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Oxana [17]
3 years ago
6

PICKING BRIANLIEST FOR THE CORRECT ANSWER!!!!!!!!!!!

Mathematics
1 answer:
pogonyaev3 years ago
3 0

Answer: D

Step-by-step explanation:

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If g (x) = 1/x then [g (x+h) - g (x)] /h
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\dfrac{-1}{x(x+h)}, h\ne 0

Step-by-step explanation:

If g(x) = \dfrac{1}{x}, then g(x+h) = \dfrac{1}{x+h}. It follows that

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

  = \dfrac{-1}{x(x+h)}, h\ne 0

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What is the radius<br> and diameter
kotegsom [21]
Diameter- Is two times the radius. Circumference is the distance around the edge of the circle.

Radius- A line from the center of a circle to a point on the circle.

Hope this helped.
7 0
3 years ago
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