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Dafna1 [17]
3 years ago
5

Use calculus to find the area a of the triangle with the given vertices. (0, 0), (6, 2), (4, 8)

Mathematics
1 answer:
Vilka [71]3 years ago
7 0

The area of the triangle with vertices  (x_1,y_1),(x_2,y_2),(x_3,y_3) is

\frac{1}{2}\left|\begin{array}{ccc}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3\end{array}\right|.

Inserting numerical values, the are of the triangle with vertices (0, 0), (6, 2), (4, 8) is

\Delta = \frac{1}{2}\left|\begin{array}{ccc}1&0&0\\1&6&2\\1&4&8\end{array}\right |\\
\Delta = 0.5(48-8)\\
\Delta = 20 \;square\; units


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Answer:

x = 1, -5.56

Step-by-step explanation:

x^2 = -7x - 8

shift -7x and -8 to the other side . Remember when u shift minus changes into plus.

x^2 + 7x + 8 = 0

using quadratic equation formula

in quadratic equation one value comes positive and other comes in negative

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taking positive sign

x = (-b + \sqrt{b^2 - 4*a*c}) /2*a

x = (-7 + \sqrt{7^2 - 4*1*8} ) /2*1

x = (-7 + \sqrt{49 + 32} ) /2

x = (-7 + \sqrt{81} )/ 2

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taking negative sign

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x = (-7 - \sqrt{49 - 32} ) /2

x = -7 - \sqrt{17} / 2

x = -7 - 4.12 / 2

x = -11.12/2

x = -5.56

therefore x = 1 , - 5.56

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