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3 years ago

What is the zeros of the quadratic function y=x^2+14x+24

Mathematics

1 answer:

777dan777 [17]3 years ago

6 0

Step-by-step explanation:

${x}^{2} + 14x + 24 = 0 \\ {x}^{2} + 12x + 2x + 24 = 0 \\ x(x + 12) + 2(x + 12) = 0 \\ (x + 12)(x + 2) = 0\\ x = - 12 \\ x = - 2$

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fomenos

Answer: A. When it comes after the decimal point, as in 4.000

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bazaltina [42]

We have the equation:

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By arranging this equation in terms of x and y, we have:

$x^2-8x+y^2-6y=-24 \\ \\$

By using the method of completing the square, we have:

$x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}$

The center of this circle is:

$(h,k)=(4,3)$

So the equation that fulfills the statement is:

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Finally, the right answer is c)

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What is the perpendicular distance from the line y= 2x-3 to the point (8,3)

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Find the perpendicular line then find the intersection then find the point

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2 times what=-1
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y=(-1/2)x+7

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D= $\sqrt{16+4}$
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