5. Choose the equation that has solutions (5,7) and (8,13).
1 answer:
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Answer:
The equation of a line that runs through (-6,2) and is parallel to a line with a slope of (-1/4) is x + 4y - 2 = 0
Step-by-step explanation:
Slope of the parallel equation is m 1 = (-1/4)
<em>If the two liner are parallel the, slope of both lines are equal.</em>
⇒The slope of the equation of line = m2 = m1 = -(1/4)
The point (x0, y0) = (-6,2)
Now, by THE POINT SLOPE FORMULA: The equation of a line is given as
( y - y0) = m (x -x0)
Now, here the equation of line is given as:
or, 4y - 8 + x + 6 = 0
or, x + 4y - 2 = 0
Hence, the required line equation is x + 4y - 2 = 0
∆BOC is equilateral, since both OC and OB are radii of the circle with length 4 cm. Then the angle subtended by the minor arc BC has measure 60°. (Note that OA is also a radius.) AB is a diameter of the circle, so the arc AB subtends an angle measuring 180°. This means the minor arc AC measures 120°.
Since ∆BOC is equilateral, its area is √3/4 (4 cm)² = 4√3 cm². The area of the sector containing ∆BOC is 60/360 = 1/6 the total area of the circle, or π/6 (4 cm)² = 8π/3 cm². Then the area of the shaded segment adjacent to ∆BOC is (8π/3 - 4√3) cm².
∆AOC is isosceles, with vertex angle measuring 120°, so the other two angles measure (180° - 120°)/2 = 30°. Using trigonometry, we find
where is the length of the altitude originating from vertex O, and so
where is the length of the base AC. Hence the area of ∆AOC is 1/2 (2 cm) (4√3 cm) = 4√3 cm². The area of the sector containing ∆AOC is 120/360 = 1/3 of the total area of the circle, or π/3 (4 cm)² = 16π/3 cm². Then the area of the other shaded segment is (16π/3 - 4√3) cm².
So, the total area of the shaded region is
(8π/3 - 4√3) + (16π/3 - 4√3) = (8π - 8√3) cm²