3.1. Consider the following numbers: \( \sqrt[3]{9} ; \sqrt{9} ; \sqrt{-9} \) Which one of the above numbers is: 3.1.1. Irrational? 3.1.2. Imaginary? 3.2. Between which TWO integers does \( \sqrt{11} \) lie? 3.3. Use the long division method to calculate: \( 111111_{2} \div 1001_{2} \quad \) (leave your answer in decimal form) 3.4. Simplify the following expression fully: \( \frac{3^{x+2} .27^{x-2}}{81^{x}} \)
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The number \( \sqrt[3]{9} \) is irrational because it cannot be expressed as a ratio of two integers, while \( \sqrt{-9} \) is imaginary, representing 3i in the complex number system. Lastly, \( \sqrt{9} \) equals 3, which is a rational number. To find where \( \sqrt{11} \) lies, we can check the squares of integers: \( 3^2 = 9 \) and \( 4^2 = 16 \). Thus, \( \sqrt{11} \) is between 3 and 4, confirming that 3 < \( \sqrt{11} \) < 4. Next, for the long division of \( 111111_2 \) by \( 1001_2 \): First, convert both binary numbers to decimal; \( 111111_2 = 63 \) and \( 1001_2 = 9 \). Dividing 63 by 9 gives us a quotient of 7, which is our answer in decimal form. Lastly, we can simplify the expression: \( 3^{x+2} \cdot 27^{x-2} = 3^{x+2} \cdot (3^3)^{x-2} = 3^{x+2} \cdot 3^{3x-6} = 3^{(x+2+3x-6)} = 3^{4x - 4} \). Dividing by \( 81^x = (3^4)^x = 3^{4x} \) gives us \( 3^{4x-4} \div 3^{4x} = 3^{-4} = \frac{1}{81} \).
