Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min
Part 1)
we have
the scale is 
the distance on a map is 
we know that
The scale is equal to the distance on a map divided by the real distance
Let
x------> distance on a map
y-------> real distance
S-------> scale

In this part we have

Find the value of y

Substitute the values


Convert to kilometers

Part 2)
In this part we have

Find the value of x

Substitute the values


Convert to centimeters

therefore
the answer is
The distance is 
A^2 + b^2 = c^2...a and b are the legs and c is the hypotenuse
20^2 + 21^2 = c^2
400 + 441 = c^2
841 = c^2
sqrt 841 = c
29 = c <== third straw will be 29 cm
Answer:
Here is the representation of a histogram. This represents in a continuous distribution the probability frequency of the books that Pedro had read in the summer.
Know that the probability frequency is:

The answer is
<span>a) 1000=-16t^2+1700, implies t² = -700 /-16, and t= 6.61s
b) </span><span>970= -16t^2+1700, </span><span>implies t² = -730 /-16, and t=6.75s
c)
reasonable domain of h
h is polynomial function, so its domain is R, (all real number)
its range
the inverse of h is h^-1 = sqrt (1700- t / 16), and its domain is </span>
<span><span><span>1700- t / 16>=0, so t <1700,
the range of h is I= ]-infinity, 1700]</span> </span> </span>