Answer:
Q13. y = sin(2x – π/2); y = - 2cos2x
Q14. y = 2sin2x -1; y = -2cos(2x – π/2) -1
Step-by-step explanation:
Question 13
(A) Sine function
y = a sin[b(x - h)] + k
y = a sin(bx - bh) + k; bh = phase shift
(1) Amp = 1; a = 1
(2) The graph is symmetrical about the x-axis. k = 0.
(3) Per = π. b = 2
(4) Phase shift = π/2.
2h =π/2
h = π/4
The equation is
y = sin[2(x – π/4)} or
y = sin(2x – π/2)
B. Cosine function
y = a cos[b(x - h)] + k
y = a cos(bx - bh) + k; bh = phase shift
(1) Amp = 1; a = 1
(2) The graph is symmetrical about the x-axis. k = 0.
(3) Per = π. b = 2
(4) Reflected across x-axis, y ⟶ -y
The equation is y = - 2cos2x
Question 14
(A) Sine function
(1) Amp = 2; a = 2
(2) Shifted down 1; k = -1
(3) Per = π; b = 2
(4) Phase shift = 0; h = 0
The equation is y = 2sin2x -1
(B) Cosine function
a = 2, b = -1; b = 2
Phase shift = π/2; h = π/4
The equation is
y = -2cos[2(x – π/4)] – 1 or
y = -2cos(2x – π/2) - 1
No i dont think so but i could be wrong i will double check to be sure though
Another way is to note that there are <span><span>(<span>104</span>)</span><span>(<span>104</span>)</span></span> (“10 choose 4”) ways to select 4 balls from a collection of 10. If 4 of those 10 balls are “special” in some way (in this case, “special” = “red”), then the number of ways to choose 4 special balls is <span><span>(<span>44</span>)</span><span>(<span>44</span>)</span></span>. (The factor of <span><span>(<span>60</span>)</span><span>(<span>60</span>)</span></span> is included to convey that, after choosing 4 special balls, we choose none of the 6 non-special balls.) This line of reasoning gives the second expression.
Answer:
∑ (n = 1 to 5) (2·(-4)^(n - 1))
S5 = 2·((-4)^5 - 1)/((-4) - 1) = 410
This can help you to understand it
