ABC ~ A’B’C’
So, AB=A’B’ , BC=B’C’ , AC = A’C’
Given, AB=15 and A’C’=4
So, AC = 4 , A’B’ = 15
Answer:
D
Step-by-step explanation:
Solution:-
The standard sinusoidal waveform defined over the domain [ 0 , 2π ] is given as:
f ( x ) = sin ( w*x ± k ) ± b
Where,
w: The frequency of the cycle
k: The phase difference
b: The vertical shift of center line from origin
We are given that the function completes 2 cycles over the domain of [ 0 , 2π ]. The number of cycles of a sinusoidal wave is given by the frequency parameter ( w ).
We will plug in w = 2. No information is given regarding the phase difference ( k ) and the position of waveform from the origin. So we can set these parameters to zero. k = b = 0.
The resulting sinusoidal waveform can be expressed as:
f ( x ) = sin ( 2x ) ... Answer
The population 150 students
The sample is 54% of students
I’m not sure but I hope this helps!
The parabola is opens upward.
a = (8 – 5)
= 3
Using the standard form:
(x – h)^2 = 4a(y – k)
(x -5)^2 = 12( y – 4)
In general form
x^2 -10x +25 =12y – 48
x^2 -10x -12y + 73 =0
