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Leno4ka [110]
2 years ago
7

What is the order of rotational symmetry of a regular octagon?

Mathematics
1 answer:
SSSSS [86.1K]2 years ago
5 0

Answer:

Answer: I do not know the answer

Step-by-step explanation:

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24. The cost to board a dog at a kennel varies
natita [175]

Answer:c=24n

Step-by-step explanation:

C~n

C=no

96=4k

Divide both sides by 4

96/4=4k/4

24=k

K=24

Our equation is c=24n

8 0
3 years ago
HELPPPPPP PLSSSSSSSSSS ASP HELPPPPPPPPPPPPPPPPPPPPPP
zhannawk [14.2K]

Answer:

if it helps please mark brainliest!

Step-by-step explanation:

essays= x

quizzes=y

2x=60

x=30 1 essay per 30 days

4y=40

y=10 1 quiz every 10 days

120/30=4

120/10= 12

so 4 essays and 12 quizzes in 120 days

if it helps please mark brainliest!

5 0
3 years ago
Solve by unfolding: a0=2, and, n>=1, a_n=7a_n-1.
vodomira [7]
We are told that the first term is 2.  The next term is 7(2) = 14; the third term is 7(14) = 98.  And so on.  So, the first term and the common ratio (7) are known.

The nth term of this geometric series is a_n = 2(7)^(n-1).

Check:  What is the first term?  We expect it is 2.  2(7)^(1-1) = 2(1) = 2.  Correct.

What is the third term?  We expect it is 98.  2(7)^(3-1) = 2(7)^2 = 98.  Right.<span />
6 0
3 years ago
The endpoints of a side of rectangle ABCD in the coordinate plane are at A(2, 7) and
FromTheMoon [43]

Answer:

ole los caracoles con mi amiga

Step-by-step explanation:

jeja

8 0
3 years ago
Find a polynomial P(x) with real coefficients having a degree 4, leading coefficient 4, and zeros 3-i and 4i.
Alisiya [41]

now, this polynomial has roots of 3-i and 4i, namely 3 - i and 0 + 4i.

let's bear in mind that a complex root never comes all by her lonesome, her sibling is always with her, the conjugate, so if 3 - i is there, 3 + i is also coming along, likewise if 0 + 4i is there, her sibling 0 - 4i is also there.

\bf \begin{cases} x=3-i\implies &x-3+i=0\\ x=3+i\implies &x-3-i=0\\ x=4i\implies &x-4i=0\\ x=-4i\implies &x+4i=0 \end{cases}\\\\[-0.35em] ~\dotfill\\\\ (x-3+i)(x-3-i)(x-4i)(x+4i)=\stackrel{y}{0} \\[2em] \underset{\textit{difference of squares}}{[(x-3)+i][(x-3)-i]}\underset{\textit{difference of squares}}{[x-4i][x+4i]}=0

\bf [(x-3)^2-i^2][x^2-(4i)^2]=y\implies [(x-3)^2-(-1)][x^2-(4^2i^2)]=0 \\[2em] [(x-3)^2-(-1)][x^2-(16(-1))]=0\implies [(x-3)^2+1][x^2+16]=0 \\[2em] [(x^2-6x+9)+1][x^2+16]=y\implies (x^2-6x+10)(x^2+16)=0 \\\\\\ x^4-6x^3+10x^2+16x^2-96x+160=0 \\\\\\ x^4-6x^3+26x^2-96x+160=0 \\\\\\ \stackrel{\textit{multiplying both sides by 4}}{4(x^4-6x^3+26x^2-96x+160)=4(0)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 4x^4-24x^3+104x^2-384x+640=y~\hfill

3 0
3 years ago
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