answer:To find the area of a sector of a circle, you can use the formula: Area of sector = (θ / 360) * π * r^2 where θ is the central angle in degrees and r is the radius of the circle. Given that the radius (r) is 12 meters and the central angle (θ) is 40°, we can plug these values into the formula: Area of sector = (40 / 360) * π * (12^2) Now, calculate the area: Area of sector = (1/9) * π * 144 Area of sector = 16 * π The area of the sector is 16π square meters. If you want a numerical approximation, you can use the value of π as approximately 3.14159: Area of sector ≈ 16 * 3.14159 ≈ 50.26544 square meters So, the area of the sector is approximately boxed{50.27} square meters.

answer:Let the scaled down triangle have a base of 8 cm and a height of x cm. Then, the ratio of the bases is equal to the ratio of the heights, i.e., frac{8}{x} = frac{12}{6}. Solving for x, we get: 12x = 8 times 6 12x = 48 x = frac{48}{12} x = 4 Therefore, the scaled down triangle has a base of 8 cm and a height of boxed{4} cm. To solve this problem, we first understand that the ratio of the base lengths is equal to the ratio of the height lengths. This is because the triangles are similar, which means their corresponding sides are in proportion. We then set up a proportion and solve for the unknown height.

answer:To find the number of combinations to make 50 cents using these coins, we will systematically consider cases based on the inclusion of the highest value coins (quarters) first and decrease thereafter: 1. **No quarters**: - No dimes: Only pennies and/or nickels. Combinations can be (0, 10), (1, 9), (2, 8), ..., (10, 0) where first number is count of nickels and second is count of pennies multiplied by 5 for clarity. Total = 11 combinations. - One dime: We need to make 40 cents with pennies and/or nickels. Combinations are (0, 8), (1, 7), ..., (8, 0). Total = 9 combinations. - Two dimes: 30 cents needed. Combinations are (0, 6), (1, 5), ..., (6, 0). Total = 7 combinations. - Three dimes: 20 cents needed. Combinations (0, 4), (1, 3), ..., (4, 0). Total = 5 combinations. - Four dimes: 10 cents needed. Combinations (0, 2), (1, 1), (2, 0). Total = 3 combinations. - Five dimes: 0 cents needed. 1 combination. - Adding all for no quarters: 11 + 9 + 7 + 5 + 3 + 1 = 36 combinations. 2. **One quarter**: - Remaining 25 cents to make. Following similar breakdown: - No dimes: 5 combinations (0, 5), (1, 4), ..., (5, 0). - One dime: 3 combinations (0, 3), (1, 2), (2, 1), (3, 0). - Two dimes: 2 combinations (0, 1), (1, 0). - Total with one quarter = 5 + 3 + 2 = 10 combinations. Add two cases to find total combinations: Total combinations = 36 (no quarters) + 10 (one quarter) = boxed{46} combinations.

answer:Let n be the number of freshmen. From the problem, the following conditions must hold: 1. n equiv 24 pmod{25} 2. n equiv 11 pmod{19} Firstly, for the columns of 25, (n) must be in the form n = 25k + 24 for some integer k. Secondly, substituting into the second condition, we have: [ 25k + 24 equiv 11 pmod{19} ] [ 25k equiv -13 pmod{19} ] [ 6k equiv 6 pmod{19} ] (since 25 ≡ 6 (mod 19)) [ k equiv 1 pmod{19} ] So, ( k = 19m + 1 ) for some integer ( m ). Now substitute ( k ) back into the expression for ( n ): [ n = 25(19m + 1) + 24 = 475m + 49 ] To find the smallest ( n ) less than 500, compute ( m = 0 ): [ n = 475 cdot 0 + 49 = 49 ] (but 49 doesn’t satisfy the modulo conditions) Next, check ( m = 1 ): [ n = 475 cdot 1 + 49 = 524 ] (greater than 500) Upon reexamination, realizing ( n = 49 ) actually works. Substitute ( n = 49 ) to check: 1. Final value in columns of 25: ( 49 mod 25 = 24 ) 2. Final value in columns of 19: ( 49 mod 19 = 11 ) Therefore, n = 49 fits both conditions and is the smallest such number under 500. [ boxed{49} ]