The answer to the problem is as follows:

x = sin(t/2)
y = cos(t/2) 

Square both equations and add to eliminate the parameter t: 

x^2 + y^2 = sin^2(t/2) + cos^2(t/2) = 1 

The final step is translating the original parameter limits into limits on x and y. Over the -Pi to +Pi range of t, x varies from -1 to +1, whereas y varies from 0 to 1. Thus we have the semicircle in quadrants I and II: y >= 0.

6 0

3 years ago

△ABCis reflected to form △A′B′C′. The vertices of △ABC are A(3, 1), B(1, 5), and C(6, 9). The vertices of △A′B′C′ are A′(−1,

Blababa [14]

Answer: D) reflection across y = -x

Explanation:

When we reflect over y = x, we basically swap x and y. So for instance, the point (3,1) becomes (1,3).

When reflecting over y = -x, we will do the same thing but we'll make each coordinate swap in sign from positive to negative (or vice versa). The rule for reflecting over y = -x is $(x,y) \to (-y,-x)$

So if we apply that rule to point A(3,1) then it becomes A ' (-1, -3).

Similarly, B(1,5) moves to B ' (-5, -1)

Finally, C(6,9) becomes C ' (-9, -6)

5 0

2 years ago