PLEASE HELLLPP verify 3 (x-1) (x+1)

Mathematics

1 answer:

svetoff [14.1K]3 years ago

6 0

Answer:

3x^2 -3

Step-by-step explanation:

3 (x-1) (x+1)

FOIL the terms in parentheses

first x*x = x^2

outer x*1 =x

inner -1 *x = -x

last -1 *1 = -1

Add them together

x^2 -x+x -1 = x^2 -1

3 ( x^2 -1)

Now multiply by 3

3x^2 -3

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Read 2 more answers

Lim t^4 - 6 / 2t^2 - 3t + 7

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to x approaching 2)

Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$