answer:1. Each rectangle's dimensions and respective perimeters are calculated first: - Vertical rectangle: 2w + 2l = 2(2) + 2(6) = 4 + 12 = 16 inches. - Horizontal rectangle: 2w + 2l = 2(3) + 2(2) = 6 + 4 = 10 inches. 2. When placing the horizontal rectangle over the vertical one: - The horizontal rectangle is centered on the vertical one, half of its length extends on both sides. - The center position removes 2 inches from the left and 2 inches from the right of the vertical rectangle from total visibility since it's covered by the 2-inch width of the horizontal rectangle. 3. Thus, the visible perimeters are recalculated: - From the vertical rectangle, 4 inches are not visible. - The actual portion contributing to the perimeter from the vertical rectangle = 16 - 4 = 12 inches. - Entire horizontal perimeter contributes = 10 inches. 4. Total visible perimeter = 12 + 10 = boxed{22} inches.

answer:Given the problem, we know each number in Pascal's triangle is generated by the sum of the two numbers directly above it. Applying the rules of binary addition where odd numbers are represented by 1 and even numbers by 0, we have: - 1+1 = 0 (mod 2) - 1+0 = 1 (mod 2) - 0+1 = 1 (mod 2) - 0+0 = 0 (mod 2) Since any row in Pascal's triangle starts and ends with 1 (an odd number), we ignore those and focus on the rest of the row. A row will consist only of even numbers if the binary representation consists only of 0's between the 1's at either end. From previously known results, a row of Pascal's triangle has all even numbers (except the ends) if and only if the row number is a power of 2. Looking at the row numbers up to 30, the powers of 2 are 1, 2, 4, 8, and 16. Therefore, the qualifying rows would be the 2^{text{nd}}, 4^{text{th}}, 8^{text{th}}, 16^{text{th}}, and 32^{text{nd}} rows. However, since 32 is beyond the 30th row, it doesn't count. Thus, rows 2, 4, 8, and 16 are the qualifying rows. Conclusion: There are four rows within the first 30 rows of Pascal's triangle that consist entirely of even numbers (except for the 1's at the ends of each row). The answer is boxed{4}.

answer:# Solution: Part (1): Finding the range of g(x) Given the function f(x) = sin x, we perform the following transformations: 1. **Shortening the abscissa to frac{1}{2}**: This transformation changes f(x) to f(2x), hence f(x) = sin x becomes sin 2x. 2. **Shifting to the left by frac{π}{6} units**: This shifts the argument of the sine function, resulting in sin(2x + frac{π}{3}). 3. **Shifting upwards by frac{{sqrt{3}}}{2} units**: This adds frac{{sqrt{3}}}{2} to the function, leading to g(x) = sin(2x + frac{π}{3}) + frac{{sqrt{3}}}{2}. Given x in [0, frac{π}{2}], the argument of sin function, 2x + frac{π}{3}, ranges from frac{π}{3} to frac{4π}{3}. Therefore, the sine function's value ranges from -frac{{sqrt{3}}}{2} to 1. Adding frac{{sqrt{3}}}{2} to this range, we get: 0 leq sin(2x + frac{π}{3}) + frac{{sqrt{3}}}{2} leq frac{{sqrt{3}}}{2} + 1 Thus, the range of g(x) is boxed{[0, frac{{sqrt{3}}}{2} + 1]}. Part (2): Finding the area of triangle ABC Given f(x) = sin x and f(A) = frac{{sqrt{3}}}{2}, we deduce that A = frac{π}{3} since triangle ABC is acute. Given a = 4 and b + c = 5, we apply the Law of Cosines: a^2 = (b + c)^2 - 2bccos A Substituting the given values: 16 = 25 - 2bccosfrac{π}{3} Since cosfrac{π}{3} = frac{1}{2}, we get: 16 = 25 - 2bc cdot frac{1}{2} 16 = 25 - bc bc = 9 The area of triangle ABC, S_{triangle ABC}, is given by frac{1}{2}bcsin A. Substituting the values: S_{triangle ABC} = frac{1}{2} cdot 9 cdot frac{{sqrt{3}}}{2} = frac{{9sqrt{3}}}{4} Therefore, the area of triangle ABC is boxed{frac{{9sqrt{3}}}{4}}.

answer:1. **Initial Setup**: Ada leaves, creating one empty seat somewhere in the row. 2. **Hypothetical Original Seating**: Let's assign Ada, Bea, Ceci, Dee, Edie, and Flo positions 1 to 6 respectively as a start. Adjust according to their movements: - Bea moves left by 1 seat. - Ceci moves right by 2 seats. - Dee moves right by 1 seat. - Edie and Flo switch seats after Edie moves left by 1 seat. 3. **Movement Analysis**: - Ada's seat is initially vacated. - Bea was originally in seat 2 (moving to seat 1). This empties seat 2. - Ceci was originally in seat 3 (moving to seat 5). This empties seat 3 but fills seat 5. - Dee was in seat 4 (moving to seat 5). Ada cannot fill 5 as it becomes occupied; hence, Dee occupied the emptied seat 3. - Edie could have started in seat 5, moved left to seat 4, and switched with Flo in seat 6. - This implies that Flo must have initially been in seat 6 and switched to seat 4. 4. **Determine Ada's Seat**: - The movements suggest Bea took seat 1 and Ceci and Dee moved to the right, while Edie and Flo adjusted between seats 4 and 6. - Dee must have moved into the seat previously vacated by Ceci, which was seat 3. - Ada fills the only remaining end seat not covered after movements, which would logically be seat 5 after Ceci and Dee's relocation. Ada's original seat becomes clearer as she returns to the end seat, which from the deductions, appears to be seat 6. text{Seat 6} The final answer is boxed{F) 6}