Ok this one is easier than you might think. The line equals 180 degrees since it is a straight line, therefore you need to do 180-27 to get what y equals. Y=153

6 0

3 years ago

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Determine the zeroes of the polynomial

Anna71 [15]

Answer:

3,7/6

Step-by-step explanation:

$(\sqrt{x^2-4x+3} )+(\sqrt{x^{2} -9} )-(\sqrt{4x^2-14x+6} )\\=(\sqrt{x^2-x-3x+3} )+(\sqrt{(x^2-3^2})-(\sqrt{4x^2-2x-12x+6})\\ =(\sqrt{x(x-1)-3(x-1)} )+\sqrt{(x+3)(x-3)}-\sqrt{2x(2x-1)-6(2x-1)} \\=\sqrt{(x-1)(x-3)}+\sqrt{(x+3)(x-3)} -\sqrt{2(2x-1)(x-3)} \\=\sqrt{x-3} (\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} )\\$

$\sqrt{x-3} =0~gives~x=3\\or~\sqrt{x-1} +\sqrt{x+3} -\sqrt{2(2x-1)} =0\\or~ \sqrt{x-1} +\sqrt{x+3} =\sqrt{2(2x-1)} \\squaring\\x-1+x+3+2\sqrt{x-1} \sqrt{x+3} =2(2x-1)\\2x+2+2\sqrt{(x-1)(x+3)} =4x-2\\2\sqrt{x^2-x+3x-3} =2x-4\\\sqrt{x^2+2x-3} =x-2\\again ~squaring\\x^2+2x-3=x^2-4x+4\\\\2x+4x=4+3\\6x=7\\x=\frac{7}{6}$

8 0

3 years ago

3. Determine whether the set of numbers can be the measures of the sides of a triangle. If so, classify the triangle as acute, o

julia-pushkina [17]

Answer:

all of them are acute

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3 0

3 years ago

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