Algebra Questions from Dec 01,2024

Browse the Algebra Q&A Archive for Dec 01,2024, featuring a collection of homework questions and answers from this day. Find detailed solutions to enhance your understanding.

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QUESTÃO 11 (VALOR 0,04 cada total) Determine o valor de: \begin{tabular}{l|l} a) \( O \mathrm{mmc}(5 a, 5 b) \), sabendo que \( \circ \mathrm{mmc}(a, b)=25 \). & b) O mdc \( \left(\frac{a}{4}, \frac{b}{4}\right) \), sabendo que o mdc \( (a, b)=64 \). \\ \( \mathrm{mmc}(5 a, 5 b)= \) & \( \mathrm{mdc}\left(\frac{a}{4}, \frac{b}{4}\right)= \)\end{tabular} Hans will rent a car for the weekend. He can choose one of two plans. The first plan has an initial fee of and costs an additional per mile driven. The second plan has an initial fee of and costs an additional per mile driven. For what amount of driving do the two plans cost the same? QUESTÃO 11 (VALOR 0,04 cada total) \begin{tabular}{l|l} Determine o valor de: \\ \( \begin{array}{ll}\text { a) } O \mathrm{mmc}(6 a, 6 b) \text {, sabendo que } \circ \mathrm{mmc}(a, b)=36 . & \text { b) O } \mathrm{mdc}\left(\frac{a}{3}, \frac{b}{3}\right) \text {, sabendo que o } \mathrm{mdc}(a, b)=81 \\ \mathrm{mmc}(6 a, 6 b)= & \end{array} \)\end{tabular} Solving a word problem using a system of linear equations of the form y ... A phone company offers two monthly plans. Plan A costs plus an additional for each minute of calls. Plan B costs plus an additional for each minute of calls. For what amount of calling do the two plans cost the same? \begin{tabular}{|c|c|}\hline System B & The system has no solution. \\ \( \begin{array}{c}(x, y)=(\square, \square\end{array} \) \\ \( 4 x=4+y \) & The system has a unique solution: \\ \( -4 x+y=-4 \) & The system has infinitely many solutions. \\ They must satisfy the following equation: \\ \( y=\square \)\end{tabular} 32) Which of the following describes what to do to the first number in each ordered pair shown above to obtain the corresponding second number? 4291. tenglamani yeching. 5. tenglamani yeching. Multiply. Simplify your answer as much as possible.
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