2-8 Practice Literal Equations and Dimensional Analysis Solve each equation or formula for the variable indicated. 1. = for 2. 6w = 2z, for d rt, r y w – 3. + 4y = 3t, for 4. 9s 5g = for mx x s – –4u, + 3c = 2x, for 2p = for 5. 6. ab b kx t, x – 7. + = + for 8. + = for a a r, m g d, h 9. + = for 10. = for v x, y q k, a – = for = for 11. 12. h, x c, b 13. 2w = 7w 2, for 14. 3ℓ + = 5 + 5ℓ, for y w y – – ℓ The formula for Ohm’s Law is = where represents voltage measured in volts, represents 15. ELECTRICITY E IR, E I current measured in amperes, and represents resistance measured in ohms. R a. Solve the formula for R. b. Suppose a current of 0.25 ampere through a resistor connected to a 12-volt battery. What is the resistance in the circuit?
2-8 Practice Literal Equations and Dimensional Analysis Solve each equation or formula for the variable indicated. 1. d = rt, for r 2. 6w – y = 2z, for w 3. mx + 4y = 3t, for x 4. 9s – 5g = –4u, for s 5. ab + 3c = 2x, for b 6. 2p = kx – t, for x 7. + a = a + r, for m 8. + g = d, for h 9. + v = x, for y 10. – q = k, for a 11. = h, for x 12. = c, for b 13. 2w – y = 7w – 2, for w 14. 3ℓ + y = 5 + 5ℓ, for ℓ 15. ELECTRICITY The formula for Ohm’s Law is E = IR, where E represents voltage measured in volts, I represents current measured in amperes, and R represents resistance measured in ohms. a. Solve the formula for R. b. Suppose a current of 0.25 ampere through a resistor connected to a 12-volt battery. What is the resistance in the circuit?
day_15-_2-8_hw_practice.docx