Answer:
The absolute brightness of the Cepheid star after a period of 45 days is -5.95
Step-by-step explanation:
Since the absolute magnitude or brightness of a Cepheid star is related to its period or length of its pulse by
M = –2.78(log P) – 1.35 where M = absolute magnitude and P = period or length of pulse.
From our question, it is given that P = 45 days.
So, M = –2.78(log P) – 1.35
M = –2.78(log 45) – 1.35
M = –2.78(1.6532) – 1.35
M = -4.60 - 1.35
M = -5.95
So, the absolute magnitude or brightness M of a Cepheid star after a period P of 45 days is -5.95
F(x)=3x/2 for 0≤x≤2
<span>.....=6 - 3x/2 for 2<x≤4 </span>
<span>g(x) = -x/4 + 1 for 0≤x≤4 and g'(x)=-1/4 </span>
<span>so h(x)= f(g(x)) = (3/2)(-¼x+1)=-3x/8 + 3/2 for 0≤x≤2 </span>
<span>for x=1, h'(x)=-3/8 so h'(1)=-3/8 </span>
<span>When x=2, g(2)=1/2 so h'(2)=g'(2)f '(1/2)= -(1/4)(3/2)=-3/8 </span>
<span>When x=3, h(x)=6 - (3/2)(1 - x/4) = 9/2 +3x/8 </span>
<span>h'(x)=3/8 so h'(3) = 3/8</span>
Answer:

Step-by-step explanation:
A line is a one-dimensional figure that has no thickness and extends infinitely in both directions.
A line segment is a line that has two end-points and a ray is a line that has one end-point.
Given equation is 
To find : a line that lies entirely in the set defined by the given equation.
Solution:
Take 
Check:

Therefore,
satisfy the given equation.
Answer:
y = 0
Step-by-step explanation:
Since there is no indication of a shift in the graph of f(x) = 3cos(-0.25x), then the midline must be the normal midline for a cosine or sine function which is: y = 0.
Answer:
here's the solution :-
Step-by-step explanation:
i hope it helped...