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hram777 [196]
3 years ago
9

Find the equation of the line in slope-intercept form. e through (−5, 6) with slope = 3

Mathematics
1 answer:
love history [14]3 years ago
5 0

Answer:

Step-by-step explanation:

y - 6 = 3(x + 5)

y - 6 = 3x + 15

y = 3x + 21

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7 0
2 years ago
XY is a diameter of a circle and Z is a point on the circle such that ZY=6. If the area of the triangle XYZ is 18 square root 3
nataly862011 [7]
<h2>Answer:</h2>

4π

<h2>Step-by-step explanation:</h2>

As shown in the diagram, triangle XYZ is a right triangle. Therefore, its area (A) is given by:

A = \frac{1}{2} x b x h      -------------(i)

Where;

A = 18\sqrt{3}

b = XZ = base of the triangle

h = YZ = height of the triangle = 6

<em>Substitute these values into equation(i) and solve as follows:</em>

18\sqrt{3} =  \frac{1}{2} x b x 6

18\sqrt{3} =  3b

<em>Divide through by 3</em>

6\sqrt{3} =  b

Therefore, b = XZ = 6\sqrt{3}

<em>Now, assume that the circle is centered at O;</em>

Triangle XOZ is isosceles, therefore the following are true;

(i) |OZ| = |OX|

(ii) XZO = ZXO = 30°

(iii) XOZ + XZO + ZXO = 180°   [sum of angles in a triangle]

=>  XOZ + 30° + 30° = 180°

=>  XOZ + 60° = 180°

=>  XOZ = 180° - 60°

=>  XOZ = 120°

Therefore we can calculate the radius |OZ| of the circle using sine rule as follows;

\frac{sin|XOZ|}{XZ} = \frac{sin|ZXO|}{OZ}

\frac{sin120}{6\sqrt{3} } = \frac{sin 30}{OZ}

\frac{\sqrt{3} /2}{6\sqrt{3} } = \frac{1/2}{|OZ|}

\frac{1}{12}  = \frac{1}{2|OZ|}

\frac{1}{6} = \frac{1}{|OZ|}

|OZ| = 6

The radius of the circle is therefore 6.

<em>Now, let's calculate the length of the arc XZ</em>

The length(L) of an arc is given by;

L = θ / 360 x 2 π r          ------------------(ii)

Where;

θ = angle subtended by the arc at the center.

r = radius of the circle.

In our case,

θ = ZOX = 120°

r = |OZ| = 6

Substitute these values into equation (ii) as follows;

L = 120/360 x 2π x 6

L = 4π

Therefore the length of the arc XZ is 4π

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