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Vinil7 [7]
3 years ago
6

Solve the equation a=m-n for the variable n

Mathematics
1 answer:
Bogdan [553]3 years ago
7 0

Answer:

m-a = n

Step-by-step explanation:

a=m-n

Subtract m from each side

a-m = m-n-m

a-m = -n

Multiply each side by -1

-a+m = n

m-a = n

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Problem 5

<h3>Answer:   6i</h3>

------------------

Explanation:

We have these four identities

  • i^0 = 1
  • i^1 = i
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Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.

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=============================================

Problem 6

<h3>Answer:  -3i</h3>

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Explanation:

We'll use the ideas mentioned in problem 5

46/4 = 11 remainder 2

i^46 = i^2 = -1

sqrt(-9) = 3i

The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i

In other words, here is one way you could write out the steps

-i^{46}*-\sqrt{-9}\\\\-i^{2}*-3i\\\\i^{2}*3i\\\\-1*3i\\\\-3i\\\\

=============================================

Problem 7

<h3>Answer:   -1</h3>

------------------

Work Shown:

i^10 = i^2 because 10/4 = 2 remainder 2

i^19 = i^3 because 19/4 = 4 remainder 3

i^7 = i^3 because 7/4 = 1 remainder 3

Again, all we care about are the remainders.

i^{10}+i^{19} - i^{7}\\\\i^{2}+i^{3} - i^{3}\\\\i^{2}\\\\-1

=============================================

Problem 8

<h3>Answer:    -1 + i</h3>

------------------

Work Shown:

i^22*i^6 = i^(22+6) = i^28

Earlier in problem 5, we found that i^28 = i^0 = 1

So,

i^1 - \left(i^{22}*i^{6}\right)\\\\i^1 - i^{28}\\\\i^1 - i^{0}\\\\i - 1\\\\-1+i

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