$f(x)=\frac{1}{\sqrt{x} }\\f'(x)=-\frac{1}{2} x^{-\frac{3}{2}}\\\\f'(4)=-\frac{1}{2} 4^{-\frac{3}{2} }=-\frac{1}{2} \frac{1}{4^\frac{3}{2} } \\=-\frac{1}{2} *\frac{1}{8}=-\frac{1}{16} \\when x=4\\y=f(x)=\frac{1}{\sqrt{4} } =\frac{1}{2} \\\\eq.~ of ~tangent ~line~is\\y=-\frac{1}{2} -\frac{1}{16} (x-4)$

3 0

3 years ago

The volume, V, of any cube with a side length, s. can be determined using the forma

Tpy6a [65]

Answer:

V=12.167

Step-by-step explanation:

V=s^3

V=2.3x2.3x2.3

V=12.167

Since it is cubic centimeters, then it is written as 12.167cm^2

4 0

3 years ago

What is the slope of the line with points A and B?

Zanzabum

Answer:

you will have to get the graph :\

Step-by-step explanation:

sorry

4 0

3 years ago