answer:First, rewrite the equation for clarity: 7x^2 + mx - 6 = 0. According to Vieta's formulas, the product of the roots (alpha beta) of a quadratic equation ax^2 + bx + c = 0 is given by frac{c}{a}. For our equation, it would be frac{-6}{7}. Given that one of the roots (alpha) is -3, we can find the other root (beta) by: [ alpha beta = frac{-6}{7} ] [ -3 cdot beta = frac{-6}{7} ] [ beta = frac{-6}{7} div -3 = frac{-6}{7} cdot frac{-1}{3} = frac{6}{21} = frac{2}{7} ] Thus, the other root is boxed{frac{2}{7}}.

answer:Starting with the given condition: [ frac{1}{x} + frac{1}{y} = frac{1}{15} ] Multiplying through by xy and rearranging, we obtain: [ xy - 15x - 15y = 0 ] To apply Simon's Favorite Factoring Trick, add 225 to both sides: [ xy - 15x - 15y + 225 = 225 ] Factorizing gives: [ (x-15)(y-15) = 225 ] We seek the minimal x+y. This occurs when x-15 and y-15 are factor pairs of 225 that are close to each other in value. The factor pairs of 225 are (1, 225), (3, 75), (5, 45), (9, 25), (15, 15). Since x neq y, we ignore (15, 15). The pair (9, 25) gives the smallest sum: [ x = 9 + 15 = 24, quad y = 25 + 15 = 40 ] Thus, x+y = 24 + 40 = boxed{64}.

answer:1. **Define variables and setup:** Let's consider a two-digit number ( n ) where ( a ) is the tens digit and ( b ) is the units digit. Thus, we can represent ( n ) as: [ n = 10a + b ] Since both ( a ) and ( b ) are digits between 1 and 9 (inclusive), the reverse of ( n ) can be written as: [ text{reverse of } n = 10b + a ] 2. **Sum of the number and its reverse:** When we add ( n ) and its reverse, we get: [ (10a + b) + (10b + a) ] This can be simplified by combining like terms: [ = 10a + b + 10b + a = 11a + 11b = 11(a + b) ] Hence, the sum of a number and its reverse is always a multiple of 11. 3. **List and analyze given options:** We now examine the provided options to check which one is not a multiple of 11: - ( 44 ) - ( 99 ) - ( 121 ) - ( 165 ) - ( 181 ) We will test the divisibility by ( 11 ) for each: 4. **Check each option for divisibility by 11:** - ( 44 div 11 = 4 quad text{(integer, hence multiple of 11)} ) - ( 99 div 11 = 9 quad text{(integer, hence multiple of 11)} ) - ( 121 div 11 = 11 quad text{(integer, hence multiple of 11)} ) - ( 165 div 11 = 15 quad text{(integer, hence multiple of 11)} ) - ( 181 div 11 = 16.454545ldots quad text{(not an integer, hence not a multiple of 11)} ) 5. **Conclusion:** Since the sum of a number and its reverse must be a multiple of 11, and 181 is not a multiple of 11, we conclude: [ boxed{E} ]

answer:- Let the number of girls be 4k, and the number of boys be 3k. - Including the teacher, the total count becomes 4k + 3k + 1 = 43. - Simplify and solve the equation: 7k + 1 = 43, then 7k = 42, thus k = 6. - The number of girls is 4k = 4 times 6 = boxed{24}.