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2 years ago

Simplify the expression to a + bi form: (-2 - 6i)?

Mathematics

1 answer:

mixer [17]2 years ago

6 0

The expression (-2 - 6i)-(-2-4i) to a + bi form is 0 - 2i.

Complete question.

Simplify the expression to a + bi form:

(-2 - 6i)-(-2-4i)

Square root of any negative number are expressed as a complex number. For example i = √-1

Complex numbers are generally written in the format z = x+iy

Given the expression (-2 - 6i)-(-2-4i)), in expansion:

(-2 - 6i)-(-2-4i)

= -2 - 6i + 2+4i

Collect the like terms

= (-2 + 2) - 6i + 4i

= 0 - 2i

Therefore the expression (-2 - 6i)-(-2-4i) to a + bi form is 0 - 2i.

Learn more on complex number here: brainly.com/question/12375854

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7% of what lenght is 200 ft?

Nastasia [14]

If you would like to solve 7% of what length is 200 ft, you can calculate this using the following steps:

7% of what length is 200 ft
7% * x = 200
7/100 * x = 200 /*100/7
x = 200 * 100 / 7
x = 2857 ft

Result: 7% of 2857 ft is 200 ft.

5 0

2 years ago

Read 2 more answers

The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there a

Anettt [7]

Answer:

1) The size of the colony after 4 days is 6553 mosquitoes

2) After $t = 4.9\ days$

Step-by-step explanation:

To answer this question you must use the growth formula

$N = N_0e ^ {kt}$

Where

$N_0$ is the initial population of mosquitoes = 1000

t is the time in days

k is the growth rate

N is the population according to the number of days

We know that when t = 1 and $N_0=1000$ then N = 1600

Then we use these values to find k.

$1600 = 1000e ^{k(1)}\\\\\frac{1600}{1000} = e ^ k\\\\ln(\frac{1600}{1000}) = k\\\\k = 0.47$

Now that we know k we can find the size of the colony after 4 days.

$N = 1000e ^ {0.47(4)}\\\\N = 6553\ mosquitoes$

To know how long it should take for the population to reach 10,000 mosquitoes we must do N = 10000 and solve for t.

$10000 = 1000e ^ {0.47t}\\\\10 = e ^ {0.47t}\\\\ln(10) = 0.47t\\\\t = \frac{ln(10)}{0.47}\\\\t = 4.9\ days$

6 0

2 years ago

Twenty-seven is smaller than five times a number increased by seven

mixas84 [53]

Answer:

27(less than symbol) 5x+7

Step-by-step explanation:

5 times a number is 5x

increase means add

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2 years ago

3 tenths +4 tenths +__________tenths

algol [13]

Well four plus seven would equal seven tenths.

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2 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

2 years ago