answer:1. **Identify the given information:** - Total points scored by Hannah during the season: 312 points - Average (mean) points scored per game: 13 points per game 2. **Set up the relationship and formula:** The formula for the average (mean) is: [ text{Average} = frac{text{Total points}}{text{Number of games}} ] Rearrange this formula to find the number of games: [ text{Number of games} = frac{text{Total points}}{text{Average points per game}} ] 3. **Substitute the given values into the formula:** [ text{Number of games} = frac{312 text{ points}}{13 text{ points per game}} ] 4. **Perform the division:** [ text{Number of games} = frac{312}{13} ] [ text{Number of games} = 24 ] 5. **Conclusion:** Hannah played a total of 24 games during the basketball season. [ boxed{24} ]

answer:: 1. Let's start by simplifying the given equations one by one: Since ( x ) and ( y ) are real positive numbers, let's substitute: [ 27^y = a, quad 4^x = b ] 2. From the third equation: [ 27^y - 4^x leq 1 ] We get: [ a - b leq 1 ] Which implies: [ a leq b + 1 ] 3. Substituting this into the first equation: [ frac{1}{4^x} + frac{1}{27^y} = frac{5}{6} ] Becomes: [ frac{1}{b} + frac{1}{a} = frac{5}{6} ] Since: [ frac{1}{b} + frac{1}{a} leq frac{1}{b} + frac{1}{b+1} ] Thus: [ frac{1}{b} + frac{1}{b + 1} leq frac{5}{6} ] 4. Solving: [ frac{1}{b} + frac{1}{b+1} = frac{b + 1 + b}{b(b+1)} = frac{2b + 1}{b(b+1)} ] Thus: [ frac{2b + 1}{b(b+1)} leq frac{5}{6} ] Cross multiplying, we get: [ 6(2b+1) leq 5b(b+1) ] Which expands and simplifies to: [ 12b + 6 leq 5b^2 + 5b ] [ 0 leq 5b^2 - 7b - 6 ] 5. Solving the quadratic inequality ( 5b^2 - 7b - 6 geq 0 ): [ 5b^2 - 7b - 6 = 0 ] Using the quadratic formula ( b = frac{-B pm sqrt{B^2 - 4AC}}{2A} ): [ b = frac{7 pm sqrt{49+120}}{10} ] [ b = frac{7 pm sqrt{169}}{10} ] [ b = frac{7 pm 13}{10} ] Thus: [ b = 2 quad text{or} quad b = -frac{3}{5} ] Since ( b = 4^x > 0 ): [ b geq 2 ] Which means: [ x geq frac{1}{2} ] We conclude that: [ 4^x geq 2 ] 6. Now, consider the second constraint: [ log_{27} y - log_4 x geq frac{1}{6} ] Similarly to convert: [ log_{27} y = frac{log y}{log 27} quad text{and} quad log_4 x = frac{log x}{log 4} ] Since ( x geq frac{1}{2} ), we get: [ log_{27} y geq frac{1}{6} + log_4 frac{1}{2} ] [ log_4 frac{1}{2} = log_4 (4^{-1}) = -frac{1}{2} ] Thus: [ log_{27} y geq frac{1}{6} - frac{1}{2} = -frac{1}{3} ] Resulting in: [ y geq 27^{-frac{1}{3}} = frac{1}{3} ] 7. Returning to the first constraint: [ frac{5}{6} = 4^{-x} + 27^{-y} ] Assuming ( x = frac{1}{2} ) and ( y = frac{1}{3} ): [ 4^{-frac{1}{2}} + 27^{-frac{1}{3}} = frac{1}{2} + frac{1}{3} = frac{3}{6} + frac{2}{6} = frac{5}{6} ] 8. All conditions are satisfied if: [ (x, y) = left( frac{1}{2}, frac{1}{3} right) ] Concluding with the pair: [ boxed{left(frac{1}{2}, frac{1}{3}right)} ]

answer:# Problem: The gardener grows white and red flowers: peonies and roses. In his garden, there are 301 flowers, among which there are 135 roses and 182 red flowers. What is the minimum number of red peonies that can grow in the gardener's garden? 1. **Identify the given values:** - Total number of flowers: (301) - Number of roses: (135) - Number of red flowers: (182) 2. **Define variables:** - Let (R_r) denote the number of red roses. - Let (R_p) denote the number of red peonies. - Let (W_r) denote the number of white roses. - Let (W_p) denote the number of white peonies. 3. **Establish the relationship between the variables:** - The total number of flowers is given by: [ R_r + R_p + W_r + W_p = 301 ] - The number of roses (both red and white) is: [ R_r + W_r = 135 ] - The number of red flowers (both roses and peonies) is: [ R_r + R_p = 182 ] 4. **Determine the red peonies:** - We need to find the number of red peonies (R_p). - We know from the number of red flowers equation that: [ R_p = 182 - R_r ] - The maximum possible number of red roses (R_r) is equal to the total number of roses: [ R_r leq 135 ] - To find the minimum number of red peonies, (R_p), we substitute (R_r = 135) into the equation for (R_p): [ R_p = 182 - 135 = 47 ] 5. **Verify the result:** - We assume: - Red peonies (R_p = 47) - Red roses (R_r = 135) - White roses (W_r = 0) - White peonies (W_p = 301 - (R_r + R_p + W_r) = 301 - (135 + 47 + 0) = 119) - Check the conditions: - Total number of flowers is (135 + 47 + 0 + 119 = 301). - Number of red flowers is (135 + 47 = 182). The conditions of the problem are satisfied with this setup. Thus, the minimum number of red peonies is [ boxed{47} ]

answer:1. **Define the variables and given conditions:** Let ( OA = OB = r ) and ( AB = x ). We are given that ( angle OAB = 36^circ ) and ( angle OBA = 72^circ ). We need to prove that ( OP = AB ) given ( OP^2 = OB cdot PB ). 2. **Use the Law of Sines in (triangle OAB):** [ frac{OA}{sin angle OBA} = frac{AB}{sin angle OAB} ] Substituting the known angles: [ frac{r}{sin 72^circ} = frac{x}{sin 36^circ} ] 3. **Simplify using known trigonometric values:** [ sin 72^circ = 2 sin 36^circ cos 36^circ ] [ cos 36^circ = frac{sqrt{5} + 1}{4} ] [ sin 36^circ = frac{sqrt{10 - 2sqrt{5}}}{4} ] Therefore: [ sin 72^circ = 2 left( frac{sqrt{10 - 2sqrt{5}}}{4} right) left( frac{sqrt{5} + 1}{4} right) = frac{sqrt{10 - 2sqrt{5}} (sqrt{5} + 1)}{8} ] 4. **Simplify the ratio:** [ frac{r}{frac{sqrt{10 - 2sqrt{5}} (sqrt{5} + 1)}{8}} = frac{x}{frac{sqrt{10 - 2sqrt{5}}}{4}} ] [ r = x cdot 2 cos 36^circ ] [ r = x cdot 2 cdot frac{sqrt{5} + 1}{4} = x cdot frac{sqrt{5} + 1}{2} ] This gives us: [ r = x cdot frac{sqrt{5} + 1}{2} quad text{(1)} ] 5. **Use the given condition ( OP^2 = OB cdot PB ):** [ OP^2 = r(r - OP) ] [ OP^2 = r^2 - r cdot OP ] Rearrange into a quadratic equation: [ OP^2 + r cdot OP - r^2 = 0 ] 6. **Solve the quadratic equation:** [ OP = frac{-r pm sqrt{r^2 + 4r^2}}{2} ] [ OP = frac{-r pm sqrt{5r^2}}{2} ] [ OP = frac{-r pm rsqrt{5}}{2} ] [ OP = r cdot frac{-1 pm sqrt{5}}{2} ] Since ( OP ) must be positive: [ OP = r cdot frac{sqrt{5} - 1}{2} ] 7. **Relate ( OP ) to ( AB ):** From (1): [ r = x cdot frac{sqrt{5} + 1}{2} ] Substitute ( r ) into the expression for ( OP ): [ OP = x cdot frac{sqrt{5} + 1}{2} cdot frac{sqrt{5} - 1}{2} ] Simplify: [ OP = x cdot frac{(sqrt{5} + 1)(sqrt{5} - 1)}{4} ] [ OP = x cdot frac{5 - 1}{4} ] [ OP = x cdot 1 ] [ OP = x ] 8. **Conclusion:** [ OP = AB ] The final answer is ( boxed{ OP = AB } )