Select all the correct answers.
1 answer:
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<em>Greetings from Brasil...</em>
In a trigonometric function
F(X) = ±UD ± A.COS(Px + LR)
UD - move the graph to Up or Down (+ = up | - = down)
A - amplitude
P - period (period = 2π/P)
LR - move the graph to Left or Right (+ = left | - = right)
So:
A) F(X) = COS(X + 1)
standard cosine graph with 1 unit shift to the left
B) F(X) = COS(X) - 1 = -1 + COS(X)
standard cosine graph with 1 unit down
C) F(X) = COS(X - 1)
standard cosine graph with shift 1 unit to the right
D) F(X) = SEN(X - 1)
standard Sine graph with shift 1 unit to the right
Observing the graph we notice the sine function shifted 1 unit to the right, then
<h3>Option D</h3><em>(cosine star the curve in X and Y = zero. sine start the curve in Y = 1)</em>
The square of a prime number is not prime.
a) let x ∈ R, If x ∈ {prime numbers}, then ∉{prime numbers}
there says that if x is a real and x is in the set of the prime numbers, then the square of x isn't in the set of prime numbers.
b) Prove or disprove the statement.
ok, if x is a prime number, then x only can be divided by himself. Now is easy to see that = x*x can be divided by himself and x, then x*x is not a prime number, because can be divided by another number different than himself
<h2>Explanation:</h2><h2></h2>
Hello! Remember you have to write clear questions in order to get good and exact answers. Here, I'll assume the function as:
The y-intercept of a function is the point at which the graph of the function touches the y-axis. This occurs when we set . In other words, we define the y-intercept (let's call it
as:
Setting in our function we have:
So <em>in this context the y-intercept is -16</em>