Solution: a g. p has a first term of a, a common ratio of r and its 6th term is 768. another g. p has a first term of a, a common ratio of 6r and its 3rd term is 3456. evaluate a and r.

1234567891011 match the reasons with the statements in the proof. given: j | | k m 1 = m 3 prove: l | | m 1. j||k, m∠3 = m∠1 if lines are ||, then corresponding angles are =. 2. m∠1 = m∠2 if alternate interior angles are =, then lines are ||. 3. m∠2 = m∠3 substitution 4. l||m given

A33 gram sample of a substance that's used to detect explosives has a k-value of 0.1473. n = noekt no = initial mass (at time t = 0) n = mass at time t k = a positive constant that depends on the substance itself and on the units used to measure time t = time, in days

Polygon abcde is reflected to produce polygon a′b′c′d′e′. what is the equation for the line of reflection?