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anonymous

  • one year ago

Use the Pythagorean Theorem to find the length of the missing side of a right triangle whose legs equal √2 and 2√15. a right triangle with legs as 2 square root of 15 and square root of 2. A. 64 B. 4 + 2√15 C. 8 D. None of the above

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  1. mathstudent55
    • one year ago
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    The Pythagorean Theorem is \(a^2 + b^2 = c^2\) where a and b are the lengths of the legs, and c is the length of the hypotenuse.

  2. anonymous
    • one year ago
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    ok....

  3. mathstudent55
    • one year ago
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    Replace a and b with the given lengths of te legs and find c, the length of the hypotenuse.

  4. mathstudent55
    • one year ago
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    |dw:1436938653653:dw|

  5. mathstudent55
    • one year ago
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    Do you see how the given leg lengths are substituted into a and b?

  6. anonymous
    • one year ago
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    yes.

  7. mathstudent55
    • one year ago
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    Good. Now you need to square the two terms on the left side. When you square a product, you square every factor of the product. |dw:1436939091163:dw|

  8. mathstudent55
    • one year ago
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    |dw:1436939186380:dw|

  9. nincompoop
    • one year ago
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    it is good to be able to memorize the pythagorean theorem, but it will be better if you learned the concept and its geometrical underpinnings so you may be able to derive any solution to any problem that relates to it.

  10. nincompoop
    • one year ago
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    |dw:1436940135801:dw|

  11. nincompoop
    • one year ago
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    |dw:1436940404114:dw|

  12. nincompoop
    • one year ago
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    ** just a slight change in variables from what mathstudent55 provided you ** The Pythagorean theorem states that the longest leg or side namely hypotenuse, r, is the square root of the squared sum of the adjacent side, x, and opposite side, y. \(\huge r = \sqrt{x^2+y^2} \) The reason I use r, x and y, is because these notations carry over into analytical geometry and physics. And if you haven't already noticed, if I squared both sides of the equation it look the same as what you were provided. \(\large r^{\color{red}{2}} = (\sqrt{x^2+y^2} )\color{red}{^2} \rightarrow r^2 =x^2 + y^2\)

  13. nincompoop
    • one year ago
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    |dw:1436941361148:dw|

  14. nincompoop
    • one year ago
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    and to just give you an idea as to the application that I am talking about http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html x's and y's never go away for some reason so we might as well accept and become familiar with them

  15. nincompoop
    • one year ago
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    http://imgur.com/a/VTMUq#10

  16. nincompoop
    • one year ago
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    http://imgur.com/a/VTMUq#10

  17. nincompoop
    • one year ago
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    why circle, pi and quadrants are important http://www.analyzemath.com/unitcircle/unit_circle_applet.html

  18. anonymous
    • one year ago
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    @nincompoop you've lost me

  19. nincompoop
    • one year ago
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    do you read?

  20. nincompoop
    • one year ago
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    Math is not a passive activity like what you did with mathstudent55. It requires you to actively seek out to understand patterns from simple to complex and what i just did is drew you what pythagorean theorem in its very backbone then added a few fundamental trigonometry information (and vocabulary) as you will encounter them very soon. The x, y and r are just variables and they are arbitrary chosen so they can be replaced with a, b and c respectively. As I explained, I picked those variables because they keep reappearing in other areas of mathematical and physical studies, which you will be encountering a lot. As an example, the variable r, x and y are used in a circle's standard form. \(\Longrightarrow (x-h)^2 + (y-k)^2 = r^2 \) http://www.mathwarehouse.com/geometry/circle/equation-of-a-circle.php Should you ask why would you care about a circle when you're dealing with a triangle. The idea is how one concept can carry over into another as you may have already noticed that the r in a circle radius may be used as the hypotenuse of a right triangle.

  21. anonymous
    • one year ago
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    i gave up. i don't get this at all

  22. nincompoop
    • one year ago
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    what part don't you get? we can begin to what you know about math

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