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3 years ago

Given x^2 + y^2 = 16, findname:center:Domain:Range:x-inty-int

Mathematics

2 answers:

Dennis_Churaev [7]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

The equation of a circle centred at the origin is

x² + y² = r² ( r is the radius )

x² + y² = 16 ← is the equation of a circle

centre (0, 0 ) and r = $\sqrt{16}$ = 4

The circle therefore intercepts the x- axis at (4, 0 ) and (4, 0 )

and intercepts the y- axis at (0, - 4 ) and (0, 4 )

Domain is - 4 ≤ x ≤ 4

range is - 4 ≤ y ≤ 4

x- intercepts are x = - 4, x = 4

y- intercepts are - 4, 4

rodikova [14]3 years ago

3 0

Answer:

The equation of a circle centred at the origin is

x² + y² = r² ( r is the radius )

x² + y² = 16 ← is the equation of a circle

centre (0, 0 ) and r = \sqrt{16}16 = 4

The circle therefore intercepts the x- axis at (4, 0 ) and (4, 0 )

and intercepts the y- axis at (0, - 4 ) and (0, 4 )

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Given x^2 + y^2 = 16, findname:center:Domain:Range:x-inty-int​

Given x^2 + y^2 = 16, findname:center:Domain:Range:x-inty-int