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lesya692 [45]
3 years ago
12

Sam sent an Email to 3 friends on Monday . Each of the friends then sent an Email to 3 friends on Tuesday . On Wednesday , each

or those friends then sent an Email to 3 friends . Write the prime factorization of the numbers of Emails that were sent on Wednesday
Mathematics
2 answers:
vodomira [7]3 years ago
6 0

Answer:

There were 3 x 3 x 3 emails sent on wednesday. (Or 3³)

Step-by-step explanation:

Sam sent an email to 3 friends on monday, then on tuesday each of this friends sent an email to 3 friends. We can see that:

  • friend 1: sends 3 mails on tuesday
  • friend 2: sends 3 mails on tuesday
  • friend 3: sends 3 mails on wednesday.

Therefore, on tuesday, 9 mails were sent to 9 different friends.

On wednesday, each of these 9 different friends sent an email to other 3 friends:

If 9 people send an email to other 3 friends, there will be 27 emails sent on wednesday.

The problem asks us to give the prime factorization of these 27 mails, so writing 27 in its prime factorization we get 27 = 3 x 3 x 3 = 3³.

aleksandrvk [35]3 years ago
5 0
3 + 3 + 3 = 9
Prime Factorization.

3 x 3
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A cheetah runs 300 feet in 2.29 seconds. How would you convert the cheetahs speed to miles per hour?
xeze [42]

plain and short, an hour has 60 minutes, and a minute has 60 seconds, therefore an hour has 60*60 or 3600 seconds, and we can express that  rate as hr/3600s or the other way around 3600s/hr, depending on which version we need.

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\bf \cfrac{300 ft}{2.29 s}\cdot \cfrac{3600 s}{hr}\cdot \cfrac{mi}{5280 ft}~\hspace{10em}\stackrel{\textit{using minutes as well}}{\cfrac{300 ft}{2.29 s}\cdot \cfrac{60 min}{hr}\cdot \cfrac{60 s}{min}\cdot \cfrac{mi}{5280 ft}}

6 0
3 years ago
I need help with 12 13 and 14​
nikdorinn [45]

Answer: Lines \frac{}{BC} and \frac{}{EF} are different lengths.

Step-by-step explanation:

The distance formula is \sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }, and you can use this formula to solve for the lengths of both lines \frac{}{BC} and \frac{}{EF}.

For line \frac{}{BC}, let x_{1} = the x at point B, or 1, and let x_{2} = the x at point C, or 2.

Now, let y_{1} = the y at point B, or 4, and let y_{2} = the y at point C, or -1.

Now, solve the formula to find the length \frac{}{BC} = \sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\.

\frac{}{BC} = \sqrt{(1-2)^{2} +(4-(-1))^2

\frac{}{BC} = \sqrt{(-1)^{2} +(4+1)^2

\frac{}{BC} = \sqrt{1 +5^2

\frac{}{BC} = \sqrt{(1+25)

\frac{}{BC} = \sqrt{26} \\

Now, for line \frac{}{EF}, let x_{1} = the x at point E, or -4, and let x_{2} = the x at point F, or -1.

Let y_{1} = the y at point E, or -3, and let y_{2} = the y at point F, or 1.

Now, solve the formula to find the length \frac{}{EF} = \sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\.

\frac{}{EF} = \sqrt{(-4-(-1))^{2} +(-3-1)^2

\frac{}{EF} = \sqrt{(-4+1)^{2} +(-4)^2

\frac{}{EF} = \sqrt{(-3)^2+16

\frac{}{EF} = \sqrt{(9+16)

\frac{}{EF} = \sqrt{25}

\frac{}{EF} = 5

Now, look back at \frac{}{BC}. The two lines have different lengths, so you have now justified the fact that they are not the same.

Questions 13 and 14 would be solved in much the same way- but please let me know if you want me to show the work for those as well!

7 0
3 years ago
N(A)=10<br>n(B) = 12<br>n(AUB) = ?​
earnstyle [38]

Answer:

n(AUB) = 33

Step-by-step explanation:

U is the 21'st letter of the alphabet so

9 + 1 + 2 + 21 =33

n(AUB) = 33

3 0
3 years ago
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