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Diano4ka-milaya [45]
3 years ago
12

The smallest number by which 192 should be multiplied to make it a perfect cube is __

Mathematics
1 answer:
algol [13]3 years ago
3 0

Answer: 9

Step-by-step explanation:

192 * 9 = 1728

Cube root of 1728 is 12

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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)(
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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

So, the given expression in standard form is 2+i.

Let us compare with standard form a+bi so we get a =2, b =1.

\therefore The standard form of

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

is: \bold{2+i}

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