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Ne4ueva [31]
3 years ago
15

Complete the equation for the circle with center (6, 2) and radius 8.

Mathematics
1 answer:
zzz [600]3 years ago
5 0

Can you give more info?

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Solve the equation. −3(y+5)=15
e-lub [12.9K]

Divide both sides by -3

y + 5 = -15/3

Simplify 15/3 to 5

y + 5 = -5

Subtract 5 from both sides

y = -5 - 5

Simplify -5 - 5 to -10

y = -10

8 0
3 years ago
Read 2 more answers
Joyce is helping to make wreaths for her Women’s Club to sell at a local bazaar. She will be making the bows for each wreath. Sh
jok3333 [9.3K]

Answer:

She can make 34 bows with 10 feet of ribbon left

Step-by-step explanation:

1.5x34=51

61-51=10 feet of ribbons left

51/1.5=x

4 0
2 years ago
How many cubic inches? Please help!?!
iragen [17]

1575 cubic inches

This is the answer to your question.

7 0
3 years ago
Suppose that a password for a computer system must have at least 8, but no more than 12, characters, where each character in the
uranmaximum [27]

Part a)

There are 52 letters (26 lowercase and 26 uppercase), 10 digits, and 6 symbols. There are 52+10+6 = 68 different characters to choose from.

  • If there are 8 characters for this password, then we have 68^8 = 4.5716 * 10^14 different passwords possible.
  • If there are 9 characters, then we have 68^9 = 3.1087 * 10^16 different passwords
  • If there are 10 characters, then we have 68^10 = 2.1139 * 10^18 different passwords
  • If there are 11 characters, then we have 68^11 = 1.4375 * 10^20 different passwords
  • If there are 12 characters, then we have 68^12 = 9.7748 * 10^21 different passwords

Adding up those subtotals gives

68^8+68^9+68^10+68^11+68^12 = 9.9207 * 10^21

different passwords possible.

<h3>Answer: Approximately 9.9207 * 10^21 </h3>

======================================================

Part b)

Let's find the number of passwords where we don't have a special symbol

There are 52+10 = 62 different characters to pick from

  • If there are 8 characters for this password, then we have 62^8 = 2.1834 * 10^14 different passwords possible.
  • If there are 9 characters, then we have 62^9 = 1.3537 * 10^16 different passwords
  • If there are 10 characters, then we have 62^10 = 8.3930 * 10^17 different passwords
  • If there are 11 characters, then we have 62^11 = 5.2037 * 10^19 different passwords
  • If there are 12 characters, then we have 62^12 = 3.2263 * 10^21 different passwords

Adding those subtotals gives

62^8+62^9+62^10+62^11+62^12 = 3.2792 * 10^21

different passwords where we do not have a special character. Subtract this from the answer in part a) above

( 9.9207 * 10^21)  - (3.2792 * 10^21) = 6.6415 * 10^21

which represents the number of passwords where we have one or more character that is a special symbol. I'm using the idea that we either have a password with no symbols, or we have a password with at least one symbol. Adding up those two cases leads to the total number of passwords possible.

<h3>Answer: Approximately 6.6415 * 10^21</h3>

======================================================

Part c)

The answer from part a) was roughly 9.9207 * 10^21

It will take about 9.9207 * 10^21  nanoseconds to try every possible password from part a).

Divide 9.9207 * 10^21  over 1*10^9 to convert to seconds

(9.9207 * 10^21 )/(1*10^9) = 9,920,700,000,000

This number is 9.9 trillion roughly.

It will take about 9.9 trillion seconds to try every password, if you try a password per second.

------

To convert to hours, divide by 3600 and you should get

(9,920,700,000,000)/3600 = 2,755,750,000

So it will take about 2,755,750,000 hours to try all the passwords.

------

Divide by 24 to convert to days

(2,755,750,000)/24= 114,822,916.666667

which rounds to 114,822,917

So it will take roughly 114,822,917 days to try all the passwords.

------

Then divide that over 365 to convert to years

314,583.334246576

which rounds to 314,583

It will take roughly 314,583 years to try all the passwords

------------------------------

<h3>Answers:</h3>
  • 9.9 trillion seconds
  • 2,755,750,000 hours
  • 114,822,917 days
  • 314,583 years

All values are approximate, and are roughly equivalent to one another.

5 0
3 years ago
Two number cubes each have sides that are labeled 1 to 6. Jude rolls the 2 number cubes. What is the probability that the sum of
AleksAgata [21]

Answer: 3

This is the answer if you think its wrong i don't know its correct i got it right

3 0
3 years ago
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