With some simplification,
cos⁻¹(x) - 5 cos⁻¹(x) - π/3 = 2π/3
becomes
-4 cos⁻¹(x) = π
or
cos⁻¹(x) = -π/4
However, the inverse cosine function has a range of 0 ≤ cos⁻¹(x) ≤ π, so there are no solutions to this equation.
Answer:
78%
Step-by-step explanation:
70 = 0.50 (62) + 0.50 (x)
70 = 31 + 0.5x
39 = 0.5x
x = 78
You must get at least a 78% on the test to have at least 70% in the class.
Ya, calculus and related rates, such fun!
everything is changing with respect to t
altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min
area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min
base=db/dt
alright
area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt
ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b
so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min
therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt
the base is decreasing at 1.6cm/min
