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icang [17]
4 years ago
8

Which relation is a function? {(1, 3), (2, 6), (3, 9), (4, 12)} {(3, 4), (3, 3), (1, 2), (1, 5)} {(1, 4), (1, 3), (1, 2), (1, 1)

} {(3, 4), (2, 4), (5, 1), (3, 5)}
Mathematics
1 answer:
leva [86]4 years ago
5 0

Answer:

{(1, 3), (2, 6), (3, 9), (4, 12)}

Step-by-step explanation:

The x values do not repeat, therefore this is a function.

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Nikitich [7]

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elena-14-01-66 [18.8K]

The positive solution to the quadratic equation 2 x^{2}+3 x-8=0 is x = 1.39

<h3><u>Solution:</u></h3>

Given quadratic equation is 2 x^{2}+3 x-8=0

<em><u>The general quadratic equation is of form:</u></em>

a x^{2}+b x+c=0

Now comparing the general equation with the given equation we get

a = 2 , b = 3 and c = -8

<em><u>The formula to determine roots of the quadratic equation is:</u></em>

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

On plugging in vlaues, we get

x=\frac{-3 \pm \sqrt{3^{2}-4 \times 2 \times(-8)}}{2 \times 2}

\mathrm{x}=\frac{-3 \pm \sqrt{9-(-64)}}{4}

On solving we get,

\begin{array}{l}{\mathrm{x}=\frac{-3 \pm \sqrt{9+64}}{4}} \\\\ {\mathrm{x}=\frac{-3 \pm \sqrt{73}}{4}}\end{array}

\mathrm{x}=\frac{-3+\sqrt{73}}{4} \text { OR } \mathrm{x}=\frac{-3-\sqrt{73}}{4}

x = 1.39  OR  x = -2.89  

Hence , the positive solution to the quadratic equation  is x = 1.39

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4 years ago
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