Answer:
(1.37) AUB = { 1,2,3,4,5,6}
(1.38) AUC = { 1,2,3,4,5 }
(1.39)BUC = { 1,2,3,4,5,6}
(1.40) { 2,4 }
(1.41) { 1,3,5 }
(1.42) { phi }
(1.43) AU(BUC) = { 1,2,3,4,5,6 }
(1.44) { phi }
(1.45) {1,2,3,4,5}
(1.46) { 1,2,3,4,5 }
Find the period of the function. y=3 sin x/8
1 answer:
Answer:
The period of given function is
So, Option B is correct.
Step-by-step explanation:
In this question we need to find the period of the function y= 3 sin x/8
The formula used to find period of function is:
We need to know the value of b.
To find the value of b we compare the standard equation with the equation of function given.
Standard Equation: y = a sin(bx - c) +d
Given Equation: y= 3 sin(x/8)
Comparing we get:
a= 3
b= 1/8
c= 0
d=0
So, we get the value of b i.e 1/8. Putting it in the formula to find period of given function.
Solving,
So, the period of given function is
You might be interested in
The equation of a parabola with a directrix at y = -3 and a focus at (5 , 3) is y = one twelfth (x - 5)² ⇒ 1st answer
Step-by-step explanation:
The form of the equation of the parabola is (x - h)² = 4p(y - k), where
- The vertex of the parabola is (h , k)
- The focus is (h , k + p)
- The directrix is at y = k - p
∵ The focus of the parabola is at (5 , 3)
- Compare it with the 2nd rule above
∴ h = 5
∴ k + p = 3 ⇒ (1)
∵ The directrix is at y = -3
- By using the 3rd rule above
∴ k - p = -3 ⇒ (2)
Solve the system of equations to find k and p
Add equations (1) and (2) to eliminate p
∴ 2k = 0
- Divide both sides by 2
∴ k = 0
- Substitute the value of k in equation (1) to find p
∵ 0 + p = 3
∴ p = 3
Substitute the values of h , k , and p in the form of the equation above
∵ (x - 5)² = 4(3)(y - 0)
∴ (x - 5)² = 12 y
- Divide both sides by 12
∴ (x - 5)² = y
- Switch the two sides
∴ y = (x - 5)²
The equation of a parabola with a directrix at y = -3 and a focus at (5 , 3) is y = (x - 5)²
Learn more:
you can learn more about the quadratic equations in brainly.com/question/8054589
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