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What are the release dates for The Bold and the Beautiful - 1987 1-6179?
The Bold and the Beautiful - 1987 1-6179 was released on: USA: 20 October 2011
What is the distance between Munich Germany and Boston?
The distance between Munich in Germany and Boston in the United States is 3839 miles. This is equal to 6179 kilometers.
What is the value of a 1 dollar silver certificate series 1935C in fair condition with a blue seal with an H 6179 on front right lower side of bill?
About $2.
Where is the Iron Mission Museum Foundation Inc in Cedar City Utah located?
The address of the Iron Mission Museum Foundation Inc is: 635 N Main Street, Cedar City, UT 84721-6179
What has the author Albert de Caluwe written?
Albert de Caluwe has written: 'Un Livre des morts sur bandelette de momie (Bruxelles, Muse es royaux d'Art et d'Histoire E. 6179)'
What are the ten youngest counties in Georgia?
150. Wheeler 6179 151. Stewart 5252 152. Baker 4074 153. Schley 3766 154. Echols 3754 155. Clay 3357 156. Quitman 2598 157. Glascock 2556 158. Webster 2390 159. Taliaferro 2077
What is 222444 divisible by?
1, 2, 3, 4, 6, 9, 12, 18, 36, 37, 74, 111, 148, 167, 222, 333, 334, 444, 501, 666, 668, 1002, 1332, 1503, 2004, 3006, 6012, 6179, 12358, 18537, 24716, 37074, 55611, 74148, 111222, 222444
How many miles is Paris from Washington DC?
The distance is straight path from one place to another place. There might be slight difference between the actual distance and the above mentioned distance because of the route chosen.The distance between the above mentioned places is 3839 miles approximately.
What are the factors of 222444?
The factors of 424380 are: 1, 2, 3, 4, 5, 6, 10, 11, 12, 15, 20, 22, 30, 33, 44, 55, 60, 66, 110, 132, 165, 220, 330, 643, 660, 1286, 1929, 2572, 3215, 3858, 6430, 7073, 7716, 9645, 12860, 14146, 19290, 21219, 28292, 35365, 38580, 42438, 70730, 84876, 106095, 141460, 212190, 424380
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Math sample papers -math olympaid?
These questions are from the Australasian Maths Olympiad Website.http://www.apsmo.info/APSMO_Home.phpA. (Time: 3 minutes)What is the value of:268 + 1375 + 6179 - 168 - 1275 - 6079 ?B. (Time: 5 minutes)Each of 8 boxes contains one or more marbles. Each box contains a different number of marbles, except for two boxes which contain the same number of marbles. What is the smallest total number of marbles that the 8 boxes could contain?C. (Time: 5 minutes)Find the whole number which is:less than 100;a multiple of 3;a multiple of 5;odd, and such that,the sum of its digits is odd.D. (Time: 6 minutes)Takeru has four 1 centimetre long blocks, three 5 centimetre long blocks, and three 25 centimetre long blocks. By joining these blocks to make different total lengths, how many different lengths of at least 1 centimetre can Takeru make?E. (Time: 6 minutes)The figure below is made up of 5 congruent squares. The perimeter of the figure is 72 cm. Find the number of square cm in the area of the figure.AnswersA. - 300METHOD 1: Make a simpler problem...Notice that each number being added is 100 more than one of the numbers being subtracted.The value is 100 + 100 + 100 = 300METHOD 2: Group by operation...Add the numbers 268 + 1375 + 6179 = 7822.Then add the numbers 168 + 1275 + 6079 = 7522.Finally, subtract the totals: 7822 - 7522 = 300B. - 29 marblesDraw a picture...Draw 8 boxes. Then put the smallest possible number of marbles in each box. Put 1 marble in box 1. Then put 1 marble in box 2. You can't just put 1 marble in box 3, because that would make three boxes with the same number of marbles. So put 2 marbles in box 3, 3 marbles in box 4, and so on.The smallest total number of marbles is 1+ 1 + 2 + 3 + 4 + 5 + 6 + 7 = 29 marbles.C. - 45Proceed one statement at a time. Eliminate those numbers which fail to satisfy all the conditions.WHOLE NUMBERS THAT SATISFY ALL CONDITIONSLess than 100 1, 2, 3, ..., 99Multiple of 3 3, 6, 9, ..., 99Also multiple of 5 15, 30, 45, 75, 90Odd 15, 45, 75Sum of digits is odd 45D. - 79METHOD 1: Start with a simpler problem...(a) Lengths formed by 1 cm blocks: 1, 2, 3, 4.(b) Lengths formed by remaining blocks: 5, 10, 15; 25, 30, 35, 40; 50, 55, 60, 65; 75, 80, 85, 90.(c) Each of the fifteen (b) length bocks can be combined with the four (a) lengths, thus producing 15 x 4 = 60 different amounts.TOTAL AMOUNTS:(a) 4(b) 15(c) 60GRAND TOTAL:79 different amountsMETHOD 2Number of choices for 1cm lengths, including 0, is 5: (0, 1, 2, 3, 4).Number of choices for 5cm lengths, including 0, is 4: (0, 1, 2, 3).Number of chioces for 25cm lengths, including 0, is 4: (0, 1, 2, 3).Total number of choices for all lengths is 5 x 4 x 4 = 80. However, 80 includes the choice of having none of the lengths as a choice. Since it is given that each length must be 1cm or longer, there are 80 - 1 = 79 amounts of at least 1cm.METHOD 3: Establish a maximum and then eliminate all impossibilities.Find that largest possible length that can be made with the blocks, and then subtract the number of smaller values that cannot be made. The maximum that can be made is 4 + 15 + 75 = 94cm. The 15 lengths less than 94cm that cannot be made are those that require four 5cm blocks. These are 20, 21, 22, 23, 24, 45, 46, 47, 48, 49, 70, 71, 72, 73 and 74cm lengths. The number of possible lengths is 94 - 15 = 79E. - 180 cm²Find the length of one side of the figure...Because of the common dies, all the squares are congruent to each other. The perimeter consists of 12 equal sides. The length of a side is 72 / 12 = 6cm. The area of each square is 6 x 6 = 36cm².The area of the figure is 5 x 36 = 180cm².You can buy Maths Olympiad BooksMATHS OLYMPIAD CONTEST PROBLEMSby Dr George Lenchner(Australian Edition. 2005. Reprinted with corrections 2008.)285 pagesISBN : 978-0-9757316-0-4MATHS OLYMPIAD CONTEST PROBLEMS Volume 2(Australian Edition. 2008.)Editors : R. Kalman, J. Phegan, A. Prescott320 pagesISBN : 978-0-9757316-2-8CREATIVE PROBLEM SOLVING IN SCHOOL MATHEMATICSby Dr George Lenchner(Australian Edition. 2006.)290 pagesISBN : 978-0-9757316-1-1
What is the highest record on world maths day?
In 2011, the event occurred on March 1st. 5.3 million students from over 215 countries combined to correctly answer 428,598,214 mathematics questions. The world champions of 2011 are: 4 to 7: Eric Z, Team Australia, Australia: 6,995 8 to 10: Mason F, Team New Zealand, New Zealand: 8,176 11 to 13: Kaya G, The Southport School, Australia: 8,273 14 to 18: David A, Fraser Coast Anglican College, Australia: 6,556 Top 5: 4 to 7: 1st Eric Z, Team Australia, Australia : 6995 2nd Vihangi R, Salcombe Prep School, United Kingdom : 6435 3rd Evan M, Stanley Bay School, New Zealand : 5980 4th Andrey M, Laudre San Pedro International College, Spain : 5973 5th Aditya C, Team United States, United States : 5929 8 to 10: 1st Mason F, Team New Zealand, New Zealand : 8176 2nd Sai M, India International School in Japan, Japan : 6642 3rd Edwin V, St Joseph's School(Pukekohe), New Zealand : 6547 4th Sachin Kumar M, Canadian International School Hong Kong, Hong Kong : 6534 5th Rohith N, Team Japan, Japan : 6179 11 to 13: 1st Kaya G, The Southport School, Australia : 8273 2nd David M, Aloha College - Secondary School, Spain : 7240 3rd Satvik R, St George's The English International School, Germany : 7232 4th Harish S, Team India, India : 7154 5th Moosa F, Beaconhouse School System Mandi Bahauddin, Pakistan : 6965 14 to 18: 1st David A, Fraser Coast Anglican College, Australia : 6556 2nd Tham C, Team Malaysia, Malaysia : 6502 3rd TIGER Z, Team Malaysia, Malaysia : 6188 4th Yeoh K, Team Malaysia, Malaysia : 6115 5th Edwin S, Cempaka Schools, Malaysia : 5929
