Answer:
5/3
Step-by-step explanation:
![\lim_{x \to 6} f(x) =4; and \lim_{x \to 6} g(x) =6\\ \lim_{x \to 6} \dfrac{f(x)+g(x)}{g(x)} =?](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%206%7D%20f%28x%29%20%3D4%3B%20and%20%5Clim_%7Bx%20%5Cto%206%7D%20g%28x%29%20%3D6%5C%5C%20%5Clim_%7Bx%20%5Cto%206%7D%20%5Cdfrac%7Bf%28x%29%2Bg%28x%29%7D%7Bg%28x%29%7D%20%3D%3F)
Using the laws of limits,
![\\ \lim_{x \to 6} \dfrac{f(x)+g(x)}{g(x)} = \dfrac{lim_{x \to 6}f(x)+lim_{x \to 6}g(x)}{lim_{x \to 6}g(x)}, lim_{x \to 6}g(x)\neq 0](https://tex.z-dn.net/?f=%5C%5C%20%5Clim_%7Bx%20%5Cto%206%7D%20%5Cdfrac%7Bf%28x%29%2Bg%28x%29%7D%7Bg%28x%29%7D%20%3D%20%20%5Cdfrac%7Blim_%7Bx%20%5Cto%206%7Df%28x%29%2Blim_%7Bx%20%5Cto%206%7Dg%28x%29%7D%7Blim_%7Bx%20%5Cto%206%7Dg%28x%29%7D%2C%20lim_%7Bx%20%5Cto%206%7Dg%28x%29%5Cneq%200)
Thus,
![\dfrac{lim_{x \to 6}f(x)+lim_{x \to 6}g(x)}{lim_{x \to 6}g(x)} =](https://tex.z-dn.net/?f=%5Cdfrac%7Blim_%7Bx%20%5Cto%206%7Df%28x%29%2Blim_%7Bx%20%5Cto%206%7Dg%28x%29%7D%7Blim_%7Bx%20%5Cto%206%7Dg%28x%29%7D%20%3D)
![\frac{4+6}{6} =\frac{10}{6}=\frac{5}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B4%2B6%7D%7B6%7D%20%3D%5Cfrac%7B10%7D%7B6%7D%3D%5Cfrac%7B5%7D%7B3%7D)
Two<span> trains </span>leave different<span> cities heading toward each </span>other<span> at </span>different<span> speeds. ... At the </span>same time<span>Train B, </span>traveling 60 mph<span>, leaves Eastford heading toward Westford. ... Since an equation remains true as </span>long<span> as we perform the </span>same<span> operation ... that the train's rate is 40 </span>mph<span>, which means it </span>will travel<span> 40 </span>miles<span> in </span>one<span> hour.</span>
The opposite of 7 on a number line would be -7 because if you look at 0 as a mirror, then the opposite of 7 is -7.
The coefficients would be 6 and 2.
Coefficient of k would be 6 and for n it would be 2.