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Home Question Bank Math Algebra Archive 2025 January 14

Algebra Questions from Jan 14,2025

Browse the Algebra Q&A Archive for Jan 14,2025, featuring a collection of homework questions and answers from this day. Find detailed solutions to enhance your understanding.

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ons \( -4 x+y=5 \) and \( -x+4 y=12 \) Muliply \( \frac{1}{8}(8 a-24) \) Dhe line parallel to \( 3 x+y=8 \) that passes through \( (0,-4) \) The line parallel to \( y=-3 x-2 \) that passes through \( (-2,7) \) 13. \( \left(p^{6}\right)^{3} \) 14. \( \left(q^{-4}\right)^{5} \) 15. \( 5^{3} \cdot 5^{-7} \) 16. \( -4 \cdot(-4)^{-2} \) 17. \( \frac{x^{7}}{x^{4}} \cdot x^{2} \) 18. \( \frac{v^{5} \cdot v^{3}}{v^{2}} \) 19. \( \left(-8 t^{2}\right)^{3} \) 20. \( \left(-\frac{q^{4}}{5}\right)^{-3} \) 21. \( \left(\frac{1}{3 h^{5}}\right)^{-4} \) In Exercises 22 and 23, simplify the expression. Write your answer using only \( y=-4 x-5 \) yhe line with slope 5 that passes through \( (2,-5) \) In Exercises \( 7-21 \), simplify the expression. Write your answer using only positi exponents. \( \begin{array}{lll}\text { 7. } \frac{7^{-2} m^{0}}{n^{-4}} & \text { 8. } \frac{(-9)^{0} j^{-1} k^{-4}}{2^{0}} & \text { 9. } \frac{5^{-2} w^{0}}{y^{-10}} \\ \begin{array}{lll}\text { 10. } \frac{t^{-5}}{8^{-2} s^{-3}} & \text { 11. } \frac{3^{-2} a^{-1}}{9^{-1} b^{-2} c^{0}} & \text { 12. } \frac{17 x^{0} y^{-8}}{4^{-2} z^{-6}} \\ \text { 13. }\left(p^{6}\right)^{3} & \text { 14. }\left(q^{-4}\right)^{5} & \text { 15. } 5^{3} \cdot 5^{-7} \\ \text { 19. }\left(-8 t^{2}\right)^{3} & \text { 17. } \frac{x^{7}}{x^{4} \cdot x^{2}} & \text { 18. } \frac{v^{5} \cdot v^{3}}{v^{2}}\end{array}\end{array} \begin{array}{lll}\text { 16. }-4 \cdot(-4)^{-2} & \text { 21. }\left(\frac{1}{3 h^{5}}\right)^{-4}\end{array} \) The line with slope -4 that passes through \( (0,-5) \) 2. If \( f(x)=x^{2}+5 x-7 \) and \( g(x)=x+1 \), find each of the following: \[ \begin{array}{l|l|l} & \text { a. } f(-3) \\ =-3^{2}+5(-3)-7 & \begin{array}{l} \text { b. } g(f(-2)) \\ =9-15-7 \\ =9-2)=-2^{2}+5(-2)-7 \\ \\ =9 \end{array} & \begin{array}{l} \text { c. Find } k \text { so that } g(k+3)=2 g(2 k) \\ f(-2)=4-10-7 \end{array} \\ f(K+3)+1=((2 K)+1 \\ f(-2)=-13 & K+4=4 K+2 \\ g(-13)=-13+1 & \frac{5 k}{5}=\frac{6}{5} \quad K=\frac{6}{5} \\ g(-13)=-12 & \end{array} \] e following: Find \( k \) so that \( g(k+3)=2 g(2 k) \) (6pts)
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