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Ad libitum [116K]
3 years ago
11

If 6 men can do a piece of work in 14 days, how many men are needed to do the work in 21days?

Mathematics
1 answer:
Ivenika [448]3 years ago
4 0
Total work amount
6 men * 14 days
84 men days

If it is done within 21 days, need x men
84 men days = x men * 21 days
x = 84/21 men
x = 4 men
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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
3 years ago
You accept a new job with a starting salary of $47,000. You receive a 4% raise at the start of your second year, a 5.6% raise at
Sunny_sXe [5.5K]

Answer:

1yr  47,000, 2nd yr  48,880, 3rd yr   51,617.28, 4th yr  57,346.7981 or 57,346.80

Step-by-step explanation:

1yr        47,000

2nd yr  47,000 x 4% = 48,880

3rd yr   48,880 x 5.6% = 51,617.28

4th yr   51,617.28 x 11.1% = 57,346.7981 or 57,346.80

4 0
2 years ago
AABC = AGHL Complete the congruence statement AB = ​
sweet [91]

Answer:

AB = AH

Step-by-step explanation:

Hope This Helps!!

4 0
3 years ago
Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following.
goldfiish [28.3K]

Answer: a) 0.2500

b) 0.1875

c) 0.4375

d) 1.414

e) 0, x/2

f) 1.3333

g) 0.4714

h) 2

Step-by-step explanation: see attachment below

3 0
3 years ago
(1+cos2x)/(1-cos2x) = cot^2x
sesenic [268]

We will turn the left side into the right side.

\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x

Use the identity:

\cos 2x = \cos^2 x - \sin^2 x

\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x

\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x

Now use the identity

\sin^2 x + \cos^2 x = 1 solved for sin^2 x and for cos^2 x.

\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x

\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x

\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x

\cot^2 x = \cot^2 x


8 0
3 years ago
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