Coach King is holding a free-throw shooting contest for some students in his PE class. The students participating in the contest
2 answers:
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We have that
<span>y=2sinx-cosx
y=cosx
</span>over the interval 0≤x≤pi/2-------> 0≤x≤1.57
using a graph tool
see the attached figure
the solution in the given interval is the point (pi/4, 0.707)
the y-value of the point of intersection is 0.707 -----> √2/2
the answer is the option
<span>b. (square root of 2)/2</span>
<span>y=2sinx-cosx
y=cosx
</span>over the interval 0≤x≤pi/2-------> 0≤x≤1.57
using a graph tool
see the attached figure
the solution in the given interval is the point (pi/4, 0.707)
the y-value of the point of intersection is 0.707 -----> √2/2
the answer is the option
<span>b. (square root of 2)/2</span>
Answer: The answer is (D) Reflection across the line y = -x.
Step-by-step explanation: In figure given in the question, we can see two triangles, ΔABC and ΔA'B'C' where the second triangle is the result of transformation from the first one.
(A) If we rotate ΔABC 180° counterclockwise about the origin, then the image will coincide with ΔA'B'C'. So, this transformation can take place here.
(B) If we reflect ΔABC across the origin, then also the image will coincide with ΔA'B'C' and so this transformation can also take place.
(C) If we rotate ΔABC through 180° clockwise about the origin, the we will see the image will be same as ΔA'B'C'. Hence, this transformation can also take place.
(D) Finally, if we reflect ΔABC across the line y = -x, the the image formed will be different from ΔA'B'C', in fact, it is ΔA'D'E', as shown in the attached figure. So, this transformation can not take place here.
Thus, the correct option is (D).
