 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum  MENU          Solutions for Session 8, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5 | B6| B7 | B8 | B9 |
B10 | B11 | B12 | B13    Problem B1

Here is the completed table:

 Number of teachers Amount each will receive 1 800,000 2 400,000 3 266,666.66 4 200,000 10 80,000 15 53,333.33 20 40,000 100 8,000

The last two entries will vary.   Problem B2 One rule is that the product of the input and the output is always 800,000. If M = the money each teacher receives, and T = the number of teachers, then (M) x (T) = 800,000. This is because the total prize of 800,000 is split evenly among the teachers. This also leads to a second rule: M = 800,000 / T, because the money received by each teacher is 800,000 divided by how many teachers split the prize. Additionally, T = 800,000 / M.   Problem B3 The amount of money decreases as the number of teachers increases, but it is not an exponential decay because the ratio between consecutive outputs is not constant. Looking at differences between outputs guarantees that the graph is neither linear nor quadratic. The graph is not cyclic, because there is no point where the outputs begin to repeat.    Problem B4

Here is the completed table:

 Length Width Area = length * width 50 40 2000 25 80 2000 100 20 2000 1,000 2 2000 40 50 2000 80 25 2000 2000 1 2000 0.5 4000 2000

The last four entries will vary.   Problem B5 One equation is x * y = 2,000. Note that Area = length * width is always constant at 2,000. Other possible equations are y = 2,000 / x and x = 2,000 / y.   Problem B6    Problem B7

Overall, if x is multiplied by a number, y is divided by the same number. If x is divided by a number, y is multiplied by the same number.

 a. Here is the completed table. All y-values are rounded to one decimal place.

 x y Decrease in y 20 100 -- 30 66.7 33.3 40 50 16.7 50 40 10 60 33.3 6.7 70 28.6 4.7 80 25 3.6 90 22.2 2.8 100 20 2.2

 b. As x increases by 10, y continuously decreases, but the rate of decrease lessens as x grows. c. If x doubles, y is cut in half. If x triples, y is divided by three. d. If x is very small, y must be very large, since the product of x and y is always the same. By the same argument, if x is very large, then y must be very small.   Problem B8

Let's try it:

 x y -10 -0.3 -6 -0.5 -3 -1 -2 -1.5 -0.5 -6 0.5 6 2 1.5 3 1 6 0.5 10 0.3   Problem B9 If x = 0, y is undefined, which means that no value of y will make zero times y equal to 3. Using a calculator, 3 / 0 will fail to return a number; the calculator will give an error message.   Problem B10 The graph will not cross the y-axis, because if it did, it would mean that some y-value would be assigned for x = 0 for the graph, and, as explained in the solution to Problem B9, there is no such value. Another way to look at it is to examine the behavior of the graph near x = 0. If x is a little more than zero, y is a very large, positive number, but if x is a little less than zero, y is a very large, negative number. These portions of the graph do not meet.   Problem B11 The 3 represents the area of a rectangle drawn with (0, 0) as one corner and any point on the graph as the opposite corner. Compare this to the Interactive Activity, where there were rectangles of area equaling 2,000 feet -- length times width always equaled 2,000. Here, x is the length, and y is the width.   Problem B12 In a direct variation function, an increase in x creates a proportional increase in y; if x is multiplied by 5, y is also multiplied by 5. But in inverse variation, the opposite is true: If x is multiplied by 5, y is divided by 5. The word "inverse" refers to the inverse operations of multiplication and division.   Problem B13 In this particular equation, x is the reciprocal of y, because x and y multiply together to make 1. The reciprocal is used to solve equations like 5n = 16 -- multiplying both sides of the equation by the reciprocal of 5 produces two numbers (5 and 1/5) which, when multiplied, make 1. So, multiplying by 1/5 will remove the 5 from the left side, leaving variable n by itself. This is the key to solving equations by backtracking.     Session 8: Index | Notes | Solutions | Video

© Annenberg Foundation 2017. All rights reserved. Legal Policy