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Reptile [31]
1 year ago
9

A line has a slope of -2/5 and passes through the point (10,7) . Write the equation of this line in point-slope form, slope-inte

rcept form, and standard form.
i will give brainliest if you help
Mathematics
1 answer:
qaws [65]1 year ago
6 0

(\stackrel{x_1}{10}~,~\stackrel{y_1}{7})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{7}=\stackrel{m}{-\cfrac{2}{5}}(x-\stackrel{x_1}{10})

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SIZIF [17.4K]

Angle <QAB is =15° because the opposite angles of an isosceles triangle are equal.

The length of the straight line AB = 80cm

<h3>Calculation of angle of a triangle</h3>

The angle at a point = 360°

Angle AQB= 360 - 210° = 150

But the angle that makes up a triangle= 180°

180-150= 30°

But <QAB = <QBA because triangle AQB is an isosceles triangle.

30/2 = 15°

To calculate the length of the straight line the following is carried out using the sine laws.

a/ sina, = b sinb

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8/sin15= b/sin 150

b= 8 × sin 150/sin 15

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Learn more about isosceles triangle here:

brainly.com/question/25812711

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