answer:1. **Form of the Polynomial:** The disrespectful function p(x) has leading coefficient 1, so it can be written in factored form as p(x) = (x-r)(x-s). 2. **Condition for p(p(x)) = 0:** Plugging our form in, we get: [ p(p(x)) = ((x-r)(x-s)-r)((x-r)(x-s)-s) = 0. ] The roots of this equation are (x-r)(x-s) = r and (x-r)(x-s) = s. 3. **Requirement of Exactly Three Real Roots:** By the fundamental theorem of algebra, each quadratic equation should have two roots, totaling four possible values of x. However, the problem states that this equation is satisfied by exactly three values of x. Therefore, one equation must give a double root. Without loss of generality, let the equation (x-r)(x-s) = r be the equation that produces the double root. 4. **Condition for a Double Root:** Expanding gives x^2-(r+s)x+rs-r = 0. For a double root, the discriminant must be zero: [ (r+s)^2-4(rs-r) = 0 implies r^2+2rs+s^2-4rs+4r = 0 implies r^2-2rs+s^2+4r = 0. ] 5. **Solving for r and s:** We can rewrite r^2-2rs+s^2+4r = 0 as (r-s)^2+4r = 0. Let q = sqrt{-r}, then r = -q^2 and the equation becomes: [ (2r-s)^2 + 4r = 0 implies s = 2r pm 2sqrt{-r}. ] To maximize r+s, we choose s = 2r + 2sqrt{-r} and solve for r and s: [ m = 2(-q^2+q) implies m = -2q(q-1). ] The vertex of -2q(q-1) is at q = frac{1}{2}, giving r = -left(frac{1}{2}right)^2 = -frac{1}{4} and s = 2(-frac{1}{4}) + 2left(frac{1}{2}right) = frac{3}{4}. 6. **Calculating tilde{p}(1):** Now, substituting r = -frac{1}{4} and s = frac{3}{4} into p(x) = (x-r)(x-s) and evaluating at x = 1: [ tilde{p}(1) = left(1 - left(-frac{1}{4}right)right)left(1 - frac{3}{4}right) = left(1 + frac{1}{4}right)left(frac{1}{4}right) = frac{5}{4} cdot frac{1}{4} = frac{5}{16}. ] Thus, the value of tilde{p}(1) is boxed{textbf{(A) } frac{5}{16}}.

answer:To solve the given expression step-by-step, we start by breaking it down into its individual components: 1. The first part is (-frac{1}{2})^{-2}. This can be simplified as follows: [ (-frac{1}{2})^{-2} = left(frac{1}{-frac{1}{2}}right)^2 = (2)^2 = 4 ] 2. The second part is (3-pi)^0. Any number (except for 0) raised to the power of 0 equals 1, so: [ (3-pi)^0 = 1 ] 3. The third part is |sqrt{3}-2|. This is the absolute value of sqrt{3}-2, which is: [ |sqrt{3}-2| = 2-sqrt{3} ] since sqrt{3} is approximately 1.732, and 2-sqrt{3} is positive. 4. The fourth part involves trigonometry, 2sin60°. The sine of 60 degrees is frac{sqrt{3}}{2}, so: [ 2sin60° = 2timesfrac{sqrt{3}}{2} = sqrt{3} ] Adding all these parts together: [ 4 + 1 + (2-sqrt{3}) + sqrt{3} ] [ = 4 + 1 + 2 - sqrt{3} + sqrt{3} ] [ = 4 + 1 + 2 ] [ = 7 ] Therefore, the final answer is boxed{7}.

answer:Let's denote the amount the laborer is paid for each day he works as ( x ) rupees. According to the problem, the laborer was absent for 5 days, so he worked for ( 25 - 5 = 20 ) days. For each day he was absent, he was fined 50 paise, which is ( frac{50}{100} ) rupees or ( 0.50 ) rupees. The total amount he should have received for the days he worked is ( 20x ) rupees. The total amount he was fined for the days he was absent is ( 5 times 0.50 = 2.50 ) rupees. The total amount he actually received is the amount for the days he worked minus the fines for the days he was absent, which is given as Rs. 37.50. So, we have the equation: [ 20x - 2.50 = 37.50 ] Now, we solve for ( x ): [ 20x = 37.50 + 2.50 ] [ 20x = 40.00 ] [ x = frac{40.00}{20} ] [ x = 2.00 ] Therefore, the laborer was paid Rs. boxed{2.00} for each day he worked.

answer:First, we find the derivative of y=x^2, which is y'=2x. This derivative represents the slope of the tangent line at any point on the curve. At the point (a_k, a_k^2), the slope of the tangent line is 2a_k. The equation of the tangent line can be written as: [y - a_k^2 = 2a_k(x - a_k)] Since the tangent line intersects the x-axis at y=0, we substitute y=0 into the equation and solve for x: [0 - a_k^2 = 2a_k(x - a_k)] [0 = 2a_kx - 2a_k^2 - a_k^2] [2a_k^2 = 2a_kx] [x = a_k + frac{a_k^2}{2a_k}] [x = a_k + frac{a_k}{2}] Given that this x-coordinate is a_{k+1}, we have: [a_{k+1} = a_k + frac{a_k}{2}] [a_{k+1} = frac{3}{2}a_k] Since a_1 = 16, we can find the subsequent terms: [a_2 = frac{3}{2} cdot 16 = 24] [a_3 = frac{3}{2} cdot 24 = 36] [a_4 = frac{3}{2} cdot 36 = 54] [a_5 = frac{3}{2} cdot 54 = 81] Therefore, the sum a_1 + a_3 + a_5 is: [16 + 36 + 81 = boxed{133}]