Question

question
find each product. then determine which conclusion can be drawn based on the results.
\\(\sqrt{3} \cdot \sqrt{3}\\)
\\(\sqrt{5} \cdot \sqrt{7}\\)
\\(\sqrt{2} \cdot \sqrt{18}\\)
\\(\sqrt{2} \cdot \sqrt{6}\\)
\\(\circ\\) the product of two irrational numbers is always irrational.
\\(\circ\\) the product of two irrational numbers can be rational or irrational.
\\(\circ\\) the product of two irrational numbers cannot be determined.
\\(\circ\\) the product of two irrational numbers is always rational.

part d

question
select all the expressions that result in a product that is a rational number.
\\(\frac{4}{3} \cdot \frac{12}{3}\\) \\(\frac{32}{4} \cdot \frac{35}{4}\\)
\\(\frac{\sqrt{3}}{2} \cdot \frac{22}{7}\\) \\(\sqrt{11} \cdot \frac{2}{3}\\)

Explanation:

First Set of Problems (Finding Products and Conclusion)

Step 1: Calculate \(\boldsymbol{\sqrt{3} \cdot \sqrt{3}}\)

Using the property \(\sqrt{a} \cdot \sqrt{a} = a\) (for \(a \geq 0\)), we have \(\sqrt{3} \cdot \sqrt{3} = 3\) (rational).

Step 2: Calculate \(\boldsymbol{\sqrt{5} \cdot \sqrt{7}}\)

Using the property \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\) (for \(a,b \geq 0\)), we get \(\sqrt{5 \cdot 7} = \sqrt{35}\) (irrational).

Step 3: Calculate \(\boldsymbol{\sqrt{2} \cdot \sqrt{18}}\)

First, simplify \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\). Then \(\sqrt{2} \cdot 3\sqrt{2} = 3 \cdot (\sqrt{2} \cdot \sqrt{2}) = 3 \cdot 2 = 6\) (rational).

Step 4: Calculate \(\boldsymbol{\sqrt{2} \cdot \sqrt{6}}\)

Using \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\), we have \(\sqrt{2 \cdot 6} = \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\) (irrational).

From these results, we see that the product of two irrational numbers (since \(\sqrt{3},\sqrt{5},\sqrt{7},\sqrt{2},\sqrt{6},\sqrt{18}\) are irrational) can be rational (like \(3, 6\)) or irrational (like \(\sqrt{35}, 2\sqrt{3}\)). So the conclusion is: The product of two irrational numbers can be rational or irrational.

Part D: Select Expressions with Rational Product

Expression 1: \(\boldsymbol{\frac{4}{3} \cdot \frac{12}{3}}\)

Multiply the numerators and denominators: \(\frac{4 \cdot 12}{3 \cdot 3} = \frac{48}{9} = \frac{16}{3}\) (rational, since it's a fraction of integers).

Expression 2: \(\boldsymbol{\frac{32}{4} \cdot \frac{35}{4}}\)

Simplify \(\frac{32}{4} = 8\), then \(8 \cdot \frac{35}{4} = 2 \cdot 35 = 70\) (rational).

Expression 3: \(\boldsymbol{\frac{\sqrt{3}}{2} \cdot \frac{22}{7}}\)

Multiply: \(\frac{\sqrt{3} \cdot 22}{2 \cdot 7} = \frac{11\sqrt{3}}{7}\) (irrational, because of \(\sqrt{3}\)).

Expression 4: \(\boldsymbol{\sqrt{11} \cdot \frac{2}{3}}\)

This is \(\frac{2\sqrt{11}}{3}\) (irrational, because of \(\sqrt{11}\)).

So the expressions with rational products are \(\frac{4}{3} \cdot \frac{12}{3}\) and \(\frac{32}{4} \cdot \frac{35}{4}\).

Final Answers

• For the first part (conclusion): The product of two irrational numbers can be rational or irrational.
• For Part D: \(\frac{4}{3} \cdot \frac{12}{3}\), \(\frac{32}{4} \cdot \frac{35}{4}\)

Answer:

First Set of Problems (Finding Products and Conclusion)

Step 1: Calculate \(\boldsymbol{\sqrt{3} \cdot \sqrt{3}}\)

Using the property \(\sqrt{a} \cdot \sqrt{a} = a\) (for \(a \geq 0\)), we have \(\sqrt{3} \cdot \sqrt{3} = 3\) (rational).

Step 2: Calculate \(\boldsymbol{\sqrt{5} \cdot \sqrt{7}}\)

Using the property \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\) (for \(a,b \geq 0\)), we get \(\sqrt{5 \cdot 7} = \sqrt{35}\) (irrational).

Step 3: Calculate \(\boldsymbol{\sqrt{2} \cdot \sqrt{18}}\)

First, simplify \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\). Then \(\sqrt{2} \cdot 3\sqrt{2} = 3 \cdot (\sqrt{2} \cdot \sqrt{2}) = 3 \cdot 2 = 6\) (rational).

Step 4: Calculate \(\boldsymbol{\sqrt{2} \cdot \sqrt{6}}\)

Using \(\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}\), we have \(\sqrt{2 \cdot 6} = \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\) (irrational).

From these results, we see that the product of two irrational numbers (since \(\sqrt{3},\sqrt{5},\sqrt{7},\sqrt{2},\sqrt{6},\sqrt{18}\) are irrational) can be rational (like \(3, 6\)) or irrational (like \(\sqrt{35}, 2\sqrt{3}\)). So the conclusion is: The product of two irrational numbers can be rational or irrational.

Part D: Select Expressions with Rational Product

Expression 1: \(\boldsymbol{\frac{4}{3} \cdot \frac{12}{3}}\)

Multiply the numerators and denominators: \(\frac{4 \cdot 12}{3 \cdot 3} = \frac{48}{9} = \frac{16}{3}\) (rational, since it's a fraction of integers).

Expression 2: \(\boldsymbol{\frac{32}{4} \cdot \frac{35}{4}}\)

Simplify \(\frac{32}{4} = 8\), then \(8 \cdot \frac{35}{4} = 2 \cdot 35 = 70\) (rational).

Expression 3: \(\boldsymbol{\frac{\sqrt{3}}{2} \cdot \frac{22}{7}}\)

Multiply: \(\frac{\sqrt{3} \cdot 22}{2 \cdot 7} = \frac{11\sqrt{3}}{7}\) (irrational, because of \(\sqrt{3}\)).

Expression 4: \(\boldsymbol{\sqrt{11} \cdot \frac{2}{3}}\)

This is \(\frac{2\sqrt{11}}{3}\) (irrational, because of \(\sqrt{11}\)).

So the expressions with rational products are \(\frac{4}{3} \cdot \frac{12}{3}\) and \(\frac{32}{4} \cdot \frac{35}{4}\).

Final Answers

• For the first part (conclusion): The product of two irrational numbers can be rational or irrational.
• For Part D: \(\frac{4}{3} \cdot \frac{12}{3}\), \(\frac{32}{4} \cdot \frac{35}{4}\)