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anonymous

  • one year ago

What is the value of x in the circle below

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  1. anonymous
    • one year ago
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  2. mathstudent55
    • one year ago
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    x is the radius of the circle, right?

  3. anonymous
    • one year ago
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    I think yes

  4. mathstudent55
    • one year ago
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    Yes, it is. Here is another radius. All radii are congruent. |dw:1438042883506:dw|

  5. mathstudent55
    • one year ago
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    can you find the upper x? |dw:1438042948351:dw|

  6. anonymous
    • one year ago
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    hmm i want to say 18?

  7. mathstudent55
    • one year ago
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    Notice we are dealing with a right triangle, and we can call upon our friend, Mr. Pythagoras, to help us.

  8. anonymous
    • one year ago
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    16^2 + 4^2 = x

  9. mathstudent55
    • one year ago
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    Use the Pythagorean theorem. x is the radius of the circle and is the hypotenuse of the triangle.

  10. anonymous
    • one year ago
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    Nice job at explaining @mathstudent55

  11. mathstudent55
    • one year ago
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    @omgitsjc Do you know how to use the Pythagorean theorem?

  12. anonymous
    • one year ago
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    no not really

  13. mathstudent55
    • one year ago
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    |dw:1438043205414:dw|

  14. mathstudent55
    • one year ago
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    In a right triangle, we call the two legs (the sides that form the right angle) legs a and b. The longest side, opposite the right angle, is the hypotenuse. We call the hypotenuse c.

  15. mathstudent55
    • one year ago
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    The Pythagorean theorem states that \(a^2 + b^2 = c^2\) is true for every right triangle.

  16. anonymous
    • one year ago
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    is it 16.49?

  17. mathstudent55
    • one year ago
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    Look at the last picture above. We have a = 4; b = 16; we need c \(4^2 + 16 ^2 = c^2\) \(c = \sqrt {272} \) \(c \approx 16.49\)

  18. mathstudent55
    • one year ago
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    Correct.

  19. mathstudent55
    • one year ago
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    Remember to round off to the nearest tenth.

  20. mathstudent55
    • one year ago
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    @Nixy Thanks.

  21. anonymous
    • one year ago
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    Thank you!

  22. mathstudent55
    • one year ago
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    You're welcome.

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