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Ede4ka [16]
3 years ago
6

WILL MARK BRAINLIEST IF CORRECT❤️❤️ Simplify the expression -5+i/2i Step by step

Mathematics
2 answers:
solniwko [45]3 years ago
8 0

Answer:

For complex numbers,

a + bi and a - bi

they have the interesting property that if you add them you get the real number 2a

and if you multiply them , because of the difference of square pattern, you get a^2 - b^2 i^2

But since i^2 = -1, we end up with a real number as a product.

e.g. 6 - 5i and 6 + 5i are conjugates of each other

sum = 6-5i + 6+5i = 12

product = 36 - 25i^2

= 36 -(-25) = 61

Your question is even easier, since the denominator is a monomial instead of a binomial, so we just have to multiply by i/i

Also I believe, according to the answer, that you have a typo, and you meant

(-5+i)/(2i)

= (-5+i)/(2i) *i/i

= (-5i + i^2)/2i^2)

= (-5i +i^2)/-2

= (-5i - 1)/-2

= (1 + 5i)/2 or they way they have it: 1/2 + 5i/2

Step-by-step explanation:

PLZ MARK ME BRAINLYIST

zhenek [66]3 years ago
5 0

Answer:

For complex numbers,

a + bi and a - bi

they have the interesting property that if you add them you get the real number 2a

and if you multiply them , because of the difference of square pattern, you get a^2 - b^2 i^2

But since i^2 = -1, we end up with a real number as a product.

e.g. 6 - 5i and 6 + 5i are conjugates of each other

sum = 6-5i + 6+5i = 12

product = 36 - 25i^2

= 36 -(-25) = 61

Your question is even easier, since the denominator is a monomial instead of a binomial, so we just have to multiply by i/i

Also I believe, according to the answer, that you have a typo, and you meant

(-5+i)/(2i)

= (-5+i)/(2i) *i/i

= (-5i + i^2)/2i^2)

= (-5i +i^2)/-2

= (-5i - 1)/-2

= (1 + 5i)/2 or they way they have it: 1/2 + 5i/2

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Consider the following series. 1 9 1 18 1 27 1 36 1 45 Determine whether the series is convergent or divergent. Justify your ans
Vedmedyk [2.9K]

Answer:

<em>Diverges; the limit of the terms, an, is not 0 as n goes to infinity</em>

Step-by-step explanation:

<u>Convergence of Infinite Series</u>

The series is given as the terms

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For a series to be convergent, the limit of the general term an when n goes to infinity must be 0. In such cases, the sum of all terms tends to a fixed value.

Our series has two clearly different sequences, which define it as a piecewise general term:

If n is odd, then

a_n=1

Otherwise, an is an arithmetic series starting from 9 with a common difference of 9, that is

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If we take the limit as n goes to infinity, we don't know which one should be selected, thus we must assume that any of them is a correct option, therefore both of them should converge.

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But it goes worse because of the limit of the second piece

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The answer is

<em>Diverges; the limit of the terms, an, is not 0 as n goes to infinity</em>

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