Tell the kind of symmetry this figure has: point, line, plane. (Select all that apply.)

Mathematics

2 answers:

mamaluj [8]3 years ago

6 0

Answer with explanation:

We know that the symmetry of a figure means that when a figure is moved, flipped or reflected along any of the line, point or plane than it retain'sit's shape or structure.

Now we are given a geometrical shape as:

CIRCLE

Now when the circle is rotated along a point then it's shape is retained.

so,circle is symmetrical about a point.

a circle is symmetrical about it's center.

same is the case of a line.

a circle is symmetrical is about it's diameter.

and a plane symmetry is for a three dimensional figure.

Hence, circle is symmetrical about Point and a line.

e-lub [12.9K]3 years ago

5 0

Since a circle is simple it has no points but it can be folded many times so symmetry is really unlimited for the circle.It is not a plane because it is round and does not have sides.

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sladkih [1.3K]

Answer:

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Step-by-step explanation:

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3 years ago

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lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

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.

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