Capacity of the tank = 45 gallon
At the beginning of Oct., amount of oil in the tank = 45*0.8 = 36 gallons
At the end of Oct., amount of oil in the tank = 45*1/3 = 15 gallons
So, amount of oil used during Oct = (36-15) = 21 gallons
Answer:
270
Step-by-step explanation:
First, tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>), so if cos(<em>θ</em>) = 3/5 > 0 and tan(<em>θ</em>) < 0, then it follows that sin(<em>θ</em>) < 0.
Recall the Pythagorean identity:
sin²(<em>θ</em>) + cos²(<em>θ</em>) = 1
Then
sin(<em>θ</em>) = -√(1 - cos²(<em>θ</em>)) = -4/5
and so
tan(<em>θ</em>) = (-4/5) / (3/5) = -4/3
The remaining trig ratios are just reciprocals of the ones found already:
sec(<em>θ</em>) = 1/cos(<em>θ</em>) = 5/3
csc(<em>θ</em>) = 1/sin(<em>θ</em>) = -5/4
cot(<em>θ</em>) = 1/tan(<em>θ</em>) = -3/4
Answer:
(−4, 12), (1, −3)
Step-by-step explanation:
y = −3x
x^2 −4 = y
so
x^2 −4 = −3x
x^2 + 3x - 4 = 0
(x + 4)(x - 1) = 0
x + 4 = 0; x = -4
x - 1 = 0; x = 1
when x = 1, y = -3(1) = -3
when x = -4, y = -3(-4) = 12
Answer
(-4 , 12), (1 , -3)
The formula is
A=p (1+r/k)^kt
A future value 3000
P present value 150
R interest rate 0.025
T time?
3000=150 (1+0.025/12)^12t
Solve for t
3000/150=(1+0.025/12)^12t
Take the log
Log (3000/150)=log (1+0.025/12)×12t
12t=Log (3000/150)÷log (1+0.025/12)
T=(log(3,000÷150)÷log(1+0.025÷12))÷12
T=119.95 years
