Answer:
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The bisector of angle APQ passes through O and this is illustrated below.
<h3>How to illustrate the information?</h3>
From the information given, the center is O. and the circle passes through O and cuts at K.
In this case, it should be noted that the circles are equal according to the SAS test.
Here, AOB + APQ = 180° (Linear pair)
2AOB = 180
AOB = 90.
Therefore, the bisector of angle APQ passes through O.
Learn more about bisector on:
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Multiply both sides by 3 to get:
3r = sqrt(A)
Square both sides:
9r^2 = A
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
