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the ceiling of katie’s living room is a square that is 12 ft long on each side. to decorate for a party, she plans to hang crepe paper around the perimeter of the ceiling and then from each corner to the opposite corner. katie can buy rolls that each contain 10 ft of crepe paper. what is the minimum number of rolls she should buy? show your work.

The minimum number of rolls she should buy is 9 rolls

Step-by-step explanation:

* Lets study the information in the problem

- Katie wants to put the crepe paper around the perimeter of the

 ceiling which shaped a square of side length 12 feet

- And also from each corner to the opposite corner

- She needs the length of the 4 sides of the square and the length

 of its 2 diagonals

* Lets find the length of the diagonal of the square

∵ The two adjacent sides of the square formed two legs of a right

   angle triangle and the diagonal joining the endpoints of the legs

  is the hypotenuse of the triangle

- Use Pythagoras theorem to find the length of the diagonal

∴ The length of the diagonal = √(s² + s²) √(2s²) = s√2

∵ The length of the side of the square = 12 feet

∴ The length of the diagonal = 12√2

* Now lets find the length of the crepe papers she needs

∵ She needs the length of the 4 sides of the square and the length

  of its 2 diagonals

∴ The length of crepe papers = 12 + 12 + 12 + 12 + 12√2 + 12√2 = 81.94 feet

∵ Each roll of the crepe papers contain 10 feet

- To find the number of rolls divide the length of the crepe papers by 10

∴ The number of rolls = 81.94 ÷ 10 = 8.194

* She must to buy 9 rolls to have enough crepe papers to decorate

 her ceiling

* V.I.N:

- If she decide to buy 8 rolls, some part of ceiling will not decorate

 because the 8 rolls have 80 feet only and she needs 81.94 feet

it costs $20 to interview each person for a survey. find the cost to obtain a +- 3% margin of error.

step-by-step explanation:

copy and paste, put it in google

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: