How old is diophantus when he died? please include some equation?
Q. Hos old is diophantus when he died? please include some equation.? In diophantus epitaph the following is scripted:"A 6th of his life was spent in childhood, a 12th as a youth, and 7th as a bachelor. 5 years after his marriage, his son was born but died 4 years after his own death at half his age"
Asked by Aldrin D - Sat Feb 14 01:20:15 2009 - - 1 Answers - 0 Comments
A. https://nrich.maths.org/d iscus/messages/114352/114 482.html?1164199566
Answered by Chuck N - Sun Feb 15 01:21:34 2009
Q. Hos old is diophantus when he died? please include some equation.? In diophantus epitaph the following is scripted:"A 6th of his life was spent in childhood, a 12th as a youth, and 7th as a bachelor. 5 years after his marriage, his son was born but died 4 years after his own death at half his age"
Asked by Aldrin D - Sat Feb 14 01:20:15 2009 - - 1 Answers - 0 Comments
A. https://nrich.maths.org/d iscus/messages/114352/114 482.html?1164199566
Answered by Chuck N - Sun Feb 15 01:21:34 2009
Solve the diophantus's problem I 27 by the method of I 28?
Q. To find two numbers such that their sum and product are given. Diophantus gives the sum as 20 and the product as 96. To find two numbers such that their sum and product are given. Diophantus gives the sum as 20 and the product as 96. I understand that the answer is 12 and 8. I need to see the work in the diophantus problem I27 form!!!
Asked by John H - Mon May 25 15:41:21 2009 - - 1 Answers - 0 Comments
Q. To find two numbers such that their sum and product are given. Diophantus gives the sum as 20 and the product as 96. To find two numbers such that their sum and product are given. Diophantus gives the sum as 20 and the product as 96. I understand that the answer is 12 and 8. I need to see the work in the diophantus problem I27 form!!!
Asked by John H - Mon May 25 15:41:21 2009 - - 1 Answers - 0 Comments
Who is really the father of agebra, Al-Khwarizmi or diophantus?
Q. Who is really the father of agebra, Al-Khwarizmi or diophantus?
Asked by JHINX - Thu Apr 19 05:30:27 2007 - - 2 Answers - 0 Comments
A. Abu Ja'far Muhammad ibn Musa Al-Khwarizmi, (780 850 CE), is the father of Algebra and the grandfather of computer science. He was the popularizer of Arabic numerals, adopter of zero (the symbol, that is) and the decimal system, astronomer, cartographer, in briefs an encyclopedic scholar.
Answered by Zain - Thu Apr 19 06:18:42 2007
Q. Who is really the father of agebra, Al-Khwarizmi or diophantus?
Asked by JHINX - Thu Apr 19 05:30:27 2007 - - 2 Answers - 0 Comments
A. Abu Ja'far Muhammad ibn Musa Al-Khwarizmi, (780 850 CE), is the father of Algebra and the grandfather of computer science. He was the popularizer of Arabic numerals, adopter of zero (the symbol, that is) and the decimal system, astronomer, cartographer, in briefs an encyclopedic scholar.
Answered by Zain - Thu Apr 19 06:18:42 2007
Hos old is diophantus when he died? please include some equation.?
Q. In diophantus epitaph the following is scripted:"A 6th of his life was spent in childhood, a 12th as a youth, and 7th as a bachelor. 5 years after his marriage, his son was born but died 4 years after his own death at half his age"
Asked by Aldrin D - Sun Feb 1 01:30:21 2009 - - 1 Answers - 0 Comments
Q. In diophantus epitaph the following is scripted:"A 6th of his life was spent in childhood, a 12th as a youth, and 7th as a bachelor. 5 years after his marriage, his son was born but died 4 years after his own death at half his age"
Asked by Aldrin D - Sun Feb 1 01:30:21 2009 - - 1 Answers - 0 Comments
purchasing a copy of Diophantus's Arithmetica?
Q. yes, the book that fermat wrote his last theorem in a column out of THIS book! so im interested of purchasing a copy of this book, reprinted of course anyone know where i can buy this? preferably in store, but can also order online help please!
Asked by alyssaaa - Tue Dec 30 23:00:35 2008 - - 1 Answers - 0 Comments
A.
Answered by Mielu istetz - Wed Dec 31 02:34:09 2008
Q. yes, the book that fermat wrote his last theorem in a column out of THIS book! so im interested of purchasing a copy of this book, reprinted of course anyone know where i can buy this? preferably in store, but can also order online help please!
Asked by alyssaaa - Tue Dec 30 23:00:35 2008 - - 1 Answers - 0 Comments
A.
Answered by Mielu istetz - Wed Dec 31 02:34:09 2008
After the death about 290A.D.of Diophantus a famous Greek mathematician someone described his life as a puzzle
Q. he was a boy for 1/6 of his life after 1/12 more, he aquired a beard after another 1/7, he married in the fifth year after marriage his son was born the son lived half as many years as his father diophantus died 4 years after his son how old was diophantus when he died? how did u get dat answer? best answer plz
Asked by mark y - Wed Jul 11 15:29:57 2007 - - 3 Answers - 0 Comments
A. This is a simultaneous equation question. Firstly you know the guy dies in 290AD (i hope thats what it means). therefore you know he was born in 290AD-x (x is his age). 1/6 of his life is therefore 1/6x so he was a boy for 1/6x, and then after 1/12x he grew a beard. His age at this point was 1/6x+1/12x, and when he gets married his age is 1/6x+1/12x+1/7x which is simplified to 33/84x. After 5 more years he has a child so he has a child at 33/84x+5. The son lives for 1/2 of his fathers life so he lives for 1/2x. We know that the father dies in 290AD so his son dies in 286AD. Finally, the year in which the son dies is therefore: The year of Diophantus birth+ his life up until he gives birth to his child+the lifespna of the child. His… [cont.]
Answered by Danny G - Wed Jul 11 16:46:54 2007
Q. he was a boy for 1/6 of his life after 1/12 more, he aquired a beard after another 1/7, he married in the fifth year after marriage his son was born the son lived half as many years as his father diophantus died 4 years after his son how old was diophantus when he died? how did u get dat answer? best answer plz
Asked by mark y - Wed Jul 11 15:29:57 2007 - - 3 Answers - 0 Comments
A. This is a simultaneous equation question. Firstly you know the guy dies in 290AD (i hope thats what it means). therefore you know he was born in 290AD-x (x is his age). 1/6 of his life is therefore 1/6x so he was a boy for 1/6x, and then after 1/12x he grew a beard. His age at this point was 1/6x+1/12x, and when he gets married his age is 1/6x+1/12x+1/7x which is simplified to 33/84x. After 5 more years he has a child so he has a child at 33/84x+5. The son lives for 1/2 of his fathers life so he lives for 1/2x. We know that the father dies in 290AD so his son dies in 286AD. Finally, the year in which the son dies is therefore: The year of Diophantus birth+ his life up until he gives birth to his child+the lifespna of the child. His… [cont.]
Answered by Danny G - Wed Jul 11 16:46:54 2007
how long did diophantus live? (math problem)?
Q. let x be the number of years he lived. find how long he lived by using the facts below: 1/6 of his life was spent in boyhood 1/12 of his life was spent in youth after 1/7 more of his life past, he got married five years after getting married, he had a son his son lived 1/2 as long as diophantus lived the son died 4 years before diophantus died
Asked by jon n - Wed Oct 28 18:57:05 2009 - - 1 Answers - 0 Comments
Q. let x be the number of years he lived. find how long he lived by using the facts below: 1/6 of his life was spent in boyhood 1/12 of his life was spent in youth after 1/7 more of his life past, he got married five years after getting married, he had a son his son lived 1/2 as long as diophantus lived the son died 4 years before diophantus died
Asked by jon n - Wed Oct 28 18:57:05 2009 - - 1 Answers - 0 Comments
How do you solve the difference between two known squares with a known solution?
Q. How do you solve x^2-y^2=60 Thanks! (It has to do with Diophantus II 10)
Asked by thebassgirl - Mon Apr 20 14:55:32 2009 - - 1 Answers - 0 Comments
A. Remember how to factor a difference of squares: (x - y)(x + y) = 60 Assuming that x and y have to be positive integers, you have the following ways to factor 60: 1 x 60 2 x 30 3 x 20 4 x 15 5 x 12 6 x 10 If you add the two equations: x - y = a x + y = b 2x = a + b So only the ones that add to even numbers will work: 2 x 30 6 x 10 Try solving each one in turn: x + y = 30 x - y = 2 2x = 32 x = 16 y = 14 x + y = 10 x - y = 6 2x = 16 x = 8 y = 2 Answers: (16,14) and (8,2) Double-check: 16 - 14 = 256 - 196 = 60 8 - 2 = 64 - 4 = 60
Answered by Puzzling - Mon Apr 20 15:04:10 2009
Q. How do you solve x^2-y^2=60 Thanks! (It has to do with Diophantus II 10)
Asked by thebassgirl - Mon Apr 20 14:55:32 2009 - - 1 Answers - 0 Comments
A. Remember how to factor a difference of squares: (x - y)(x + y) = 60 Assuming that x and y have to be positive integers, you have the following ways to factor 60: 1 x 60 2 x 30 3 x 20 4 x 15 5 x 12 6 x 10 If you add the two equations: x - y = a x + y = b 2x = a + b So only the ones that add to even numbers will work: 2 x 30 6 x 10 Try solving each one in turn: x + y = 30 x - y = 2 2x = 32 x = 16 y = 14 x + y = 10 x - y = 6 2x = 16 x = 8 y = 2 Answers: (16,14) and (8,2) Double-check: 16 - 14 = 256 - 196 = 60 8 - 2 = 64 - 4 = 60
Answered by Puzzling - Mon Apr 20 15:04:10 2009
After the death about 290 A.D of Diophantus a famous Greek mathematician, someone described his life a puzzle?
Q. he was a boy for 1/6 of his life after 1/12 more, he aquired a beard after another 1/7, he married in the fifth year after marriage his son was born the son lived half as many years as his father diophantus died 4 years after his son how old was diophantus when he died?\
Asked by mark y - Wed Jul 11 14:34:50 2007 - - 2 Answers - 0 Comments
A. he was 84 i answered this one on the other question.
Answered by Danny G - Wed Jul 11 16:52:19 2007
Q. he was a boy for 1/6 of his life after 1/12 more, he aquired a beard after another 1/7, he married in the fifth year after marriage his son was born the son lived half as many years as his father diophantus died 4 years after his son how old was diophantus when he died?\
Asked by mark y - Wed Jul 11 14:34:50 2007 - - 2 Answers - 0 Comments
A. he was 84 i answered this one on the other question.
Answered by Danny G - Wed Jul 11 16:52:19 2007
Quadratic equations by diophantus. express it in modern terms?
Q. to find two numbers such that their sum and the sum of their squares are given numbers. diophantus takes the sum to be 20 and the sum of the squares to be 208. he also points out that to solve this it is necessary that double the sum of the squares exceed the square of the sum by a square. express the problem in modern terms, solve it and explain why diophantus needed to add a condition.
Asked by MathKid - Wed Mar 11 02:17:58 2009 - - 2 Answers - 0 Comments
A. Suppose you are given two numbers m,n. can you find a,b such that m = a + b and n = a^2 + b^2 ? The numbers mentioned are natural numbers. One problem is that this cannot be done for every pair (m,n) since n = a^2 + b^2 is not always possible. Numbers which are not the sum of two squares are 1,3,4,6,7,9,11,12,14,15,1 6,19,.. so that they are quite numerous. Even if n were chosen to be the sum of two squares a^2 + b^2, the other number m may not be writable in the form a+b. An example is (m,n)=(3,10) For the specific problem we have a+b = 20, a^2 + b^2 = 208, (a+b)^2 =a^2+2ab+b^2=400 subtracting, 2ab=192,ab=96,recall a+b=20,b=20-a a(20-a)=96, we get a^2 - 20a -96 = 0 and (a,b)=(8,12) or else (12,8) check: 8+12=20, 8^2+12^2=208. So… [cont.]
Answered by knashha - Sat Mar 14 14:04:37 2009
Q. to find two numbers such that their sum and the sum of their squares are given numbers. diophantus takes the sum to be 20 and the sum of the squares to be 208. he also points out that to solve this it is necessary that double the sum of the squares exceed the square of the sum by a square. express the problem in modern terms, solve it and explain why diophantus needed to add a condition.
Asked by MathKid - Wed Mar 11 02:17:58 2009 - - 2 Answers - 0 Comments
A. Suppose you are given two numbers m,n. can you find a,b such that m = a + b and n = a^2 + b^2 ? The numbers mentioned are natural numbers. One problem is that this cannot be done for every pair (m,n) since n = a^2 + b^2 is not always possible. Numbers which are not the sum of two squares are 1,3,4,6,7,9,11,12,14,15,1 6,19,.. so that they are quite numerous. Even if n were chosen to be the sum of two squares a^2 + b^2, the other number m may not be writable in the form a+b. An example is (m,n)=(3,10) For the specific problem we have a+b = 20, a^2 + b^2 = 208, (a+b)^2 =a^2+2ab+b^2=400 subtracting, 2ab=192,ab=96,recall a+b=20,b=20-a a(20-a)=96, we get a^2 - 20a -96 = 0 and (a,b)=(8,12) or else (12,8) check: 8+12=20, 8^2+12^2=208. So… [cont.]
Answered by knashha - Sat Mar 14 14:04:37 2009
can you please give me a poem about the contributions of diophantus? (mathematician)?
Q. can you please give me a poem about the contributions of diophantus? (mathematician)?
Asked by asteeg - Mon Jun 19 04:42:30 2006 - - 1 Answers - 0 Comments
A. here's a haiku for ya: O Diophantus A mathematician Not much rhymes with that
Answered by ScarletSky - Mon Jun 19 05:49:59 2006
Q. can you please give me a poem about the contributions of diophantus? (mathematician)?
Asked by asteeg - Mon Jun 19 04:42:30 2006 - - 1 Answers - 0 Comments
A. here's a haiku for ya: O Diophantus A mathematician Not much rhymes with that
Answered by ScarletSky - Mon Jun 19 05:49:59 2006
diophantus of alexandria?
Q. what problem was he trying to solve? how did he come upon his solutions? what methods did he use? what techniques did he develop to solve the problem? please help answer these questions!!! THANKS
Asked by jenn - Sat Feb 16 18:18:00 2008 - - 1 Answers - 0 Comments
A. What problem was he trying to solve? * the solution of algebraic equations and on the theory of numbers.
Answered by Georgia Peach - Sat Feb 16 18:24:32 2008
Q. what problem was he trying to solve? how did he come upon his solutions? what methods did he use? what techniques did he develop to solve the problem? please help answer these questions!!! THANKS
Asked by jenn - Sat Feb 16 18:18:00 2008 - - 1 Answers - 0 Comments
A. What problem was he trying to solve? * the solution of algebraic equations and on the theory of numbers.
Answered by Georgia Peach - Sat Feb 16 18:24:32 2008
Is Aryabhatta is the first to find general solutions to equations?
Q. Babylonians and Diophantus found special ad hoc solutions. Aryabhatta found the general solution to linear indeterminate equations
Asked by RRon - Sat Feb 9 09:10:59 2008 - - 4 Answers - 0 Comments
A. Aryabhata tried different place value systems until he utter the words "place to place is ten times in value"...Thats the origin of the decimal place value system. Aryabhata had an empty place for zero. Then Brahmagupta added zero and negative numbers. Brahmagupta and Mahavira played with these newly found numbers and developed arithmetic as we know it today. Diophantos found solutions to algebraic equations in a case by case basis. No general solutions in his work. Aryabhata was the first to find a general solution to first degree equations. (498 AD). Brahmagupta (628 AD) found a general solution to second degree equations. It took another thousand years to find a general solution to the third degree equations by Cardon of Italy in 1560… [cont.]
Answered by Mano - Sat Feb 9 11:26:55 2008
Q. Babylonians and Diophantus found special ad hoc solutions. Aryabhatta found the general solution to linear indeterminate equations
Asked by RRon - Sat Feb 9 09:10:59 2008 - - 4 Answers - 0 Comments
A. Aryabhata tried different place value systems until he utter the words "place to place is ten times in value"...Thats the origin of the decimal place value system. Aryabhata had an empty place for zero. Then Brahmagupta added zero and negative numbers. Brahmagupta and Mahavira played with these newly found numbers and developed arithmetic as we know it today. Diophantos found solutions to algebraic equations in a case by case basis. No general solutions in his work. Aryabhata was the first to find a general solution to first degree equations. (498 AD). Brahmagupta (628 AD) found a general solution to second degree equations. It took another thousand years to find a general solution to the third degree equations by Cardon of Italy in 1560… [cont.]
Answered by Mano - Sat Feb 9 11:26:55 2008
How Smart are You #2???
Q. Another question/puzzle i need help with: Diophantus The Greek mathmetician Diophantus, who lived in the second century A.D. was the first person to replace unknowns by letters(variables). There is a famous problem that you can use to calculate how long he lived. Diophantus passed one sixth of his life in childhood, one twelfth in youth, and one seventh more as a young bachelor. Five years after his marriage, he had a son who died four years before his father. The son was half the age when he died. How old was Diophantus when he died.
Asked by jersmine14 - Sun Sep 9 18:59:20 2007 - - 4 Answers - 0 Comments
A. x/6 + x/12 + x/7 + 5 + x/2 + 4 = x 25x/28 + 9 = x 25x + 252 / 28 = x 3x = 252 x = 84 He was 84 when he died.
Answered by lilsweetsarah4 - Sun Sep 9 20:33:21 2007
Q. Another question/puzzle i need help with: Diophantus The Greek mathmetician Diophantus, who lived in the second century A.D. was the first person to replace unknowns by letters(variables). There is a famous problem that you can use to calculate how long he lived. Diophantus passed one sixth of his life in childhood, one twelfth in youth, and one seventh more as a young bachelor. Five years after his marriage, he had a son who died four years before his father. The son was half the age when he died. How old was Diophantus when he died.
Asked by jersmine14 - Sun Sep 9 18:59:20 2007 - - 4 Answers - 0 Comments
A. x/6 + x/12 + x/7 + 5 + x/2 + 4 = x 25x/28 + 9 = x 25x + 252 / 28 = x 3x = 252 x = 84 He was 84 when he died.
Answered by lilsweetsarah4 - Sun Sep 9 20:33:21 2007
Need help with a math "riddle" please?
Q. Diophantus's youth lasted 1/6 of his life.He grew a beard after 1/12 more. After 1/7 more of his life, he married. 5 yrs later he had a son. The son live 1/2 as long as his dad, and Diophatus died 4 yrs after his son. How many years did Diophantus live? Please show the math >Thanks !!!
Asked by weisenheimer - Tue Mar 24 17:37:14 2009 - - 2 Answers - 0 Comments
A. Let X be his age. Let Y be his son's age. (1/6)X + (1/12)X + (1/7)X + 5 + Y + 4 = 2Y X = 2Y Solve the simultaneous equations: .393X + 9 = Y X = 2Y X= 2(.393X + 9) X = 84 He lived to be 84. Y=X/2=42
Answered by Wesley J - Tue Mar 24 17:54:55 2009
Q. Diophantus's youth lasted 1/6 of his life.He grew a beard after 1/12 more. After 1/7 more of his life, he married. 5 yrs later he had a son. The son live 1/2 as long as his dad, and Diophatus died 4 yrs after his son. How many years did Diophantus live? Please show the math >Thanks !!!
Asked by weisenheimer - Tue Mar 24 17:37:14 2009 - - 2 Answers - 0 Comments
A. Let X be his age. Let Y be his son's age. (1/6)X + (1/12)X + (1/7)X + 5 + Y + 4 = 2Y X = 2Y Solve the simultaneous equations: .393X + 9 = Y X = 2Y X= 2(.393X + 9) X = 84 He lived to be 84. Y=X/2=42
Answered by Wesley J - Tue Mar 24 17:54:55 2009
I need math Help!!Please?
Q. Diophantus passed 1/6 of his life in childhood, 1/12 in youth, and 1/7 more as a bachelor, five years after his marriage was born a son who died four years before his father at half the age at which his father died. What was Diophantus' age when he died?
Asked by DDELUCA - Sun Apr 8 14:22:08 2007 - - 6 Answers - 0 Comments
A. 84 is the LCM for 6, 12, and 7. By luck, this number works out. Childhood 84/6 = 14 Youth = 84/12 = 7 (age 15 - 21) Bachelor = 84/7 = 12 (age 22 - 33) Married at 33 Had child at 38. (33 + 5) Son died at 42 (half of 84) when father was 38 + 42 or 80 Diophantus died at 84 I am not sure how you'd solve this one algebraically.
Answered by suesysgoddess - Sun Apr 8 14:42:09 2007
Q. Diophantus passed 1/6 of his life in childhood, 1/12 in youth, and 1/7 more as a bachelor, five years after his marriage was born a son who died four years before his father at half the age at which his father died. What was Diophantus' age when he died?
Asked by DDELUCA - Sun Apr 8 14:22:08 2007 - - 6 Answers - 0 Comments
A. 84 is the LCM for 6, 12, and 7. By luck, this number works out. Childhood 84/6 = 14 Youth = 84/12 = 7 (age 15 - 21) Bachelor = 84/7 = 12 (age 22 - 33) Married at 33 Had child at 38. (33 + 5) Son died at 42 (half of 84) when father was 38 + 42 or 80 Diophantus died at 84 I am not sure how you'd solve this one algebraically.
Answered by suesysgoddess - Sun Apr 8 14:42:09 2007
Solve this riddle. Now!?
Q. Diophantus s childhood lasted one-sixth of his life. His beard grew after one-twelfth more. He married after one-seventh more. His son was born five years later. The son lived to half his father s age. Diophantus died four years after his son. How old was Diophantus when he died? Show your working out, please.
Asked by Scotch Whisky - Tue Dec 1 10:03:32 2009 - - 3 Answers - 0 Comments
A. let D = Diophantus' lifetime in years Childhood=D/6 When Beard grew= D/12+D/6=3D/12=D/4 When married=D/4+D/7=11D/28 When son born=11D/28+5 Son lived for D/2. So now we know birth and lifespan of son relative to Dio. Figure out son's death date relative to Dio's age: D/2+(11D/28+5)= 25D/28+5. We know 4 years after "25D/28+5" was Dio's death at age D. 25D/28+9=D 9=3/28D 84=D 84 years old.
Answered by (Insert Name) - Tue Dec 1 20:44:25 2009
Q. Diophantus s childhood lasted one-sixth of his life. His beard grew after one-twelfth more. He married after one-seventh more. His son was born five years later. The son lived to half his father s age. Diophantus died four years after his son. How old was Diophantus when he died? Show your working out, please.
Asked by Scotch Whisky - Tue Dec 1 10:03:32 2009 - - 3 Answers - 0 Comments
A. let D = Diophantus' lifetime in years Childhood=D/6 When Beard grew= D/12+D/6=3D/12=D/4 When married=D/4+D/7=11D/28 When son born=11D/28+5 Son lived for D/2. So now we know birth and lifespan of son relative to Dio. Figure out son's death date relative to Dio's age: D/2+(11D/28+5)= 25D/28+5. We know 4 years after "25D/28+5" was Dio's death at age D. 25D/28+9=D 9=3/28D 84=D 84 years old.
Answered by (Insert Name) - Tue Dec 1 20:44:25 2009
ProBlem In ALGEBRA!!age problem?
Q. It is told that the age of Diophantus, a Greek mathematician, may be calculated from his epitaph. His eitaph read as follows: "Diophantus spent one-sixthe of his life in childhood, one-twelfth in youth and one-seventh as a bachelor. Five years after his marraige was born a son who died four years before his father died at half his father's age." How old was Diophantus when he died?
Asked by Jonah K.E. - Wed Dec 10 09:26:39 2008 - - 1 Answers - 0 Comments
Q. It is told that the age of Diophantus, a Greek mathematician, may be calculated from his epitaph. His eitaph read as follows: "Diophantus spent one-sixthe of his life in childhood, one-twelfth in youth and one-seventh as a bachelor. Five years after his marraige was born a son who died four years before his father died at half his father's age." How old was Diophantus when he died?
Asked by Jonah K.E. - Wed Dec 10 09:26:39 2008 - - 1 Answers - 0 Comments
Are you good challange's? find out here?
Q. Here is a famous problem about Diophantus, a Greek mathematician from the third century. (Let x be the number of years Diophantus lived. Find how long he lived by using the following facts about him to write and solve an equation.) ; One sixth of his life was spent in boyhood. One twelfth of his life was spent as a youth. After 1/7 more of his life passed, he got married. five years after getting married, he had a son. His son lived 1/2 as long as Diophantus lived. the son died 4 years before Diophantus dies. Let's see what you come up with!
Asked by Mouse - Tue Dec 1 18:37:53 2009 - - 1 Answers - 0 Comments
A. (x/6+x/12+x/7+5)+x/2+4=x 1/84(14+7+12+42)x + 9 = x 75/84 x + 9 = x x(1-75/84) = 9 9/84 x = 9 x= 84
Answered by Flavio D - Tue Dec 1 18:56:41 2009
Q. Here is a famous problem about Diophantus, a Greek mathematician from the third century. (Let x be the number of years Diophantus lived. Find how long he lived by using the following facts about him to write and solve an equation.) ; One sixth of his life was spent in boyhood. One twelfth of his life was spent as a youth. After 1/7 more of his life passed, he got married. five years after getting married, he had a son. His son lived 1/2 as long as Diophantus lived. the son died 4 years before Diophantus dies. Let's see what you come up with!
Asked by Mouse - Tue Dec 1 18:37:53 2009 - - 1 Answers - 0 Comments
A. (x/6+x/12+x/7+5)+x/2+4=x 1/84(14+7+12+42)x + 9 = x 75/84 x + 9 = x x(1-75/84) = 9 9/84 x = 9 x= 84
Answered by Flavio D - Tue Dec 1 18:56:41 2009
Can You Answer This Math Question?
Q. Diophantus was a Greek mathematician who lived in the third century. Most of what is known about Diophantus's life comes from an algebraic riddle from around the early sixth century. The riddle states: Diophantus's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Diophantus died four years after his son. How many years did Diophantus live? Show or explain how you got your answer.
Asked by MaRRiUM - Wed Dec 9 20:50:37 2009 - - 1 Answers - 0 Comments
A. x= his age 1/6*x+1/12*x+1/7x+5+1/2*x +4=x Now multiply by 84(LCD) 14x+7x+12x+420+42x+336=84 x or 75x+756=84x or -9x=-756 or x=84 The dude lived to be 84 years old
Answered by Brian - Wed Dec 9 21:04:29 2009
Q. Diophantus was a Greek mathematician who lived in the third century. Most of what is known about Diophantus's life comes from an algebraic riddle from around the early sixth century. The riddle states: Diophantus's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and Diophantus died four years after his son. How many years did Diophantus live? Show or explain how you got your answer.
Asked by MaRRiUM - Wed Dec 9 20:50:37 2009 - - 1 Answers - 0 Comments
A. x= his age 1/6*x+1/12*x+1/7x+5+1/2*x +4=x Now multiply by 84(LCD) 14x+7x+12x+420+42x+336=84 x or 75x+756=84x or -9x=-756 or x=84 The dude lived to be 84 years old
Answered by Brian - Wed Dec 9 21:04:29 2009
From Yahoo Answer Search: 'diophantus'
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