A lot of teachers don't teach SohCahToa anymore! this acronym will really help you. I'd suggest looking up how to use it, it'll make trigonometry WAY easier!
Answer:
Step-by-step explanation:
Notice when x increases 1, y is 4 times the previous one, so
the function is like 
To determine the constant C, put any pair of (x, y)
Use x = 0, y = 0.2, so
0.2 = 
= C * 1 = C
then
-43 - 4r = 3 - 27r
Add 27r to both sides. -43 - 4r + 27r = 3 -27r + 27r or -43 + 23r = 3
Add 43 to both sides. -43 + 43 +23r = 3 + 43 or 23r = 46.
Divide both sides by 23 to get r by itself. 23r / 23 = 46 / 23 or r = 2
r = 2
Answer:
6x²y(9 - 2y)
Step-by-step explanation:
Step 1: Factor out GCF
6(9x²y - 2x²y²)
Step 2: Factor out <em>x²</em>
6x²(9y - 2y²)
Step 3: Factor out <em>y</em>
6x²y(9 - 2y)
And we have our factored answer.
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
