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3 years ago

Which statement is always true about a regular polygon?

Mathematics

2 answers:

polet [3.4K]3 years ago

8 0

Answer:

All polygons have straight lines

Step-by-step explanation:

ruslelena [56]3 years ago

8 0

For a polygon to be 'regular' it must have all sides the same length and all interior angles the same.

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За^n(а^n + a^n-1) what is it??

solniwko [45]

Distribute the 3a^n to all the other values then solve...
(3a^n*a^n)+(3a^n*a^n)+(3a^n*-1)
4a^2n+4a^2n-3a^n=
8a^2n-3a^n

8 0

4 years ago

4in 10in use the dimension to find valuemos the candle round to the nearest hundreds

lilavasa [31]

Answer:

Umm let me see

Step-by-step explanation:

6 0

3 years ago

Find the values of a and b so that the solution of the linear system is (-9,1).

Diano4ka-milaya [45]

Answer:

Step-by-step explanation:

we can write -9 instead of x and y=1 instead of 1

so we write solution again

-9a+1b=-31

-9a-1b=-41

-18a=-72

a=4

we should write 4 instead of a

-9(4)+1b=-31

-36+b=-31

b=5

a=4

6 0

3 years ago

VEEL

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago

Which of the following functions is graphed below?

erastovalidia [21]

Could you please attach the graph itself because the question is not proper

3 0

3 years ago