In order to find out where your holes and asymptotes are is to factor both the top and the bottom of that rational function. The numerator factors to (x+2)(x-2) and the denominator factors to x(x+1)(x-2). So since there is an (x-2) in both the numerator and denominator, that is called a removable discontinuity which we also know as a hole. The other factors in the denominator, the x and the (x+1) are vertical asymptotes, or values of x that make the rational function undefined (you're NEVER allowed to have a 0 in the denominator of a fraction!). So your correct choice is c. The way you find the y coordinate for the hole is to plug in 2 for x and solve it for y. No biggie.
Answer:
when f(x) is -1, f(x) = -3
when f(x) is 0, f(x) = 1
when f(x) is 3, f(x)= 13
so is -3, 1 and 13.
you basically replace x with the number and then evaluate
hope is helpful!!
Answer:
47031790b
Step-by-step explanation:
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Answer: 
Step-by-step explanation:
Given
The hypotenuse is 13 units
One of the lengths is 12 units
The marked angle is x
from the figure, we can write
![\Rightarrow \cos x=\dfrac{12}{13}\\\\\Rightarrow x=\cos^{-1}[\dfrac{12}{13}]\\\\\Rightarrow x=22.62^{\circ}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Ccos%20x%3D%5Cdfrac%7B12%7D%7B13%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Ccos%5E%7B-1%7D%5B%5Cdfrac%7B12%7D%7B13%7D%5D%5C%5C%5C%5C%5CRightarrow%20x%3D22.62%5E%7B%5Ccirc%7D)
