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3 years ago

How do i solve this using imaginary numbers? PLEASE HELP

Mathematics

1 answer:

Tasya [4]3 years ago

8 0

Answer:

14 -2i

Step-by-step explanation:

9 + sqrt(-4) - ( -5 + sqrt(-16))

We know the sqrt of a negative number is i * sqrt(number)

9 + isqrt(4) - ( -5 + isqrt(16))

9 + 2i - (-5 + 4i)

Distribute the negative sign

9+2i +5 -4i

Combine like terms

14 -2i

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The area of Lake Erie is about 7.1538 times 10 to the power of 8 square miles. The area of the Indian Ocean is about 9.3214 time

andrew-mc [135]

Answer: Should be 7.6745982363164 times larger

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3 years ago

I need help, I dont know how to convert ratios into decimals and percentages [EX: write the ratio 5 to 8 in a decimal/percent]

Anon25 [30]

Fractions are just unsimplified numbers. For example 5/8 is just 5 divided by 8 so 0.625. To find the percent from that you move the decimal two places to the right making it 62.5%. Hope this helps!

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3 years ago

I don’t understand these 3 questions and I need help.

rjkz [21]

Not sure about the first one but the second one is option c (or -3). You just have to factor and find the solutions (so it would be -7 and 4 and then u just add them)

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6 0

3 years ago

Read 2 more answers

5b - 5 - 126 - 7<br> I need steps please and thank you

MA_775_DIABLO [31]

Answer:

5b - 138

Step-by-step explanation:

Alright let's break it down.

First, you can see that each constant (5,126,7) have negatives in front of them. SO you are going to subtract each one of them.

When subtracting negatives it's basically just adding them together. How to do it is simply adding:

5 + 126 + 7

Then you get 138.

BUT, it was negative numbers. So it's actually -138.

Then bring back the 5b and your answer is:

5b - 138

Mark brainliest if you can :D

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3 years ago

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Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$

$A = \frac{L}{k} - \frac{L}{k} e^{kt}$

So, D = R - A =

$D = \frac{L}{k} - \frac{L}{k} - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}$

when t = 0(at initial time), the initial value of D =

$D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}$

4 0

3 years ago