answer:# Part 1: Jordan Normal Form of ( e^{d/dt} ) 1. **Operator Analysis**: The operator ( frac{d}{dt} ) acts on quasi-polynomials ( e^{lambda t} p(t) ) where ( p(t) ) is a polynomial of degree less than 5. For a polynomial ( p(t) ), we have: [ frac{d}{dt}(e^{lambda t} p(t)) = e^{lambda t} (lambda p(t) + p'(t)) ] This can be written as: [ frac{d}{dt} = lambda text{id} + f ] where ( f(p) = p'(t) ). 2. **Exponential of the Operator**: We need to find ( expleft(frac{d}{dt}right) ). Using the linearity of the exponential function for operators, we get: [ expleft(frac{d}{dt}right) = exp(lambda text{id} + f) = e^{lambda} exp(f) ] 3. **Nilpotent Operator ( f )**: The operator ( f ) (differentiation) is nilpotent with index 6 because the highest degree of ( p(t) ) is 4, and the 5th derivative of any polynomial of degree less than 5 is zero. Thus, ( f^6 = 0 ). 4. **Jordan Form of ( f )**: Since ( f ) is nilpotent with index 6, in any basis, ( f ) admits a Jordan form similar to ( J_6 ), the Jordan nilpotent matrix of order 6: [ J_6 = begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 0 & 0 & 1 & 0 & 0 & 0 0 & 0 & 0 & 1 & 0 & 0 0 & 0 & 0 & 0 & 1 & 0 0 & 0 & 0 & 0 & 0 & 1 0 & 0 & 0 & 0 & 0 & 0 end{pmatrix} ] 5. **Exponential of ( f )**: The Jordan form of ( exp(f) ) is ( I_6 + J_6 ), where ( I_6 ) is the identity matrix of order 6. 6. **Combining Results**: Therefore, the Jordan form of ( expleft(frac{d}{dt}right) ) is: [ e^{lambda} (I_6 + J_6) = e^{lambda} I_6 + e^{lambda} J_6 ] # Part 2: Jordan Normal Form of ( text{ad}_A ) 1. **Operator Definition**: The operator ( text{ad}_A ) is defined as ( B mapsto [A, B] ), where ( A ) is a diagonal matrix. This can be written as: [ text{ad}_A = A otimes I - I otimes A^T ] 2. **Diagonalizability**: If ( A ) is diagonalizable with spectrum ( {lambda_i}_i ), then ( text{ad}_A ) is also diagonalizable. 3. **Spectrum of ( text{ad}_A )**: The spectrum of ( text{ad}_A ) is given by the set ( {lambda_i - lambda_j}_{i,j} ). The final answer is ( boxed{ e^{lambda} I_6 + e^{lambda} J_6 } ) for the first part and the spectrum ( {lambda_i - lambda_j}_{i,j} ) for the second part.

answer:If John needs 20 nails in total for the house wall and the large planks together need 15 nails, then the small planks would need the remaining number of nails. So, the number of nails needed for the small planks together would be: Total nails - Nails for large planks = Nails for small planks 20 nails - 15 nails = 5 nails Therefore, the small planks together need boxed{5} nails.

answer:Let the angle of inclination of the line be alpha, Since the line sqrt {3}x-y-3=0 can be rewritten as y= sqrt {3}x-3, Therefore, the slope of the line is k= sqrt {3}, Therefore, tanalpha= sqrt {3}; Given that 0°leqalpha<180°, Therefore, alpha=60°, Hence, the correct choice is: boxed{C}. By deriving the slope of the line from the equation of the line, we can find the tangent value of the angle of inclination, and thus determine the angle of inclination. This question tests the ability to determine the angle of inclination of a line based on its equation, which is a basic problem.

answer:Use the discriminant approach to determine when the quadratic equations do not have two distinct real solutions: 1. **Discriminants Analysis**: - For ( x^2 + 2bx + c = 0 ), discriminant ( Delta_1 = (2b)^2 - 4c = 4b^2 - 4c ). We require ( Delta_1 leq 0 ) or ( b^2 - c leq 0 ), thus ( c geq b^2 ). - For ( x^2 + 2cx + b = 0 ), discriminant ( Delta_2 = (2c)^2 - 4b = 4c^2 - 4b ). We require ( Delta_2 leq 0 ) or ( c^2 - b leq 0 ), thus ( b geq c^2 ). 2. **Combine and Simplify Inequalities**: [ left{ begin{array}{l} c geq b^2 b geq c^2 end{array} right. ] This system of inequalities is typically satisfied when ( b = c = 1 ), because any larger values would cause one of the inequalities to be violated. 3. **Verify Feasible Pairs**: - Test ( b = 1 ), ( c = 1 ): Both inequalities ( 1 geq 1^2 ) and ( 1 geq 1^2 ) hold true. - Increase ( b ) or ( c ) leads to violation of at least one inequality, as ( b^2 ) and ( c^2 ) grow faster than ( b ) and ( c ) respectively. 4. **Conclusion**: The only feasible pair is ( (b, c) = (1, 1) ). Therefore, the answer is 1. The final answer is boxed{textbf{(B) } 1}