Algebra - Desklib

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∗10−3 t
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 ¿+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,
T m= 35−5
2 =15
Time period for one complete cycle t p= 12+12=24
So the period = 2 π
tp
= π
12
So putting it all together,
H=T m + Asin (tp∗t− π
2 )
That is , H=15+ 20 sin ( π
12 t− π
2 )
a)
b)
given,
H= A +Bsin (Ct + D )