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Patterns, Functions, and Algebra
Session 5 Part A Part B Part C Part D Part E Homework
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Session 5 Materials:



Solutions for Session 5, Homework

See solutions for Problems: H1 | H2 | H3 | H4

Problem H1

Earlier we found that Achilles ran 13 1/2 miles in 1 1/2 hours. Therefore, we have to find a way to get the tortoise to the 13 1/2-mile mark after 1 1/2 hours. The tortoise walks at 1 mile per hour, so it can walk 1 1/2 miles in that time. The remaining distance must be its head start: 13 1/2 - 1 1/2 = 12 miles.

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Problem H2

Use the same logic as in Problem H1. Achilles runs 27 miles in the 3 hours, therefore the tortoise needs a head start that will get it to the 27-mile mark after 3 hours. Because it walks at 2 miles per hour, it can walk 6 miles in 3 hours, so the head start is 27 - 6 = 21 miles. An algebraic equation could also be used for this problem.

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Problem H3

It works because the roof is a straight line, and therefore it has a constant rate of change. By carefully measuring the 1st and 2nd support, the carpenter has calculated a rate of change: (change in height of support) / (distance between supports). Since this rate is constant, and the distance between supports stays the same, the change in the support's height must also be constant. This is identical to predicting the next number in the output of a linear function; in this case, the output drops by 3 for every new support.

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Problem H4

It can be done by solving the algebraic equation for the other variable, using the technique of "undoing" that was first used in Session 3. For the equation d = 3t + 2, start by subtracting 2 from each side to produce d - 2 = 3t. Then divide both sides by 3, so that the equation is (d - 2) / 3 = t. If a linear function can be undone, the result will always be a new, linear function. The only linear functions which cannot be undone are constant functions like y = -7. See Session 5, Problem D7 and Session 3, Problems E9-E12.

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