What is the correct name of the polygon below

Mathematics

2 answers:

Zigmanuir [339]3 years ago

7 0

Since it has 8 sides, it is an octagon.
This table will help
triangle (or trigon)--3
quadrilateral (or tetragon)--4
pentagon--5
hexagon--6
heptagon--7
octagon--8
nonagon (or enneagon)--9
decagon--10
hendecagon (or undecagon)--11
dodecagon (or duodecagon)--12
tridecagon (or triskaidecagon)--13
tetradecagon (or tetrakaidecagon)--14
pentadecagon (or pentakaidecagon)--15
hexadecagon (or hexakaidecagon)--16
heptadecagon (or heptakaidecagon)--17
octadecagon (or octakaidecagon)--18
enneadecagon (or enneakaidecagon)--19
icosagon--20
icositetragon (or icosikaitetragon)--24
triacontagon--30
tetracontagon (or tessaracontagon)--40
pentacontagon (or pentecontagon)--50.
Please mark me as brainliest. Thanks!!

babunello [35]3 years ago

5 0

Octagon...it has 8 sides

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Review your answers to Question 2. What happens to the coordinates of a shape when the shape reflects about the x-axis? What hap

mixas84 [53]

Answer:

When the shape reflects about the x-axis, the x-coordinate remains unchanged but the y-coordinate becomes negative. When the shape reflects about the y-axis, the y-coordinate remains the same but the x-coordinate becomes negative.

Step-by-step explanation:

Sample answer from Edmentum

4 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$