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mihalych1998 [28]
3 years ago
5

What is the closest perfect square greater than 54

Mathematics
1 answer:
Anna35 [415]3 years ago
6 0

Answer:

64

Step-by-step explanation:

Perfect squares are integers multiplied by themselves.

  • 2 times 2  = 4
  • 3 times 3  = 9
  • 4 times 4  = 15

The closest perfect squares to 54 are 49 (7^2) and 64 (8^2).

49 is less than 54, so that's ruled out.

Therefore, the closest perfect square to 54 that is greater than it is 64.

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Find the major axis for the ellipse <br> x² + 16y2-96y + 128 = 0
Softa [21]

The major axis for the ellipse, x² + 16y² - 96y + 128 = 0 is the x-axis

To answer the question, we need to write it in the standard form of the equation of an ellipse

<h3>Equation of an ellipse</h3>

The equation of an ellipse centered at  (h,k) is

(x - h)²/a² + (y - k)²/b² (1) where a > b and the major axis is parallel to the x axis

Given x² + 16y² - 96y + 128 = 0, we convert it into the standard equation of an ellipse.

So, x² + 16y² - 96y + 128 = 0

Dividing through by 16, we have

x²/16 + 16y²/16  - 96y/16 + 128/16 = 0/16

x²/16 + y² - 6y + 8 = 0

Completing the square in y by adding and subtracting (-6/2)² = (-3)²

x²/16 + y² - 6y + (-3)² - (-3)² + 8 = 0

x²/16 + (y - 3)² - 9 + 8 = 0

x²/16 + (y - 3)² - 1 = 0

x²/16 + (y - 3)² = 1

x²/4² + (y - 3)²/1² = 1  (2)

Comparing equations (1) and (2), we have that a = 4 and b = 1.

Since a = 4 > b = 1, the major axis for the ellipse is the x-axis

So, the major axis for the ellipse, x² + 16y² - 96y + 128 = 0 is the x-axis

Learn more about ellipse here:

brainly.com/question/26679189

#SPJ1

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