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Talja [164]
3 years ago
7

What is the smallest degree of rotation that will map a regular 18-gon onto itself?

Mathematics
2 answers:
Korolek [52]3 years ago
6 0
Because the 18-gon is symmetrical, you can rotate it by 360/18 or 20 degrees. This is because when you rotate it by 20 degrees(to the left for explanation matters), any angle will now be in the position of its angle to its left.
neonofarm [45]3 years ago
3 0

Answer:

The answer is 20

Step-by-step explanation:

360 will be divided by 18 which will give you 20. For those of you who had the same question but instead of "18-gon" you had "20-gon" if you divide 360 by 20 your answer will be 18.

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Help !!!!!!!!!!!!!!!!!!!!!!!!!!
Blizzard [7]

Step-by-step explanation:

so 5/9(86-32) then you take

86-32=(54)

now it look like C=5/9=(54)

C=30

so c=30 is the answer :)

4 0
3 years ago
What is 5.316 - 1.942 (show ur work)
Fiesta28 [93]

Answer:

3.374

Step-by-step explanation:

\mathrm{Write\:the\:numbers\:one\:under\:the\:other,\:line\:up\:the\:decimal\:points.}

\mathrm{Add\:trailing\:zeroes\:so\:the\:numbers\:have\:the\:same\:length.}

\begin{matrix}\:\:&5&.&3&1&6\\ -&1&.&9&4&2\end{matrix}

\mathrm{Subtract\:each\:column\:of\:digits,\:starting\:from\:the\:right\:and\:working\:left}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:6-2=4

\frac{\begin{matrix}\:\:&5&.&3&1&\textbf{6}\\ -&1&.&9&4&\textbf{2}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\textbf{4}\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}

\frac{\begin{matrix}\:\:&5&.&3&\textbf{1}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\mathrm{The\:bottom\:number\:is\:larger\:than\:the\:upper\:number.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}

\mathrm{The\:top\:digit\:is\:not\:bigger\:than\:the\:bottom\:one.\:\:Try\:to\:'borrow'\:a\:digit\:from\:the\:left.}

\frac{\begin{matrix}\:\:&5&.&\textbf{3}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Borrow\:}1\mathrm{\:from\:}5\mathrm{.\:\:The\:remainder\:is\:}4

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&10&\:\:&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&3&1&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{\:\:}&\:\:&\:\:&\:\:&4\end{matrix}}

\mathrm{Add\:}1\mathrm{\:ten\:to\:}3:\quad \:10+3=13

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{13}&\:\:&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{3}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Borrow\:}1\mathrm{\:from\:}13\mathrm{.\:\:The\:remainder\:is\:}12

\frac{\begin{matrix}\:\:&4&\:\:&\textbf{12}&10&\:\:\\ \:\:&\linethrough{5}&.&\textbf{\linethrough{13}}&1&6\\ -&1&.&\textbf{9}&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\textbf{\:\:}&\:\:&4\end{matrix}}

\mathrm{Add\:}1\mathrm{\:ten\:to\:}1:\quad \:10+1=11

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{\:\:}&4\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:11-4=7

\frac{\begin{matrix}\:\:&4&\:\:&12&\textbf{11}&\:\:\\ \:\:&\linethrough{5}&.&\linethrough{13}&\textbf{\linethrough{1}}&6\\ -&1&.&9&\textbf{4}&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\textbf{7}&4\end{matrix}}

\mathrm{Place\:the\:decimal\:point\:in\:the\:answer\:directly\:below\:the\:decimal\:points\:in\:the\:terms}

\frac{\begin{matrix}\:\:&4&\textbf{\:\:}&12&11&\:\:\\ \:\:&\linethrough{5}&\textbf{.}&\linethrough{13}&\linethrough{1}&6\\ -&1&\textbf{.}&9&4&2\end{matrix}}{\begin{matrix}\:\:&\:\:&\textbf{.}&3&7&4\end{matrix}}

\mathrm{In\:the\:bolded\:column,\:subtract\:the\:second\:digit\:from\:the\:first}:\quad \:4-1=3

\frac{\begin{matrix}\:\:&\textbf{4}&\:\:&12&11&\:\:\\ \:\:&\textbf{\linethrough{5}}&.&\linethrough{13}&\linethrough{1}&6\\ -&\textbf{1}&.&9&4&2\end{matrix}}{\begin{matrix}\:\:&\textbf{3}&.&3&7&4\end{matrix}}

=3.374

Hence the correct answer is 3.374

7 0
2 years ago
5√-54 in simplest radical form
olga nikolaevna [1]

The answer is 15i radical 6

3 0
3 years ago
Read 2 more answers
According to scientific research, the probability that a person between the ages of 20 and 25 having high blood pressure, is onl
boyakko [2]

Answer:

<h2>The probability is {\frac{83}{100} }^9 \times \frac{17}{100} + {\frac{83}{100} }^8 \times {\frac{17}{100} }^2 + {\frac{83}{100} }^{10}.</h2>

Step-by-step explanation:

The probability of having high blood pressure is 17%.

We need to find the probability of 2 or fewer have high blood pressure.

Possibility 1: No one have high blood pressure.

The probability is {\frac{100 -17}{100} }^{10} = {\frac{83}{100} }^{10}.

Possibility 2: Only one have high blood pressure.

The probability is {\frac{83}{100} }^9 \times \frac{17}{100}.

Possibility 3: 2 among the 10 have the high blood pressure.

{\frac{83}{100} }^8 \times {\frac{17}{100} }^2.

Any of the possibilities can happen.

Hence, the required probability is {\frac{83}{100} }^9 \times \frac{17}{100} + {\frac{83}{100} }^8 \times {\frac{17}{100} }^2 + {\frac{83}{100} }^{10}.

4 0
3 years ago
Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work
valina [46]

Answer:

Step-by-step explanation:

Question

Find the perimeter of a triangle with vertices A(2,5) B(2,-2) C(5,-2). Round your answer to the nearest tenth and show your work.​

perimeter of a triangle = AB+AC+BC

Using the distance formula

AB = sqrt(-2-5)²+(2-2)²

AB = sqrt(-7)²

AB =sqrt(49)

AB =7

BC = sqrt(-2+2)²+(2-5)²

BC = sqrt(0+3²)

BC =sqrt(9)

BC =3

AC= sqrt(-2-5)²+(2-5)²

AC= sqrt(-7)²+3²

AC =sqrt(49+9)

AC =sqrt58

Perimeter = 10+sqrt58

5 0
3 years ago
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