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3 years ago

9

Carlin pays $ 25 for her monthly phone service and $.02 for each minute m of talk time. Which expression represents her monthly

bill?
A. 2m + 25 B. 25m + 0.02 C. .02m + 25 D. 25m + 2

1 answer:

Artemon [7]3 years ago

5 0

Answer:

I think the answer B

If not then, sorry

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Let's find the slope (m) :

$\dfrac{y2 - y1}{x2 - x1}$

$\dfrac{16 - 11}{9 - 6}$

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now, let's plug the value of coordinates of a point lying on the line ( for example : (6 , 11) and slope (m) to find the y - intercept (c) :

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Peter attends 6 lessons each week . A year he attends 52 weeks. But he missed 5 classes how many lessons did he attend during th

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Given that ABCD is a rhombus, find the value of x.<br> (X+14)

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Answer:

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Step-by-step explanation:

we know that

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2) The sum of the interior angles in any triangle must be equal to 180 degrees

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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

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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

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