Question

1. on monday, the low tide at cook inlet, alaska is $-1\frac{1}{2}$ ft. on tuesday, the low tide rises $1\frac{1}{3}$ ft. what is the new position of the tide on tuesday?

Explanation:

Step1: Convert mixed numbers to improper fractions

The initial tide level on Monday is \(-1\frac{1}{2}\) ft. Converting this to an improper fraction: \(-1\frac{1}{2}=-\frac{2\times1 + 1}{2}=-\frac{3}{2}\) ft. The rise in tide on Tuesday is \(1\frac{1}{3}\) ft. Converting this to an improper fraction: \(1\frac{1}{3}=\frac{3\times1+ 1}{3}=\frac{4}{3}\) ft.

Step2: Add the two fractions to find the new tide level

To add \(-\frac{3}{2}\) and \(\frac{4}{3}\), we need a common denominator. The least common denominator of 2 and 3 is 6. So, we rewrite the fractions: \(-\frac{3}{2}=-\frac{3\times3}{2\times3}=-\frac{9}{6}\) and \(\frac{4}{3}=\frac{4\times2}{3\times2}=\frac{8}{6}\). Now, adding them together: \(-\frac{9}{6}+\frac{8}{6}=\frac{-9 + 8}{6}=-\frac{1}{6}\) ft? Wait, no, wait. Wait, the problem says the low tide on Monday is \(-1\frac{1}{2}\) (which is \(-\frac{3}{2}\)) and it rises \(1\frac{1}{3}\) (which is \(\frac{4}{3}\))? Wait, no, maybe I made a mistake. Wait, \(-1\frac{1}{2}\) is \(-\frac{3}{2}\) and \(1\frac{1}{3}\) is \(\frac{4}{3}\)? Wait, no, let's check again. Wait, \(-1\frac{1}{2}\) is \(-\frac{3}{2}\), and the rise is \(1\frac{1}{3}\), so the new level is \(-\frac{3}{2}+\frac{4}{3}\). Let's compute that: \(-\frac{3}{2}+\frac{4}{3}=\frac{-9 + 8}{6}=-\frac{1}{6}\)? Wait, that can't be right. Wait, maybe the original problem has a typo? Wait, no, maybe I misread the numbers. Wait, the problem says "the low tide at Cook Inlet, Alaska is \(-1\frac{1}{2}\) ft. On Tuesday, the low tide rises \(1\frac{1}{3}\) ft." Wait, no, maybe the rise is \(1\frac{1}{2}\)? Wait, the user's image shows "1 1/3"? Wait, maybe it's a typo, but assuming the numbers are correct. Wait, no, let's do the calculation again. \(-1\frac{1}{2}=-\frac{3}{2}\), \(1\frac{1}{3}=\frac{4}{3}\). The common denominator is 6. So \(-\frac{3}{2}=-\frac{9}{6}\), \(\frac{4}{3}=\frac{8}{6}\). Adding them: \(-\frac{9}{6}+\frac{8}{6}=-\frac{1}{6}\). But that seems odd. Wait, maybe the rise is \(1\frac{1}{2}\)? Let's check. If the rise is \(1\frac{1}{2}\), then \(-\frac{3}{2}+\frac{3}{2}=0\). But the problem says \(1\frac{1}{3}\). Alternatively, maybe I misread the initial tide. Wait, the initial tide is \(-1\frac{1}{2}\) (which is -1.5) and the rise is \(1\frac{1}{3}\) (which is approximately 1.333). So -1.5 + 1.333 = -0.166..., which is \(-\frac{1}{6}\). So that's correct.

Wait, but maybe the problem was supposed to have the rise as \(1\frac{1}{2}\). But according to the given problem, it's \(1\frac{1}{3}\). So the calculation is \(-1\frac{1}{2}+1\frac{1}{3}=-\frac{3}{2}+\frac{4}{3}=\frac{-9 + 8}{6}=-\frac{1}{6}\) ft. But that seems like a small rise. Alternatively, maybe the initial tide is \(-1\frac{1}{3}\) and the rise is \(1\frac{1}{2}\)? No, the problem says \(-1\frac{1}{2}\) and \(1\frac{1}{3}\). So the answer is \(-\frac{1}{6}\) ft? Wait, but that seems odd. Wait, maybe I made a mistake in the signs. The low tide is \(-1\frac{1}{2}\) (below sea level), and it rises \(1\frac{1}{3}\) ft (moves up towards sea level). So the new position is \(-1\frac{1}{2}+1\frac{1}{3}\). Let's convert to decimals: \(-1.5 + 1.333... = -0.166...\), which is \(-\frac{1}{6}\) ft. So that's correct.

Answer:

The new position of the tide on Tuesday is \(\boxed{-\frac{1}{6}}\) feet.