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Elis [28]
2 years ago
8

Add the following complex numbers:

Mathematics
2 answers:
V125BC [204]2 years ago
6 0

Answer:

B. 7 - 9i

Step-by-step explanation:

1) Remove Parentheses.

2 - 8i + 5 - i

2) Add 2 and 5.

7 - 8i - i

3) Subtract i from -8i.

7 - 9i

Maurinko [17]2 years ago
3 0

Answer:

Option B

The answer is 7 – 9i

Step-by-step explanation:

(2 – 8i) + (5 – i)

2 – 8i + 5 – i

7 – 9i

Thus, The answer is 7 – 9i

<u>-TheUnknownScientist 72</u>

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What is the exact angle between the clock hands at 6:45:00 PM? Plz show work
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Answer:

90°

Step-by-step explanation:

angle on a straight line=180 and half of 180=90°

6 0
3 years ago
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How to multiply a whole number and a mixed fraction?
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Convert the mixed number into a decimal and multiply on a calculator (or by hand).
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If a person bikes 6 miles in 15 minutes, how far can she bike in 1.5 hours?
Natalka [10]

Answer:

36\:miles

Step-by-step explanation:

There is <em>one-and-a-half</em> <em>hour</em> in <em>ninety</em><em> </em><em>minutes</em>,<em> </em>so multiply 6 by 15 to get 90, then whatever is done to the bottom is also done to the top, so you <em>square</em> 6 to get 36.

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3 years ago
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

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3 years ago
how much time would it take for an airplane to reach its destination if it traveled at an average speed of 790 kilometers/hour f
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