answer:Parker started with 4 dumbbells and added 2 more, making a total of 6 dumbbells. If the total weight of the dumbbells is 120 pounds, we can find the weight of each dumbbell by dividing the total weight by the number of dumbbells. 120 pounds ÷ 6 dumbbells = 20 pounds per dumbbell Each dumbbell weighs boxed{20} pounds.

answer:**Analysis** This problem focuses on finding the value of a trigonometric function, emphasizing the knowledge of double-angle formulas and reduction formulas, and is considered a basic question. **Solution** Given the problem, we have (frac{pi}{6} < alpha+ frac{pi}{6} < frac{2pi}{3}), and (cos left(alpha+ frac{pi}{6}right)= frac{4}{5} ). Therefore, (cos left(2alpha- frac{pi}{6}right)=cos left( frac{pi}{6}-2alpharight)=cos left[ frac{pi}{2}-left( frac{pi}{3}+2alpharight)right] =sin left( frac{pi}{3}+2alpharight)=2sin left( frac{pi}{6}+alpharight)cos left( frac{pi}{6}+alpharight)= frac{24}{25} ). Thus, the answer is boxed{frac{24}{25}}.

answer:1. **Choose a basis for ( U_1 ):** Let ( {e_1, e_2, e_3} ) be a basis for ( U_1 ). Since ( dim U_1 = 3 ), we have 3 basis vectors. 2. **Extend the basis of ( U_1 ) to a basis for ( U_2 ):** Since ( U_1 subseteq U_2 ) and ( dim U_2 = 6 ), we can extend the basis ( {e_1, e_2, e_3} ) of ( U_1 ) to a basis ( {e_1, e_2, e_3, e_4, e_5, e_6} ) of ( U_2 ) by adding 3 more vectors. 3. **Extend the basis of ( U_2 ) to a basis for ( V ):** Since ( dim V = 10 ), we can further extend the basis ( {e_1, e_2, e_3, e_4, e_5, e_6} ) of ( U_2 ) to a basis ( {e_1, e_2, e_3, e_4, e_5, e_6, e_7, e_8, e_9, e_{10}} ) of ( V ) by adding 4 more vectors. 4. **Consider the linear maps ( T ) that preserve ( U_1 ) and ( U_2 ):** We need to find the dimension of the set of all linear maps ( T: V rightarrow V ) such that ( T(U_1) subseteq U_1 ) and ( T(U_2) subseteq U_2 ). 5. **Structure of the linear map ( T ):** Since ( T ) must map ( U_1 ) into itself and ( U_2 ) into itself, the matrix representation of ( T ) with respect to the chosen basis will have a block structure: [ T = begin{pmatrix} A & B & C 0 & D & E 0 & 0 & F end{pmatrix} ] where: - ( A ) is a ( 3 times 3 ) matrix representing the action of ( T ) on ( U_1 ), - ( D ) is a ( 3 times 3 ) matrix representing the action of ( T ) on the complement of ( U_1 ) in ( U_2 ), - ( F ) is a ( 4 times 4 ) matrix representing the action of ( T ) on the complement of ( U_2 ) in ( V ), - ( B ) is a ( 3 times 3 ) matrix, - ( C ) is a ( 3 times 4 ) matrix, - ( E ) is a ( 3 times 4 ) matrix. 6. **Count the number of free parameters:** - The matrix ( A ) has ( 3 times 3 = 9 ) free parameters. - The matrix ( D ) has ( 3 times 3 = 9 ) free parameters. - The matrix ( F ) has ( 4 times 4 = 16 ) free parameters. - The matrix ( B ) has ( 3 times 3 = 9 ) free parameters. - The matrix ( C ) has ( 3 times 4 = 12 ) free parameters. - The matrix ( E ) has ( 3 times 4 = 12 ) free parameters. 7. **Sum the number of free parameters:** [ 9 + 9 + 16 + 9 + 12 + 12 = 67 ] Therefore, the dimension of the set ( varepsilon ) of all such linear maps ( T ) is ( 67 ). The final answer is (boxed{67}).

answer:Let's denote the initial population as ( P ). Over 10 years, the population increases by 60% due to birth. This means the population due to birth alone would be ( P + 0.60P = 1.60P ). However, we also need to account for emigration and immigration. Over 10 years, 2000 people leave per year due to emigration, which totals ( 2000 times 10 = 20,000 ) people. Similarly, 2500 people come in per year due to immigration, which totals ( 2500 times 10 = 25,000 ) people. The net effect of emigration and immigration over 10 years is ( 25,000 - 20,000 = 5,000 ) people added to the population. So, the final population is the initial population increased by 60% plus the net effect of emigration and immigration, which is: [ 1.60P + 5,000 = 165,000 ] Now, we can solve for ( P ): [ 1.60P = 165,000 - 5,000 ] [ 1.60P = 160,000 ] [ P = frac{160,000}{1.60} ] [ P = 100,000 ] Therefore, the initial population of the area was boxed{100,000} people.