Formula is y = a(x-h)^2 + k
Where h is 1 and k is 1
f (x) = a(x-1)^2 + 1
-3 = a(0-1)^2 + 1
-3 = a(-1)^2 + 1
-3 = a(1) + 1
-3 - 1 = a
-4 = a
a = -4
A must be equal to -4
y = -4(x-1)^2 + 1
0 = -4(x-1)^2 + 1
4(x^2 - 2x + 1) - 1 = 0
4x^2 - 8x + 4 - 1 = 0
4x^2 - 8x + 3 = 0
4x^2 - 8x = -3
Divide fpr 4 each term of the equation....x^2 - 2x = -3/4
We must factor the perfect square ax^2 + bx + c which we don't have. We must follow the rule (b/2)^2 where b is -2....(-2/2)^2 =
(-1)^2 = 1 and we add up that to both sides
x^2 - 2x + 1 = -3/4 + 1
x^2 - 2x + 1 = 1/4
(x-1)^2 = 1/4
square root both sides x-1 = (+/-) 1/2
x1 = +1/2 + 1 = 3/2
x2 = -1/2 + 1 = 1/2
x-intercepts are 1/2 and 3/2, in form (3/2,0); (1/2,0)
Sqrt(15) would be irrational
Answer is B
Answer: 30 degrees
Step-by-step explanation:
Answer:
Step-by-step explanation:
sin(2x) = 2 sin(x) cos(x) cos(2x) = cos2(x) – sin2(x) = 1 – 2 sin2(x) = 2 cos2(x) – 1. Now I am not sure if this is right but I remember another similar formula. Here is the correct formula cot x = cos x / sin x
2 ) ( sin² x + cos² x ) / cos x = sec x
1/cos x = sec x
sec x = sec x
cos² x + sin² x = 1
<h3>E
xplanation:</h3>
Replace cos^2(θ) with 1-sin^2(θ), and cot(θ) with cos(θ)/sin(θ).
cos^2(θ)cot^2(θ) = cot^2(θ) - cos^2(θ)
(1 -sin^2(θ))cot^2(θ) = . . . . . replace cos^2 with 1-sin^2
cot^2(θ) -sin^2(θ)·cos^2(θ)/sin^2(θ) = . . . . . replace cot with cos/sin
cot^2(θ) -cos^2(θ) = cot^2(θ) -cos^2(θ) . . . as desired
