Subcategories

• Module 0 Day 10 Challenge Part 1

  Circumference, Pi, Diameter

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• Module 0 Day 10 Challenge Part 2

  Circumference, Pi, Diameter

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  Hello! I was confused by how to solve the mini-problem... b103dba4-0e78-4c0a-accd-7f98bc5e90eb-image.png

• Module 0 Day 10 Challenge Part 3

  Approximations for Pi

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  Hi @RZ923,

  You are totally right that the closest number that Prof. Loh can use to get the answer \(100\) times \(2\pi\) is \(628.\) Actually, \(2\pi\) is equal to \(2\times 3.14=6.28.\)

  However, Prof. Loh didn't choose the number \(710\) for the purpose of geting an answer that is close to \(100\) (but it's still a good bonus). He did it for a different reason: to illustrate the integers \(710\) and \(226,\) which we can use to get the number that will be very close to \(\pi:\) \(\frac{710}{226}=\frac{355}{113}={\color{darkred}3.14159292}\approx 3.141592654...= \pi.\)

  But you have a great observation here! It's awesome! So, keep it up 🙂

• Module 0 Day 10 Challenge Part 4

  Approximations for Pi

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  @fabulousgrizzly np! 🙂

• Module 0 Day 10 Your Turn Part 1

  Circumference of Circular Sections

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  @geniusraccoon Hmmm I think its because the three gears need to be in a triangle to move, because the gears wont work one vs one on a line, you will be missing the first and the third if u use a line

• Module 0 Day 10 Your Turn Part 2

  Angles of Circular Sections

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• Module 0 Day 10 Your Turn Part 3

  Change in Circumference Based on Change in Radius

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