answer：First, I need to determine the least common multiple (LCM) of 4 and 8 to find the smallest number that both 4 and 8 can divide into without leaving a remainder.The prime factorization of 4 is (2^2), and the prime factorization of 8 is (2^3). The LCM is found by taking the highest power of each prime number present in the factorizations. In this case, the highest power of 2 is (2^3), so the LCM of 4 and 8 is 8.Next, I need to find the smallest integer (n) such that (60n) is divisible by 8. To do this, I divide 60 by the LCM:[frac{60}{8} = 7.5]Since (n) must be an integer, I round up to the next whole number, which is 8. Finally, I verify by multiplying 60 by 8:[60 times 8 = 480]Since 480 is divisible by both 4 and 8, the minimum value of (n) is 8.

answer：To determine the coefficient of ( x^{3} ) in the expansion of ( (1 - x)(1 + x)^{6} ), I can use the binomial theorem. First, I'll expand ( (1 + x)^{6} ) using the binomial coefficients. This gives me ( 1 + 6x + 15x^{2} + 20x^{3} + dots ). Next, I'll multiply this expansion by ( (1 - x) ). This involves distributing ( 1 ) and ( -x ) across each term of the expanded ( (1 + x)^{6} ). Multiplying by ( 1 ) leaves the terms unchanged: ( 1 + 6x + 15x^{2} + 20x^{3} + dots ). Multiplying by ( -x ) shifts each term by one degree and changes their signs: ( -x - 6x^{2} - 15x^{3} - 20x^{4} - dots ).Now, I'll combine like terms to find the coefficient of ( x^{3} ). The ( x^{3} ) terms from both expansions are ( 20x^{3} ) and ( -15x^{3} ). Adding these together gives ( 5x^{3} ).Therefore, the coefficient of ( x^{3} ) in the expansion is ( 5 ).

answer：First, I recognize that γ is inversely proportional to δ, which means their product is a constant, k.Given that γ = 5 when δ = 15, I can calculate k by multiplying these values: 5 * 15 = 75.Now, to find γ when δ = 45, I use the same constant k. Setting up the equation 45 * γ = 75, I solve for γ by dividing both sides by 45: γ = 75 / 45 = 5/3.Therefore, the value of γ when δ is 45 is 5/3.

answer：First, I need to find the inverse function of ( y = 1 + 2^{-x} ). To do this, I'll start by swapping ( x ) and ( y ) in the equation, resulting in ( x = 1 + 2^{-y} ).Next, I'll isolate the exponential term by subtracting 1 from both sides, which gives me ( x - 1 = 2^{-y} ).Now, I'll solve for ( y ) by taking the logarithm base 2 of both sides. This step transforms the equation into ( log_2(x - 1) = -y ).Finally, I'll multiply both sides by -1 to solve for ( y ), obtaining the inverse function ( y = -log_2(x - 1) ).To find ( g(5) ), I'll substitute 5 into the inverse function: ( g(5) = -log_2(5 - 1) = -log_2(4) ). Since ( log_2(4) = 2 ), it follows that ( g(5) = -2 ).Therefore, the correct answer is (boxed{text{B}}).

answer：First, I need to determine the number of spruces in the forest. Since spruces make up 10% of the total 4000 trees, I can calculate this by multiplying 4000 by 0.10.Next, I'll find out how many pines there are. Pines account for 13% of the total trees, so I'll multiply 4000 by 0.13 to get the number of pines.The problem states that there are as many oaks as there are spruces and pines combined. So, I'll add the number of spruces and pines together to find the total number of oaks.Finally, to find the number of birches, I'll subtract the total number of spruces, pines, and oaks from the overall total of 4000 trees. This will give me the number of birches in the forest.

answer：First, I need to determine the agreed amount to be paid after one year of service, which I'll denote as ( A ). The servant received a uniform and Rs. 250 for 9 months of service. Since 9 months is three-fourths of a year, the servant should receive three-fourths of the agreed amount plus the uniform.I can set up the equation:[frac{3}{4}A + U = 250 + U]By subtracting ( U ) from both sides, I isolate the term with ( A ):[frac{3}{4}A = 250]To find the full agreed amount ( A ), I divide Rs. 250 by ( frac{3}{4} ):[A = frac{250}{frac{3}{4}} = 250 times frac{4}{3} = 333.overline{3}]Therefore, the agreed amount to be paid after one year of service is approximately Rs. 333.33.