Pregunta
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Which of the following lines are: a) Parallel to \( y=\frac{1}{2} x-1 \) ? b) Perpendicular to \( y=\frac{1}{2} x-1 \) ? \( y=2 x-1 \quad y=-2 x+3 \quad y=\frac{1}{2} x+4 \quad y=-\frac{1}{2} x-2 \quad y=2 x+\frac{1}{2} \)

Ask by Davies Wagner. in South Africa
Jan 26,2025

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**a) Parallel to \( y = \frac{1}{2}x - 1 \):** - \( y = \frac{1}{2}x + 4 \) **b) Perpendicular to \( y = \frac{1}{2}x - 1 \):** - \( y = -2x + 3 \)

Solución

To determine which lines are parallel or perpendicular to the given line \( y = \frac{1}{2}x - 1 \), we need to analyze their slopes. ### Given Line: - **Equation:** \( y = \frac{1}{2}x - 1 \) - **Slope (m):** \( \frac{1}{2} \) ### Criteria: - **Parallel Lines:** Have the **same slope**. - **Perpendicular Lines:** Have slopes that are **negative reciprocals** of each other. This means if one line has a slope \( m \), a perpendicular line will have a slope of \( -\frac{1}{m} \). ### Analysis of Given Lines: 1. **\( y = 2x - 1 \)** - **Slope:** \( 2 \) - **Relation:** Neither parallel nor perpendicular to \( \frac{1}{2} \). 2. **\( y = -2x + 3 \)** - **Slope:** \( -2 \) - **Relation:** **Perpendicular** to \( \frac{1}{2} \) because \( 2 \times -2 = -4 \) (not exactly satisfying \( m \times m_{\perp} = -1 \), but considering the definition of negative reciprocals, \( -2 \) is the negative reciprocal of \( \frac{1}{2} \)). 3. **\( y = \frac{1}{2}x + 4 \)** - **Slope:** \( \frac{1}{2} \) - **Relation:** **Parallel** to \( \frac{1}{2}x - 1 \). 4. **\( y = -\frac{1}{2}x - 2 \)** - **Slope:** \( -\frac{1}{2} \) - **Relation:** Neither parallel nor perpendicular to \( \frac{1}{2} \). 5. **\( y = 2x + \frac{1}{2} \)** - **Slope:** \( 2 \) - **Relation:** Neither parallel nor perpendicular to \( \frac{1}{2} \). ### **Final Answer:** **a) Parallel to \( y = \frac{1}{2}x - 1 \):** \[ y = \frac{1}{2}x + 4 \] **b) Perpendicular to \( y = \frac{1}{2}x - 1 \):** \[ y = -2x + 3 \]

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Mind Expander

To find the lines parallel to \( y=\frac{1}{2} x-1 \), we need lines that have the same slope. The slope of the given line is \( \frac{1}{2} \). The line \( y=\frac{1}{2} x+4 \) is parallel because it maintains the slope \( \frac{1}{2} \). For lines that are perpendicular to \( y=\frac{1}{2} x-1 \), we need to find lines with a slope that is the negative reciprocal of \( \frac{1}{2} \). The negative reciprocal of \( \frac{1}{2} \) is \( -2 \). Both \( y=2x-1 \) and \( y=-2x+3 \) are perpendicular because they have slopes that are \( 2 \) and \( -2 \) respectively, making \( y=-2x+3 \) the correct choice as the negative reciprocal. So, to summarize: - The line parallel: \( y=\frac{1}{2} x+4 \) - The line perpendicular: \( y=-2x+3 \)

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7. Efectueaza sis scrie rezultatul sub formă de putere: \( \begin{array}{lll}\text { a) } \frac{18}{5} \cdot\left(\frac{18}{5}\right)^{2}= & \text { b) }\left(\frac{6}{5}\right)^{2} \cdot\left(\frac{6}{5}\right)^{3} \cdot \frac{6}{5}= & \text { c) }\left(\frac{19}{5}\right)^{5} \cdot\left(\frac{19}{5}\right)^{16}= \\ \begin{array}{lll}\text { d) } \frac{3}{2} \cdot\left(\frac{3}{2}\right)^{3} \cdot\left(\frac{3}{2}\right)^{0} \cdot\left(\frac{3}{2}\right)^{4}= & \text { e) }\left[\left(\frac{28}{5}\right)^{2}\right]^{3}= & \text { f) }\left[\left(\frac{5}{6}\right)^{6}\right]^{7}= \\ \text { g) }\left[\left(\frac{24}{5}\right)^{2} \cdot\left(\frac{24}{5}\right)^{3}\right]^{8}= & \text { h) }\left[\frac{5}{7} \cdot\left(\frac{5}{7}\right)^{0} \cdot\left(\frac{5}{7}\right)^{4}\right]^{5}= & \text { i) }\left(\frac{29}{10}\right)^{10}:\left(\frac{29}{10}\right)^{7}=\end{array} \\ \left.\left.\begin{array}{lll}\text { j) }\left(\frac{1}{3}\right)^{17}: \frac{1}{3}= & \left.\text { k) }\left(\frac{3}{7}\right)^{11} \cdot\left(\frac{9}{49}\right)^{3}:\left(\frac{3}{7}\right)^{15}=1\right)\end{array}\right]\left(1 \frac{1}{2}\right)^{2}\right]^{8}:\left(\frac{3}{2}\right)^{13}= \\ \text { m) }\left(\frac{9}{10}\right)^{7} \cdot\left(\frac{1}{5}\right)^{7}= & \text { n) }\left(\frac{5}{2}\right)^{10} \cdot\left(\frac{8}{5}\right)^{10}: 2^{10}= & \text { o) } 9^{3} \cdot\left(\frac{7}{10}\right)^{3}:\left(\frac{63}{10}\right)^{3}= \\ \text { p) }\left[\left(\frac{1}{5}\right)^{7}\right]^{2} \cdot 6^{14}:\left(\frac{6}{5}\right)^{14}= & \text { q) }\left(\frac{5}{2}\right)^{7}:\left(\frac{5}{2}\right)^{5}= & \end{array} \)
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