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Onal Equations Why should restrictions on the variablo in a rational equation be listed before you begin solving the equation A. To avoid any values of the variablo that mako a dencminator zero the correct answer below B. To uvoid any negative values of the variable in a numetator C. Io avoid any values of the variasle that make a rumerator zero D. To avord any negative values of the varable in a donominator

Ask by Adkins Pierce. in the United States
Jan 26,2025

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To avoid values that make the denominator zero, list restrictions before solving the equation.

Solución

The correct answer is: **A. To avoid any values of the variable that make a denominator zero.** ### Explanation: When solving rational equations, it is crucial to identify and restrict values of the variable that would make any denominator equal to zero. This is because division by zero is undefined in mathematics, and including such values in the solution set would lead to invalid results. By listing these restrictions before solving the equation, you ensure that the solutions you find are valid and do not lead to undefined expressions.

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The Deep Dive

It's crucial to identify restrictions on the variable in a rational equation because certain values can lead to undefined scenarios, particularly with denominators. For instance, if you plug in a value that makes a denominator zero, the whole equation crumbles. So, listing those restrictions upfront helps you dodge major pitfalls and makes the solution process smoother! Moreover, thinking ahead can save you from unnecessary headaches while solving equations. If you accidentally include a value that violates your restrictions, you might find yourself getting incorrect answers or needing to backtrack. It’s like mapping out a journey before hitting the road—ensures you get to your destination without unexpected detours!

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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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