To complete the statement with inequalities, we need to find a number that is less than the cube root of 15 but also greater than a known number. To make the comparison easier, let's find an integer that, when cubed, is close to 15. Since \(2^3 = 8\) and \(3^3 = 27\), we can see that the cube root of 15 lies between 2 and 3. Thus, we can place these numbers in the inequalities to show that 2 is less than the cube root of 15 and the cube root of 15 is less than 3. The completed statement will look like this: \(2 < \sqrt[3]{15} < 3\) So you'll fill in the blanks with the numbers 2 and 3 to show that \(2 < \sqrt[3]{15} < 3\).
To solve the inequality \( 3|1 - 9b| \geq 28 \), you can follow these steps: 1. Divide both sides of the inequality by 3 to isolate the absolute value expression: \( |1 - 9b| \geq \frac{28}{3} \) 2. Recognize that an absolute value expression \( |x| \geq a \) where \( a > 0 \) can be split into two separate inequalities: \( x \geq a \) or \( x \leq -a \) In this case, \( x \) corresponds to \( 1 - 9b \) and \( a \) corresponds to \( \frac{28}{3} \). 3. Set up the two inequalities: \( 1 - 9b \geq \frac{28}{3} \) and \( 1 - 9b \leq -\frac{28}{3} \) 4. Solve each inequality separately: For \( 1 - 9b \geq \frac{28}{3} \): a) Subtract 1 from both sides: \( -9b \geq \frac{28}{3} - 1 \) \( -9b \geq \frac{28}{3} - \frac{3}{3} \) \( -9b \geq \frac{25}{3} \) b) Divide by -9, remembering to flip the inequality sign when dividing by a negative: \( b \leq -\frac{25}{27} \) For \( 1 - 9b \leq -\frac{28}{3} \): a) Subtract 1 from both sides: \( -9b \leq -\frac{28}{3} - 1 \) \( -9b \leq -\frac{28}{3} - \frac{3}{3} \) \( -9b \leq -\frac{31}{3} \) b) Divide by -9, flipping the inequality sign: \( b \geq \frac{31}{27} \) 5. Combine the two parts of the solution: \( b \leq -\frac{25}{27} \) or \( b \geq \frac{31}{27} \) This is the solution to the inequality.
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