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3 years ago

Secants BE and CF intersect at point D inside A. What is the measure of CDE?

Mathematics

1 answer:

Aleks04 [339]3 years ago

6 0

MCE = 360 - (150 + 70 + 50)
mCE = 360 - 270
mCE = 90

<CDE = 1/2(mBE + mCE)
<CDE = 1/2(150 + 90)
<CDE = 1/2(240)
<CDE = 120

answer
<CDE = 120°

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Rounded to the nearest hundredth , what is the positive solution to the quadratic equation 0=2x^2+3x-8

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>Solution:</h3>

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

The general quadratic equation is of form:

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

Now comparing the general equation with the given equation we get

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

The formula to determine roots of the quadratic equation is:

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

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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