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marin [14]
3 years ago
5

The path of a race will be drawn on a coordinate grid like the one shown below. The starting point of the race will be at (−5.5,

2). The finishing point will be at (2, −5.5).
Part A: Use the grid to determine in which quadrants the starting point and the finishing point are located. Explain how you determined the locations. (6 points)

Part B: A checkpoint will be at (5.5, 2). In at least two sentences, describe the differences between the coordinates of the starting point and the checkpoint, and explain how the points are related. (4 points)

Mathematics
1 answer:
Mazyrski [523]3 years ago
3 0
A) the starting point is in quadrant Q because the x-value is negative, while the y-value is positive. The finishing point is in quadrant S because the x-value is positive, while the y-value is negative.
B) The points are connected by a straight line, so you don't have to wander off. By the way, the checkpoint is in Quadrant P.
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Answer:

cos(a + b) = \frac{1}{12}(1-2\sqrt{30})

Step-by-step explanation:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) [Identity]

cos(a) = -\frac{1}{3}

cos(b) = -\frac{1}{4}

Since, terminal side of angle 'a' lies in quadrant 3, sine of angle 'a' will be negative.

sin(a) = -\sqrt{1-(-\frac{1}{3})^2} [Since, sin(a) = \sqrt{(1-\text{cos}^2a)}]

         = -\sqrt{\frac{8}{9}}

         = -\frac{2\sqrt{2}}{3}

Similarly, terminal side of angle 'b' lies in quadrant 2, sine of angle 'b' will be  negative.

sin(b) = -\sqrt{1-(-\frac{1}{4})^2}

         = -\sqrt{\frac{15}{16}}

         = -\frac{\sqrt{15}}{4}

By substituting these values in the identity,

cos(a + b) = (-\frac{1}{3})(-\frac{1}{4})-(-\frac{2\sqrt{2}}{3})(-\frac{\sqrt{15}}{4})

                = \frac{1}{12}-\frac{\sqrt{120}}{12}

                = \frac{1}{12}(1-\sqrt{120})

                = \frac{1}{12}(1-2\sqrt{30})

Therefore, cos(a + b) = \frac{1}{12}(1-2\sqrt{30})

5 0
3 years ago
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A. y= - 2/3 x - 4 (only this line has slope = -2/3)

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