l. The solution to 0 lt; (x + 144)(x + 67) is A ‘7?u1 B ‘_””_:.,*X C lt;_1g,~ D ‘_:u;,gt;x Use the following graph to answer question 2, y
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I dont understand how the find the solution in either of these equations.
l. The solution to 0 < (x + 144)(x + 67) is A ‘—7?u—§1—"B ‘—_““—_:.,—*XC <—_1°«—g,—~D ‘—_:u—;,—>x Use the following graph to answer question 2, y 2. The solution to 11:02 0 is (“,5 <x<~2,xER}{xI-SSxS-2,xeR}{xI-S >xorx> 2,xER}{xI—52x0r122,xER} pow?» ADLC Mathematlcs 20-1
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