Solve for z \( \begin{array}{l}z+5>15 \text { or } 5-2 z>9 \\ \text { z } z<2 \text { or } z>-2 \\ z>10 \text { or } z>-2 \\ z>-10 \text { or } z<2 \\ z>10 \text { or } z<-2 \\ \text { Submit Question } \\ \text { S }\end{array} \)

To solve the inequalities, let's break each one down. 1. For \( z + 5 > 15 \): - Subtract 5 from both sides: \( z > 10 \). 2. For \( 5 - 2z > 9 \): - Subtract 5 from both sides: \( -2z > 4 \). - Divide by -2 (remember to reverse the inequality): \( z < -2 \). Now we have: - From the first inequality: \( z > 10 \) - From the second inequality: \( z < -2 \) Thus, the solution to the first set of inequalities is \( z > 10 \) or \( z < -2 \). Next, let's analyze the second inequalities: 1. \( z z < 2 \) is likely a typo and should be rewritten more clearly. Assuming it's \( z^2 < 2 \): - This gives us \( -\sqrt{2} < z < \sqrt{2} \). For the complete set, here are our interpretations: - \( z > 10 \) or \( z < -2 \) - \( -\sqrt{2} < z < \sqrt{2} \) The final solution state covers the ranges: - Either \( z \) can be greater than 10. - Or \( z \) is less than -2, which includes some values but excludes anything between -√2 and √2. Thus, summarizing, The complete solution for z can be: \( z < -2 \) or \( z > 10\).