Find limit as x approaches 1 for x-1/ x^2-3x+2

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: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.