Multiplication puzzle

sergij07 [2.7K]

The answer is: [A]: He did not apply the distributive property correctly for 4(1 + 3i) .
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Explanation:
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Note the distributive property of multiplication:
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a*(b+c) = ab + ac.
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As such: 4*(1 + 3i) = (4*1) + (4*3i) = 4 + 12i ;
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Instead, Donte somehow incorrectly calculated:
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4*(1 + 3i) = (4*1) + 3i = 4 + 31; (and did the rest of the problem correctly);

Note: - (8 - 5i) = -8 + 5i (done correctly;
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So if Donte did not apply the distributive property correctly for 4*(1+3i)—and incorrect got 4 + 3i (as mentioned above); but did the rest of the problem correctly, he would have got:
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4+ 3i - 8 + 5i = -4 + 8i (the incorrect answer as stated in our original problem.
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This corresponds to: "Answer choice: [A]: <span>He did not apply the distributive property correctly for 4(1 + 3i)."
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6 0

3 years ago

Read 2 more answers

In a photograph, a stadium measures 7 inches across by 2 inches high. If the actual stadium measures 1000 feet across, which equ

Ivahew [28]

Answer:

x/500 = 2/8.

Step-by-step explanation:

7 0

2 years ago

mr.roxbury weighed 240 pounds when he started on a diet. over a one year period mr.roxbury lost 32% of his body weight. in the f

Annette [7]

Answer:

187.68

Step-by-step explanation:

240*.32=76.8

240-76.8=163.2

163.2*.15=24.48

163.2+24.48=187.68

3 0

2 years ago

What is 7 1/5 - 6 2/5

Artist 52 [7]

Answer:

I think , the answer is 9/5 or 1.8 in decimal

3 0

3 years ago

Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decrea

maw [93]

Answer:

a. dQ/dt = -kQ

b. $Q = 9e^{-kt}$

c. k = 0.178

d. Q = 1.063 mg

Step-by-step explanation:

a) Write a differential equation for the quantity Q of hydrocodone bitartrate in the body at time t, in hours, since the drug was fully absorbed.

Let Q be the quantity of drug left in the body.

Since the rate of decrease of the quantity of drug -dQ/dt is directly proportional to the quantity of drug left, Q then

-dQ/dt ∝ Q

-dQ/dt = kQ

dQ/dt = -kQ

This is the required differential equation.

b) Solve your differential equation, assuming that at the patient has just absorbed the full 9 mg dose of the drug.

with t = 0, Q(0) = 9 mg

dQ/dt = -kQ

separating the variables, we have

dQ/Q = -kdt

Integrating we have

∫dQ/Q = ∫-kdt

㏑Q = -kt + c

$Q = e^{-kt + c}\\Q = e^{-kt}e^{c}\\Q = Ae^{-kt} (A = e^{c})$

when t = 0, Q = 9

$Q = Ae^{-kt} \\9 = Ae^{-k0}\\9 = Ae^{0}\\9 = A\\A = 9$

So, $Q = 9e^{-kt}$

c) Use the half-life to find the constant of proportionality k.

At half-life, Q = 9/2 = 4.5 mg and t = 3.9 hours

So,

$Q = 9e^{-kt}\\4.5 = 9e^{-kX3.9}\\\frac{4.5}{9} = e^{-kX3.9}\\\frac{1}{2} = e^{-3.9k}\\$

taking natural logarithm of both sides, we have

$ln\frac{1}{2} = ln(e^{-3.9k})\\\\-ln2 = -3.9k\\k = -ln2/-3,9 \\k = -0.693/-3.9\\k = 0.178$

d) How much of the 9 mg dose is still in the body after 12 hours?

Since k = 0.178,

$Q = 9e^{-0.178t}$

when t = 12 hours,

$Q = 9e^{-0.178t}\\Q = 9e^{-0.178X12}\\Q = 9e^{-2.136}\\Q = 9 X 0.1181\\Q = 1.063 mg$

7 0

3 years ago

Multiplication puzzle​

Multiplication puzzle