Algebra

Running head: ALGEBRA 1
Algebra
Name
Institution

ALGEBRA 2
Algebra
Question 1
Part a
Q=f ( t )=300 ( 0.75 )
t
40 = A × e−kt
Taking the natural log on both sides of the equation yields:
ln ( Q ) =ln ( 300 ) +ln ( ( 0.75 )
t
40 ) =5.7038+ t
40 ln (0.75)
¿ 5.7038+ t
40 ln (0.75)
ln ( Q )=5.7038− t
40 × 0.28768=5.7038−0.007192 t
Q=e5.7038 × e−0.007192 t =300 e−0.007192t
Therefore , A=300∧k =0.007192
Part b
Initial quantity when t=0 is 300mg
Half the quantity=150mg
Q=300 ( 0.75 )
t
40 =150
( 0.75 )
t
40 =150
300 =0.5
ln ( 0.75 )
t
40 =ln (0.5)
t
40 ln ( 0.75 ) =ln (0.5)
t = 40 ln (0.5)
ln ( 0.75 ) =96.3768 minutes
Part c

ALGEBRA 3
Question 2
Angle (θ+ π
2 ) lies in the second quadrant where the cosine is negative.
cos ( θ+ π
2 )=−sinθ
The correct answer is option B
Question 3
In this case, the minimum temperature is −5 °C while the maximum temperature is 35 °C.
Mid− point= 35+5
2 =15 °C
Amplitude= 35−−5
2 =20°C
Period is1 day=24 hours
The formula for a cosine function is H= Acos ( B ( t−C ) ) + D
Where A=amplitude , 2 π
B = period , C= phase shift ,∧D=vertical translation

ALGEBRA 4
2 π
B =24 , B= π
12
The maximum temperature occurs at 4pm. That is, 12 hours after t=0 implying that the phase
shift is 12 hours.
Therefore , H=20 cos ( π
12 ( t−12 ) ) +15
The sketch is shown below:
Part b
H=20cos ( π
12 ( t−12 ) )+ 15=15+20 cos ( π t
12 −π )
¿ 15+20 cos ( π t
12 − π
2 − π
2 )
cos (θ− π
2 )=sin θ
In this case, θ= π t
12 − π
2