8th graders: click on the following link for review questions before Tuesday's test on Radicals https://sites.google.com/site/lassmath/my-forms Problem of the Week 1/7/19 Tom and Bill
Tom is standing in a hole that is 4 feet deep. Bill asks him how much deeper he is going to dig the hole. Tom replies that he will dig 4 feet 2 inches deeper and that the top of his head will then be the same distance below ground level that it is now above ground level. How tall is Tom? 12/17/18 Wire Length In the diagram, the rectangular wire grid contains 15 identical squares. The length of the grid is 10. What is the length of wire needed to construct the grid?  Solution 76. Since the length of the rectangular grid is 10, the side of each square in the grid is 10 ÷ 5 = 2. The height of the grid is therefore 6 (3 squares). There are four horizontal wires, each of length 10, and six vertical wires, each of length 6, for a total length of wire of 4 • 10 + 6 • 6 = 40 + 36 = 76. 12/10/18 How Tall Was the Tree? Lightning hit a tree one-fourth of the distance up the trunk from the ground and broke the tree so that its top landed 60 feet from its base, thus creating a triangle. How tall was the tree before it broke?  Solution Let hbe the height, in feet, of the tree before it fell. Then (h/4)2+ 602= (3h/4)2, so h2/16 + 3600 = 9h2/16. Solving for hreveals that h2/2 = 3600, so h= 12/3/18 Facetious In how many possible arrangements of the letters in the word FACETIOUS are the vowels in alphabetical order? Solution
A total of 9! (9x8x7x6x5x4x3x2x1) ways exist to arrange the letters in the word FACETIOUS (nine ways of picking the first letter exist, eight ways to pick the second exist, and so on). For each arrangement of the consonants in the word, the vowels can be arranged in 5! (5x4x3x2x1) ways. In only one of them are the vowels in alphabetical order. Hence, 1/5! of the arrangements have the vowels in alphabetical order, for a total of (9!/5!), or 3024.
11/26/18 Townhouses A contractor is asked to build a new set of townhouses in attached clusters of different sizes. He created plans for one-, two-, and three-house clusters, as shown in the diagram below. The builder used computer software to draw the line segments used to represent the houses. How many line segments are needed to draw 4 houses? 10 houses? 47 houses? nhouses? Explain your strategy.  Solution 21; 51; 236; 5n+ 1. One house takes 6 segments. Each additional house only requires 5 additional segments. Therefore, the sequence for the number of segments is 6, 11, 16, 21, 26, 31, 36, . . . . If n is the number of houses, then the nth term for this sequence is 5n + 1, or 5 times the number of houses plus 1. The 5th house would require 5(5) + 1 = 26 segments; the 10th house would require 5(10) + 1 = 51 segments; and the 47th house would require 5(47) + 1 = 236 segments. Another method is to begin with 6 segments for the first house and notice that each additional house adds 5 segments to the total. Therefore, the total number of segments is 6 + 5(n – 1) = 5n + 1. 11/12/18 Classmates' Birthdays
Your mathematics teacher displays a monthly calendar, which highlights all the birthdays of your classmates. Your mathematics class has 25 students. What is the probability that 3 or more students were born in the same month? Solution 1, or 100 percent. In a class of 24 students, it is possible for exactly 2 students to have been born in each of the 12 months. For 25 or more students, it is certain that there would be at least 1 month containing 3 or more birthdays. Therefore, the probability is 1 or 100%. 11/5/18 Cards in a Hat
Cards numbered 1 to 50 are put in a hat. What is the probability that the first two cards chosen at random have prime numbers on them? Solution 3/35, or about 0.086. There are 15 prime numbers in the range of 1 to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Remember that 1 is not a prime number. The probability of selecting a prime on the first pick is 15 out of 50; for the second pick, the probability is 14/49. The probabilities are combined by multiplying them, so 15/50 × 14/49 = 210/2450, or 3/35, or about 0.086. 10/29/18 How Many?
How many two-digit numbers exist such that when the product of its digits is added to the sum of its digits, the result is equal to the original two-digit number? Solution
Nine two-digit numbers: 19, 29, 39, 49, 59, 69, 79, 89, and 99. Any two-digit number can be represented by 10a + b. The product of the digits is ab, and the sum is a + b. When does ab + (a + b) = 10a + b? Solving, when ab = 9a, or when b = 9. |
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