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Let's go through each problem step by step. 1. **Solve the inequalities:** 1) \(\frac{1}{2}x - 5 - 0.2 \times 3x - x^2 \geq 0\): Let's simplify and solve the expression: \[ \frac{1}{2}x - 0.6x - x^2 - 5 \geq 0 \] Simplify: \[ -0.1x - x^2 - 5 \geq 0 \] This is a quadratic inequality. Consider finding the roots and analyzing intervals if needed. 2) \(3.6x + 1 > 0\): \[ 3.6x > -1 \\ x > -\frac{1}{3.6} \] 2. **Solve the inequalities using interval method:** \((1) \, (x+3)(x+7) \leq 0\) Find roots: - \(x + 3 = 0 \Rightarrow x = -3\) - \(x + 7 = 0 \Rightarrow x = -7\) The intervals to test are \((-\infty, -7)\), \([-7, -3]\), and \((-3, \infty)\). 3. **Solve the equation:** \((2) \, 5 + 4 \cdot \frac{9z - 2}{z} = 0\) Simplify and solve for \(z\): \[ 5 + \frac{36z - 8}{z} = 0 \\ 5z + 36z - 8 = 0 \\ 41z = 8 \\ z = \frac{8}{41} \] 4. **Find the values where the expression has meaning:** \(\frac{\sqrt{x+3} - 5}{x^2 - 9x + 2}\) Ensure the denominator is not zero: \[ x^2 - 9x + 2 \neq 0 \] Solve the quadratic \(x^2 - 9x + 2 = 0\) using the quadratic formula if needed. 5. **Find the domain of the function:** \(y = \frac{1}{\sqrt{x^3}}\) Ensure the expression under the square root is non-negative and not zero: \[ x^3 > 0 \rightarrow x > 0 \] If specific calculations are needed for the quadratic solutions, apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Let me know if you need calculations for any specific part!