Find midpoint of PQ with endpoints P (0,2) and Q (6,-2)

Mathematics

1 answer:

agasfer [191]2 years ago

6 0

Answer:

(3, 0)

Step-by-step explanation:

The easiest way to do this problem is to average the x-coordinates and y-coordinates.

So, (0 + 6) / 2 = 6/2 = 3 = x and (2 - 2)/2 = 0/2 = 0 = y

Therefore, the midpoint is (3,0)

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What is lim x→3 x^2 - x - 6/ x - 3

Kisachek [45]

<h3>Answer: D) 5</h3>

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Explanation:

If we plugged x = 3 into the expression, then we'd get x-3 = 3-3 = 0 in the denominator. That's not allowed. But we can simplify first

x^2-x-6 factors to (x-3)(x+2). The key here is that (x-3) is a factor. It cancels with the x-3 in the denominator

So, $\frac{x^2-x-6}{x-3} = \frac{(x-3)(x+2)}{x-3} = x+2$

Allowing us to say,

$\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}\frac{(x-3)(x+2)}{x-3}\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}(x+2)\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 3+2\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 5\\\\\\$