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3 years ago

Find the markup $111.00 percent of markup 50%

1 answer:

5 0

50% markup of $111.00 is the same as increasing 111 by half

50% of 111 = 55.5
Then add 55.5 on 111 = $166.50

Final value $166.50

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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

4 years ago

Evaluate this expression. 5^-3

sweet-ann [11.9K]

Answer:

$\Huge \boxed{\frac{1}{125}}$

Step-by-step explanation:

<h2>Order of operations</h2>

PEMDAS

Parenthesis, exponent, multiply, divide, add, and subtract from left to right.

Do exponent.

$\displaystyle \frac{1}{5^3}$

$\displaystyle 5^3=5\times5\times5=125$

$\displaystyle \frac{1}{125}=125$

$\large \boxed{\frac{1}{125}}$ , which is our answer.

Hope this helps!

4 0

3 years ago

(3b + 12 ) divided by 2 =30 Whats is b ?

denpristay [2]

b=16 , First you multiply 30 by 2 to see what sum you need for the numerator. Then you subtract the 60 that you get by the 12 and get 48. So 3 multiplied by b should give you 48. So you just divide 48 by 3 and get 16.

6 0

4 years ago

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$