Inequality(compound) Ix<1 or x>3 whats the Interval?

Mathematics

1 answer:

spin [16.1K]3 years ago

4 0

Answer:

x>3

Step-by-step explanation:

Ix<1 is unrelated to the interval as x>3 is.

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Rewrite the expression 4+<img src="https://tex.z-dn.net/?f=%5Csqrt%7B16-%284%29%285%29%7D" id="TexFormula1" title="\sqrt{16-(4)(

Inessa05 [86]

Answer:

$2+i$

Step-by-step explanation:

Given the expression:

$\dfrac{4+\sqrt{16-(4)(5)}}{2}$

To find:

The expression of above complex number in standard form $a+bi$ .

Solution:

First of all, learn the concept of $i$ (pronounced as <em>iota</em>) which is used to represent the complex numbers. Especially the imaginary part of the complex number is represented by $i$ .

Value of $i =\sqrt{-1}$ .

Now, let us consider the given expression:

$\dfrac{4+\sqrt{16-(4)(5)}}{2}\\\Rightarrow \dfrac{4+\sqrt{16-(4\times 5)}}{2}\\\Rightarrow \dfrac{4+\sqrt{16-20}}{2}\\\Rightarrow \dfrac{4+\sqrt{-4}}{2}\\\Rightarrow \dfrac{4+\sqrt{(-1)(4)}}{2}\\\Rightarrow \dfrac{4+\sqrt{(-1)}\sqrt4}{2}\\\Rightarrow \dfrac{4+\sqrt4i}{2} \ \ \ \ \ (\because \sqrt{-1} =i) \\\Rightarrow \dfrac{4+2i}{2}\\\Rightarrow 2+i$