$\displaystyle \lim_{x\rightarrow3}~\dfrac{2x^2-18}{x^2-3x} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x^2-3^2)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x-3)(x+3)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x+3)}{x} \\ \\ \\=\dfrac{2(3+3)}{3}\\ \\ \\=\dfrac{2*3*2}{3} =\Large \boxed{\sf \bf \ 4 \ }$

Thank you

8 0

3 years ago

Kelly and Adam are buying boxes filled with candy to give to friends. Kelly chooses a cone-shaped box that has radius x. Adam ch

kondaur [170]

Kelly and Adam are buying candy boxes to give to friends

Step-by-step explanation:

Kelly statement is true.

1. Kelly buy a candy box which is cone -shaped with radius x.

2.Adam buy a candy box which is square pyramid shape with same height as Kelly's box but the side length is xπ√.

3.Two candy boxes have different in shape so volumes of the boxes will be different.

6 0

3 years ago

Can someone help me please

Andrej [43]

It would be D because the equation just seems flipped, but the requirements of x are fulfilled.

3 0

3 years ago