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Any questions on High School Algebra, College Algebra, Abstract Algebra. I can help with word problems, solving equations, trigonometry, inequalities, Gaussian Elimination, Linear Algebra, groups, fields, you name it!

Ph.D. in Mathematics, specialist in Algebra. I have taught High School students and college students at three state universities.

Mathematical Association of America. American Mathematical Society.

Regular contributions to the problems section of the American Mathematical Monthly journal.

B.S., M.S., Ph.D.

User | Date | K | C | P | Comments |
---|---|---|---|---|---|

JohnnyX | 09/10/16 | 10 | 10 | 10 | Thanks for ya enormous amount of help! |

Michael | 03/31/16 | 10 | 10 | 10 | Thank you. I really struggled with this ..... |

Jim | 12/21/15 | 10 | 10 | 10 | Thanks! Once again, what seemed insoluble becomes ..... |

Muhammad Ali | 11/28/15 | 5 | 8 | 10 | |

Tressa | 02/08/15 | 10 | 10 | 10 |

The flaw is in the hidden assumption that the least such n is 3 or greater. In fact , the least such n is n = 2 If you try to carry out the argument with only two elements in S , Ti and Tj won't have

You can't give an exact value for log2 (1,000,000) using decimals or fractions because it's an irrational number. You can use the rules for logarithms to simplify it and then use a calculator. 1

You need the square roots of 3-2i and -2+i. I will find the roots of 3-2i, then leave the roots of -2+i to you. (a+bi)² = 3-2i a² - b² + 2abi = 3-2i a² - b² = 3 2ab = -2 b = -1/a

The sum of their ages is equal to twice the difference if their ages p + q = 2(p-q) The product of their ages is 675 pq = 675 Solve the two equations : p + q = 2(p-q) pq = 675 Simplify

x + y + z = 1 (x + y + z)^2 = 1 x^2 + y^2 +z^2 + 2xy + 2xz + 2yz = 1 35 + 2xy +2xz + 2yz = 1 2xy +2xz + 2yz = -34 xy + xz + yz = -17 -17 = (1)(xy + xz + yz ) = (x + y + z)(xy + xz +

Algebra

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