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Juliette [100K]
3 years ago
11

What is the rotational symmetry ??? 2, 6, 12 or infinite?

Mathematics
2 answers:
igor_vitrenko [27]3 years ago
4 0
It has an infinite rotation symmetry because no matter how you rotate the object, it look the same.
Umnica [9.8K]3 years ago
3 0

Answer: C) 12

Step-by-step explanation:

Got this right on USA test prep

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Which expression is not equivalent to the other expressions?
nexus9112 [7]
It’s -4(3x+6) because if you solve it you get a different answer then all of them all over them were the same answer except -4(3x-6)
7 0
3 years ago
One leg of a right triangle is 7 units long, and its hypotenuse is 16 units long. What is the length of the other leg? Round to
NemiM [27]

Answer:

7^{2} + x^{2} = 16^{2}

49+x^2=256

x^2=207

x=14.387=14

4 0
11 months ago
In how many ways can 5 question on a true-false test be answered?
slavikrds [6]
I belive it's 32. You have to raise 2 to the 5th power
7 0
3 years ago
The small copier makes 438 copies each day. The large copier makes 4 times as many copies each day. How many copies do both copi
atroni [7]

Answer:1752

Step-by-step explanation: So what you need to do is to multiply the number of copies that the small copier makes with the 4 because it says that it makes it 4 times what the smaller copier does.

5 0
3 years ago
If this particular arithmetic sequence, a7=34 and a15=74. What is the value of a21
Ilya [14]
This is an arithmetic sequence, meaning, to get the next term, you simply add some value to the current one, namely the "common difference" "d".

now, we know the 17th term is 34, let's go to the 15term then

34 + d
34 + d + d
34 + d + d + d
34 + d + d + d + d
34 + d + d + d + d + d
34 + d + d + d + d + d + d
34 + d + d + d + d + d + d + d
34 + d + d + d + d + d + d + d + d

now, notice, we got to the 15th term, which is 34  + d + d + d + d + d + d + d + d, or namely 34 + 8d, it just so happen we know is 74, therefore,

\bf 34+8d=74\implies 8d=40\implies d=\cfrac{40}{8}\implies \boxed{d=5}

ok, now that we know what the common difference is, let's find the first term in the sequence using the 7th term of 34,

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=7\\
d=5\\
a_7=34
\end{cases}
\\\\\\
a_7=a_1+(7-1)5\implies 34=a_1+(7-1)5
\\\\\\
34=a_1+30\implies \boxed{4=a_1}

and now let's use those 2 found folks, to get the 21st term,

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=4\\
d=5\\
n=21
\end{cases}
\\\\\\
a_{21}=4+(21-1)5\implies a_{21}=4+100\implies a_{21}=104
3 0
2 years ago
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