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const2013 [10]
4 years ago
10

What is the y-intercept of this equation 5x-6y=30

Mathematics
1 answer:
Finger [1]4 years ago
7 0
Turn the equation into linear equation which is y=-5/6x+5.so the y-intercept is 5
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<h3><em>=</em><em> </em><em>3</em><em> </em><em>×</em><em> </em><em>(</em><em>5</em><em> </em><em>-</em><em> </em><em>1</em><em>)</em><em> </em><em>+</em><em> </em><em>1</em><em>3</em></h3><h3><em>=</em><em> </em><em>3</em><em> </em><em>×</em><em> </em><em>4</em><em> </em><em>+</em><em> </em><em>1</em><em>3</em></h3><h3><em>=</em><em> </em><em>1</em><em>2</em><em> </em><em>+</em><em> </em><em>1</em><em>3</em></h3><h3><em>=</em><em> </em><em>2</em><em>5</em></h3>

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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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