answer:1. **Calculate the missing leg of the first triangle using the Pythagorean theorem:** [ text{Let } b text{ be the missing leg. Then, } b^2 = 17^2 - 15^2 = 289 - 225 = 64 Rightarrow b = sqrt{64} = 8 text{ inches}. ] 2. **Determine the scale factor between the two triangles:** [ text{Scale factor} = frac{51}{17} = 3. ] 3. **Apply the scale factor to find the corresponding shorter side in the second triangle:** [ text{Shortest side of the second triangle} = 3 times 8 = 24 text{ inches}. ] [ boxed{24 text{ inches}} ]

answer:1. Let's denote: - The distance Vintik drove forward as 2x km, - The distance Vintik drove backward as y km, - The distance Shpuntik drove forward as x km, (half of Vintik's forward distance) - The distance Shpuntik drove backward as 2y km (twice Vintik's backward distance). 2. From the problem, we know the total distance driven in both directions by Vintik is 12 km: [ 2x + y = 12 quad text{(1)} ] 3. We also know that the total fuel consumption should equal 75 liters. The fuel consumption is as follows: - Forward for Vintik and Shpuntik: 2x cdot Z + x cdot Z = 3x cdot Z, - Backward for Vintik and Shpuntik: y cdot 5 + 2y cdot 5 = 3y cdot 5. Substituting the fuel consumption rates and knowing that Z = 1 liter/km and that in total it is 75 liters: [ 3x cdot 1 + 3y cdot 5 = 75 ] Simplifying: [ 3x + 15y = 75 quad text{(2)} ] 4. We now have a system of linear equations: [ begin{cases} 2x + y = 12 3x + 15y = 75 end{cases} ] 5. We solve Equation (1) for y: [ y = 12 - 2x ] 6. Substitute y = 12 - 2x into Equation (2): [ 3x + 15(12 - 2x) = 75 ] [ 3x + 180 - 30x = 75 ] [ 180 - 27x = 75 ] [ -27x = 75 - 180 ] [ -27x = -105 ] [ x = frac{105}{27} ] [ x = frac{35}{9} ] [ x = 5 ] 7. Substitute x = 5 into y = 12 - 2x: [ y = 12 - 2 cdot 5 ] [ y = 12 - 10 ] [ y = 2 ] 8. Thus, Vintik drove 2x = 2 cdot 5 = 10 km forward and y = 2 km backward. Shpuntik drove x = 5 km forward and 2y = 2 cdot 2 = 4 km backward. 9. The total distance they need to walk back home is the total forward distance they traveled minus the total backward distance: [ text{Total distance driven forward} = 2x + x = 3x = 15 , text{km} ] [ text{Total distance driven backward} = y + 2y = 3y = 6 , text{km} ] [ text{Distance to walk back} = 15 - 6 = 9 , text{km} ] Conclusion: [ boxed{9 , text{km}} ]

answer:1. **Understand the problem constraints:** - The confectioner needs to make 43 cakes with each cake requiring less than 9 eggs. - Any number of eggs purchased, when grouped into 2, 3, 4, 5, or 6, leaves a remainder of 1. 2. **Formulate the mathematical conditions:** - Let ( n ) be the total number of eggs. Since ( n ) must be a multiple of 43 (as all 43 cakes require the same number of eggs), we write: [ n = 43k quad text{for some integer } k. ] - ( n ) also leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6. That is: [ n equiv 1 pmod{60} ] This is because the least common multiple (LCM) of 2, 3, 4, 5, and 6 is 60. - Therefore, we look for a number ( n ) that satisfies both conditions: [ n = 60m + 1 quad text{and} quad n = 43k ] where ( m ) and ( k ) are integers. 3. **Find the smallest ( n ) below 387:** - We start testing integer multiples of 60 and checking if adding 1 gives a multiple of 43: [ begin{array}{ll} 60 times 2 + 1 = 121 & text{not a multiple of 43} 60 times 3 + 1 = 181 & text{not a multiple of 43} 60 times 4 + 1 = 241 & text{not a multiple of 43} 60 times 5 + 1 = 301 & text{is a multiple of 43} 60 times 6 + 1 = 361 & text{not a multiple of 43} end{array} ] - Checking, ( 301 = 43 times 7 ): [ 301 = 43 times 7 text{ which is correct.} ] 4. **Derive the number of eggs used per cake:** - The total number of eggs used is 301 and since there are 43 cakes: [ e text{ (eggs per cake)} = frac{301}{43} = 7 ] - Each cake uses 7 eggs, which meets the condition of using less than 9 eggs per cake. # Conclusion: The smallest number of eggs that the confectioner will use to make 43 cakes, with each cake requiring exactly 7 eggs, is: [ 301 , text{eggs} ] and the eggs needed per cake is: [ boxed{7} ]

answer:Given the function y=x^3-ax^2+4d, we want it to be monotonically decreasing in the interval (0,2). For a function to be monotonically decreasing, its first derivative must be negative throughout the given interval. Let's find the first derivative of the function with respect to x: y' = 3x^2 - 2ax Now, let's set the derivative less than zero for the function to be monotonically decreasing: 3x^2 - 2ax < 0 Since we're considering the interval (0,2), x is positive, so we can divide the inequality by x without changing its direction: 3x - 2a < 0 Now, let's find the maximum value of a such that the inequality holds for all x in (0,2): 3(2) - 2a < 0 6 - 2a < 0 -2a < -6 a > 3 However, there's no upper limit on the value of a that makes the function monotonically decreasing in the given interval. Thus, the answer is: boxed{ageq3}