We have that the graph will have a solution that is mathematically given as
y=12sin(6t-\pi)+2
<h3>
Daylight time June trough to December</h3>
Question Parameters:
A city averages 14 hours of daylight in June (the longest days) and 10 in December (the shortest days).
Generally the equation for the Amplitude and vertical shift is mathematically given as
A=2
vertical shift =
vertical shift =
vertical shift =1
Therefore, the time is given as
Where
Hence,The graph will have a solution
y=12sin(6t-\pi)+2
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brainly.com/question/14375099
You're given that φ is an angle that terminates in the third quadrant (III). This means that both cos(φ) and sin(φ), and thus sec(φ) and csc(φ), are negative.
Recall the Pythagorean identity,
cos²(φ) + sin²(φ) = 1
Multiply the equation uniformly by 1/cos²(φ),
cos²(φ)/cos²(φ) + sin²(φ)/cos²(φ) = 1/cos²(φ)
1 + tan²(φ) = sec²(φ)
Solve for sec(φ) :
sec(φ) = - √(1 + tan²(φ))
Given that cot(φ) = 1/4, we have tan(φ) = 1/cot(φ) = 1/(1/4) = 4. Then
sec(φ) = - √(1 + 4²) = -√17
3 ( 2t + 5 ) = 5 t + 25
3 * 2 t + 3 * 5 = 5 t + 25
6 t + 15 = 5 t + 25
6t - 5t = 25 - 15
t = 10
Hope this helps!
Answer:
1/2
Step-by-step explanation:
In the straight line form...
y=mx+b
m is the constant of proportionality or the "K"
first polygon
ext. angle=180°-120°
=60°
n=360°/60°
n=6
second polygon
n=2(6)=12
ext. ang= 360°/n = 360°/12° = 30°
int. ang = 180°-30°= 150°
answer is C
