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

8 0

3 years ago

A building 50 feet high casts a shadow 68 feet long. Find the measure of the angle of elevation of the sun. Round answer to near

Vedmedyk [2.9K]

Answer:

36 °

Step-by-step explanation:

In this case a rectangular triangle is formed so we can use the trigonometric functions to calculate the angle, they give us the value of the two legs, therefore we can apply the tangent function.

Tan A ° = opposite / adjacent

replacing:

Tan A ° = 50/68

Tan A ° = 0.735

A ° = arc tan (0.735)

A ° = 36.32

Which means that the angle of inclination is 36°

4 0

3 years ago

Sinxcosy = 1/2 (sin(x+y)+cos (x-y))

sasho [114]

8 0

3 years ago