$f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}$

8 0

2 years ago

I missed my class today because i came in late, can someone help me? would be very apricated

snow_tiger [21]

The least number of possible bags of beads packed would be 2 because the greatest common factor is 2. To find how many of each bead is packed into each bag you would have to divide 14 and 2 which is 7, 16 and 2 which is 8, and 18 and 2 which is 9, so there will be 8 read beads, 9 orange beads, and 7 green beads.

Hope this helps!!!!

5 0

3 years ago

Is anyone here in connections academy geometry. even if your not can someone help me with the question in the images. If correct

Inessa [10]

Answer:

$13.05cm^2$