Using the greatest common factor, reduce the fraction \frac{32}{48} 48 32 to its simplest form

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a

Alja [10]

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

$Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c} = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}$

when t = 0, Q = 200 L × 1 g/L = 200 g

$Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}$

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

$2 = 200e^{(-t/100)}\\\frac{2}{200} = e^{(-t/100)}$

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0

2 years ago

Bonny's truck has a fuel tank which holds 50 gallons, and she typically drives 425 miles on a full tank. She has added a second

Usimov [2.4K]

Its C because u have to divide 425 by 50 to get the mpg. Then because the truck now has an 80 gallon tank u have to multiply 8.5 with 80

6 0

2 years ago

Read 2 more answers

When Θ = 5 pi over 6, what are the reference angle and the sign values for sine, cosine, and tangent?

stira [4]

Note that (5π)/6 radians = 150°. Therefore the given angle is in quadrant 2.

Refer to the figure shown below.

Reference angles are measured relative to the horizontal axis.
Therefore the reference angle in each quadrant is π/6 radians or 30°.
Denote the reference angle as θ'.
Then, in quadrant 1,
cos θ' = √3/2, sin θ' = 1/2, tan θ' = √3.

Because we are in quadrant 2,
sin θ' = π/6;
sin(5π/6) is positive, but cos (5π/6) and tan (5π/6) are negative.

Answer:
5π/6 is in quadrant 2.
The reference angle, θ' = π/6.
sin(5π/6) is positive, cosine and tangent are negative.

3 0

2 years ago

Read 2 more answers

A rectangle that had the perimeter of 70 and the area of 250 has enlarged; its sides became twice as long. what is the new area

aliya0001 [1]

The perimeter of the original rectangle is:
P = 2w + 2l = 70
The area of the original rectangle is:
A = w * l = 250
Then, by modifying the length of its sides we have:
Perimeter:
P '= 2 (2w) +2 (2l)
Rewriting:
P '= 2 (2w + 2l)
P '= 2P
P '= 2 (70)
P '= 140
Area:
A '= (2w) * (2l)
Rewriting:
A '= (2) * (2) (w) * (l)
A '= 4 * w * l
A '= 4 * A
A '= 4 * 250
A '= 1000
Answer:
the new area and the new perimeter are:
P '= 140
A '= 1000

3 0

2 years ago

The hypotenuse of a 45°-45°-90° triangle measures 10 StartRoot 5 EndRoot in. A right triangle is shown. The hypotenuse has a len

suter [353]

Answer:

$5\sqrt{10}\ units$