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yan [13]
3 years ago
12

What is the correct domain and range in set notation?

Mathematics
2 answers:
Neporo4naja [7]3 years ago
8 0
Domain: [0 , infinity). This is also written {x : x is real, x >= 0}

Range: (-infinity , infinity). This is also written {y: y is real}
Svetlanka [38]3 years ago
5 0

DOMAIN: [0, +6)

RANGE: (-2.5, +2.5)

REMEMBER THAT THE DOMAIN IS SET ACCORDING TO THE GRAPH’S RELATION TO THE X-AXIS (-X, +X).

REMEMBER THAT THE RANGE IS ISET ACCORDING TO THE GRAPH’S REALTION TO THE Y-AXIS (-Y, +Y).

REMEMBER THAT THE BEST WAY TO APPRECIATE THIS AID TO YOUR LEARNING TO MARK “BRAINLIEST”.

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<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%20x%5E%7B2%7D-10x%2B25%7D%2B25%2B12%5Csqrt%7Bx%7D%20%3D15%5Csqrt%7Bx%7D" id="TexFor
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\sqrt{ x^{2}-10x+25}+25+12\sqrt{x} =15\sqrt{x} \\  \\  \therefore \: \sqrt{ x^{2}-10x+ {5}^{2} }+25 =15\sqrt{x} - 12\sqrt{x}\\  \\  \therefore \: \sqrt{( x - 5)^{2}}+25 =3\sqrt{x}  \\  \\ \therefore \:  x - 5+ 25 = 3 \sqrt{x}  \\  \\ \therefore \: x + 20 = 3 \sqrt{x}  \\  \\ squaring \: both \: sides \\ (x + 20)^{2}  = ( {3 \sqrt{x} })^{2}  \\ \therefore \:  {x}^{2}  + 40x + 400 = 9x \\ \therefore \:  {x}^{2}  + 40x + 400  - 9x = 0 \\  \therefore \:  {x}^{2}  + 31x + 400  = 0 \\

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