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Algebra: Formulas, Equations, Graphs, and Calculations

   

Added on  2022-11-14

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Algebra
1)
a) Given,
Q=f(t)=300 ∗ (0.75) 𝑡
40
a)
To write the formula for Q=𝐴𝑒−𝑘𝑡, let g(t) =𝐴𝑒−𝑘𝑡
0.75-1=-0.25
Which is 25% decreasing
We need to find the points, calculate f(0), f(1)
f(0)=300*0.75=225; (0,300)
f(1)=300*(0.75) 1
40=297.85; (1,297.85)
g(t)= =𝐴𝑒−𝑘𝑡
g(1)= =𝐴𝑒−𝑘=297.85
here A=300(from Q=f(t))
so, 297.85=300=𝑒−𝑘
=𝑒−𝑘 = 297.85
300
k=7.192*10−3
b)
finding t half
given Q=300
half of Q=150
plugging this into Q=𝐴𝑒−𝑘𝑡,
150=300𝑒−7,192∗10−3𝑡
Solving for t,
T=96.377minutes

c)
here,
red curve is the graph, violet line is the halving time.
2)
Starting from (1,0), move along the unit circle until the angle θ is formed with x axis.
cosθ is the x coordinate and sinθ is the y coordinate.
cos(90+θ) lies in the second coordinate. Therefore the x coordinate is negative and y coordinate is
positive.
Cos(90+θ)=cos90cosθ-sin90sinθ
=0-sinθ
=-sinθ

So cos(90+θ)=-sinθ
That is, cos(𝜋
2+θ)=-sinθ
3)
Amplitude is the half range.
That is A=1
2 (35 − (−5)) = 20
Mean value is the average of the two extremes.
That is,
𝑇𝑚 = 35−5
2 =15
Time period for one complete cycle 𝑡𝑝= 12+12=24
So the period = 2𝜋
𝑡𝑝
= 𝜋
12
So putting it all together,
𝐻 = 𝑇𝑚 + 𝐴𝑠𝑖𝑛 (𝑡𝑝 ∗ 𝑡 − 𝜋
2)
That is , 𝐻 = 15 + 20sin( 𝜋
12 𝑡 − 𝜋
2)
a)
b)
given,
𝐻 = 𝐴 + 𝐵𝑠𝑖𝑛(𝐶𝑡 + 𝐷)
Comparing with 𝐻 = 15 + 20sin( 𝜋
12 𝑡 − 𝜋
2)
A=15

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