15/3= 5 If you are not sure for my answer, check your calculator.
Answer: 36
Step-by-step explanation:
(11−9)^2(8−5)^2
=(22)((8−5)^2)
=4(8−5)^2
=4(3^2)
=(4)(9)
=36
<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](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2-x-6%7D%7Bx-3%7D%20%3D%20%5Cfrac%7B%28x-3%29%28x%2B2%29%7D%7Bx-3%7D%20%3D%20x%2B2)
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\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%203%7D%5Cfrac%7Bx%5E2-x-6%7D%7Bx-3%7D%20%3D%20%5Clim_%7Bx%20%5Cto%203%7D%5Cfrac%7B%28x-3%29%28x%2B2%29%7D%7Bx-3%7D%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%203%7D%5Cfrac%7Bx%5E2-x-6%7D%7Bx-3%7D%20%3D%20%5Clim_%7Bx%20%5Cto%203%7D%28x%2B2%29%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%203%7D%5Cfrac%7Bx%5E2-x-6%7D%7Bx-3%7D%20%3D%203%2B2%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%203%7D%5Cfrac%7Bx%5E2-x-6%7D%7Bx-3%7D%20%3D%205%5C%5C%5C%5C%5C%5C)
The answer is (5p+6q)(5p-6q)
Steps:
Apply difference of the squares rule: x^2-y^2=(x+y)(x-y)
25p^2-36q^2=(5p+6q)(5p-6q)
Hope this helps you! (:
-PsychoChicken4040