In an arithmetic sequence, the difference between consecutive terms is constant. In formulas, there exists a number 
such that
In an geometric sequence, the ratio between consecutive terms is constant. In formulas, there exists a number such that
So, there exists infinite sequences that are not arithmetic nor geometric. Simply choose a sequence where neither the difference nor the ratio between consecutive terms is constant.
For example, any sequence starting with
Won't be arithmetic nor geometric. It's not arithmetic (no matter how you continue it, indefinitely), because the difference between the first two numbers is 14, and between the second and the third is -18, and thus it's not constant. It's not geometric either, because the ratio between the first two numbers is 15, and between the second and the third is -1/5, and thus it's not constant.
In this problem you will need to use the Pythagorean theorem (c^2=a^2+b^2).
The a and b represents the two edges, while c is the diagonal side and it is called the hypotenuse. Since you already know what the hypotenuse is and what one of the sides already are you just have to use the problem: c^2-a^2=b^2. Then if you plug the data you already have into the problem you will get 10^2-6^2=b^2. That then equals 100-36=b^2. Then you subtract and get b^2=64. Then you square root both sides and you get the answer b=8.
Answer:
78.0 kilo
Step-by-step explanation:
85.8- 93.6= 7.8
101.4-93.6= 7.8
85.8-7.8 = 78.0 kilo
Answer:
Step-by-step explanation:
The population in 2003= 47000
Since the population increase by 1200 every year,
in 2004 the population will be 47000+1200
in 2005 the population will be 47000+(1200+1200) which is the same as
47000+2(1200) where 2 is 2 years after 2003,
Therefore the population x years after 2003 is 47000+x(1200).
P= 47000+1200x
b) The population at 2009 which is 6 years after 2003 will be
47000+(1200)*6=47000+7200= 54200
The population at 2009 is 54200,
The answer is 2x^2 to the question
