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2 years ago

Which complex number has an absolute value of 5?

2 answers:

5 0

For this case we can use the Pythagorean theorem to solve the problem.
We have then for the following complex number:
$-3 + 4i
$
Using the Pythagorean theorem:
$\sqrt{(-3)^2+(4)^2}$
Rewriting we have:
$\sqrt{9+16}$
$\sqrt{25}$
$5$
Answer:
A complex number that has an absolute value of 5 is:
A)–3 + 4i

natali 33 [55]2 years ago

4 0

A. -3+4i Square both numbers, add them, then find the square root. Essentially, use the Pythagorean theorem. -3^2 + 4^2 9 + 16=25 Square root of 25 is 5

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2 years ago

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Answer:

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Explanation:

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$(a+b)^n=a^n+n\cdot a^{n-1}{\cdot b}+\frac{n(n-1)}{2}\cdot a^{n-2}\cdot b^2+------+b^n$

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$T_{r+1} =^nC_r\cdot a^{n-r}\cdot b^r$

Since, $T_{r+1}$

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After substituting the values we will get

$\frac{13!}{8!\cdot 5!}\cdot x^5\cdot (-2y)^8}$

After simplification we will get

$329472\cdot x^5\cdot y^8$ which will be the 9th term of the expansion

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