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Question Put the following equation of a line into slope-intercept form, simplifying all fractions. \[ 10 x+8 y=72 \] Answer Attempt 10 out of 2

Ask by Davies Bernard. in the United States
Jan 27,2025

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The slope-intercept form of the equation is \( y = -\frac{5}{4}x + 9 \).

Solución

To convert the equation \( 10x + 8y = 72 \) into **slope-intercept form** (\( y = mx + b \)), follow these steps: 1. **Isolate the \( y \)-term**: \[ 8y = -10x + 72 \] 2. **Solve for \( y \)** by dividing every term by 8: \[ y = \frac{-10}{8}x + \frac{72}{8} \] 3. **Simplify the fractions**: \[ y = -\frac{5}{4}x + 9 \] So, the **slope-intercept form** of the equation is: \[ y = -\frac{5}{4}x + 9 \]

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To put the equation \( 10x + 8y = 72 \) into slope-intercept form (\( y = mx + b \)), we need to solve for \( y \). Start by isolating \( y \): 1. Subtract \( 10x \) from both sides: \[ 8y = -10x + 72 \] 2. Divide every term by \( 8 \) to solve for \( y \): \[ y = -\frac{10}{8}x + \frac{72}{8} \] Simplifying gives us: \[ y = -\frac{5}{4}x + 9 \] So, the slope-intercept form of the equation is \( y = -\frac{5}{4}x + 9 \). Here you go with some fun facts! Did you know that the slope of a line represents its steepness and direction? A positive slope goes upward to the right, while a negative slope, like the one we found (-5/4), goes downward! Slope-intercept form is super handy in real-life applications, like predicting trends in data. For example, if you're looking at how the temperature changes throughout the day, you can model that change using a line. The slope gives you the rate of temperature increase or decrease, while the y-intercept indicates the temperature at a specific time!

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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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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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