1. y=√5x+3
2. x2+y2=9
3. y=√16-x2
please help me..thanks a lot :)
2. x2+y2=9
3. y=√16-x2
please help me..thanks a lot :)
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1) if it is:
y=√(5x+3)
5x + 3 ≥ 0
5x ≥ -3
x ≥ -3/5 => the domain
2) x^2 + y^2 = 9
circle radius = 3
the domain is: -3 ≤ x ≤ 3
3) y=√(16-x^2)
16 - x^2 ≥ 0
(4 - x)(4 + x) ≥ 0
-4 ≤ x ≤ 4 => the domain
y=√(5x+3)
5x + 3 ≥ 0
5x ≥ -3
x ≥ -3/5 => the domain
2) x^2 + y^2 = 9
circle radius = 3
the domain is: -3 ≤ x ≤ 3
3) y=√(16-x^2)
16 - x^2 ≥ 0
(4 - x)(4 + x) ≥ 0
-4 ≤ x ≤ 4 => the domain
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1. sqrt(5x + 3), for a sqrt to be defined, the inside function must be equal to or greater than zero.
5x + 3 >= 0, 5x => -3, so that x =>-3/5
domain [-3/5, infinity]
2. circle centred at origin with radius 3 has domain [-3, 3]
3. semicircle centred at origin with radius 4 has domain [-4, 4]
5x + 3 >= 0, 5x => -3, so that x =>-3/5
domain [-3/5, infinity]
2. circle centred at origin with radius 3 has domain [-3, 3]
3. semicircle centred at origin with radius 4 has domain [-4, 4]