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Taya2010 [7]
3 years ago
13

Please help me find X

Mathematics
1 answer:
Zanzabum3 years ago
6 0

Answer:

  x = 2

Step-by-step explanation:

The product of distances from the intersection of secants to the near and far intersections with the circle are the same. For a tangent, the near and far points of intersection with the circle are the same. This relation tells us ...

  (2√3)(2√3) = x(x +4)

  12 = x² +4x

  16 = x² +4x +4 . . . . . add the square of half the x-coefficient to complete the square

  4² = (x +2)² . . . . . . . . write as squares

  4 = x +2 . . . . . . . . . . positive square root

  2 = x . . . . . . . . . . . . . subtract 2

_____

<em>Alternate solution</em>

If you believe x to be an integer, you can look for factors of 12 that differ by 4.

  12 = 1×12 = 2×6 = 3×4

The factors 2 and 6 differ by 4, so x=2 and x+4=6.

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(6x^{-2})^2(0.5x)^4\\\\=(6^2(x^{-2})^{2})(0.5^4x^4)\ \ \ \ \ \ \ \ \ \ \ \ \ as\ (ab)^m=a^mb^m\\\\=36\times 0.5^4((x^{-2})^2x^4)\ \ \ \ \ \ \ \ \ \ as\ multiplication\ is\ associative\ a(bc)=(ab)c\\\\=36\times 0.0625(x^{-4}x^4)\ \ \ \ \ \ \ \ \ \ \ as\ (x^m)^n=x^{mn}\\\\=(36\times 0.0625)(x^{-4+4})\ \ \ \ \ \ \ \ \ \ \ as\ x^mx^n=x^{m+n}\\\\=2.25x^0\\\\=2.25\ \ \ \ \ \ \ \ \ \ \ \ as\ x^0=1\\\\(6x^{-2})^2(0.5x)^4=2.25

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