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slavikrds [6]
3 years ago
12

Find the eccentricity of the ellipse 4x^2 +16y^2 +32x-32y+16=0

Mathematics
1 answer:
Neko [114]3 years ago
7 0
       4x² + 16y² + 32x - 32y + 16 = 0
     <u>                                       - 16  - 16</u>
               4x² + 16y² + 32x - 32y = -16
               4x² + 32x + 16y² - 32y = -16
4(x² + 8x + 16) + 16(y² - 2y + 1) = -16 + 4(16) + 16(1)
                  4(x + 4)² + 16(y - 1)² = -16 + 64 + 16
                  4(x + 4)² + 16(y - 1)² = 48 + 16
                  <u>4(x + 4)²</u> + <u>16(y - 1)²</u> = <u>64</u>
                      64              64          64
                        <u>(x + 4)²</u> + <u>(y - 1)²</u> = 1
                           16            2
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<h3>How to determine the value x?</h3>

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Read more about congruent triangles at:

brainly.com/question/12413243

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<u>Complete question</u>

Two triangles, abc and cde, share a common vertex c on a grid. in triangle abc, side ab is 4x - 1, side bc is 4, side ac is 5. in triangle cde, side cd is 5, side de is x + 2, side ce is 4. If Δabc ≅ Δdec, what is the value of x? a. x = 8 b. x = 5 c. x = 4 d. x = 1 e. x = 2

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1 year ago
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