If arithmetic, then there is a common difference
if geometric, then there is a common ratio
to find the common ratio or difference
if you subtract successive terms from each other, they should have the same result if it is aritmetic
if you divide successive terms by each other, they should have the same result if it is geometric
2,3 and 5
3-2=1
5-3=2
2≠1
not aritmetic
3/2=1.5
5/3=1.66666
1.5≠1.6666
not geometric
she is wrong
I'm going to assume that you meant the following:
(x-1)
---------------
x^2-3x+2
Factoring the denominator, we get:
(x-1)
------------
(x-1)(x-2)
Notice that we have x-1 in both the numerator and the denominator. This factor cancels. HOWEVER: x can still not equal 1, which would lead to div. by zero. With the factor x-1 cancelled, we have 1 / (x-2). As x approaches 1 from either side, y approaches 1 / (1-2), or 1 / (-1), or -1. Thus, the limit exists and is -1.
Answer:
244 students.
Step-by-step explanation:
Okay, so. We know 150 people of the 7th grade is coming and 62.5% is the entire of the 7th graders coming. We would multiply.
150 * 62.5% = 93.75
We would round it up. It is over 5, so it would be 94 students. Then we would add
150 + 94 = 244
There are 244 students in 7th grade.
5/12 is the simplest form that it can be simplified to.
