3 7 10 2 5 July 8, 2019 16 32 64 128 7 10 x 3 2 6 5 3 7 3z )= 2 10 5z – 3 5 x 7 10 2 5 3 7 10 x 2 3 5 3 7 3z )+ 2 5z 10 )= 0 3 5y 2 7 10 3 5z 2 7 10 3 2 2x 5 7 10 Solve for s. – 3 5 x 7 10 2 5 7 10 blank 3 2 6 5 5 x 7 10 2 mũ 3 5 2 3 5 7 10 as a fraction 3 5x 1 2 7 10 3 7 x 28 5 2 10 2 3x 5 7 3 10
