:
Mr X
Step-by-step explanation:
his identity is a variable ; it could be anybody of the male gender.
all the other ones denote known designations, i.e, B+ grade
<span>149 times 235 is equal to </span>35,015.