answer:First, we need to find out how many coconut trees Randy has. To do this, we take half the number of mango trees and then subtract 5. Half of 60 mango trees is 60 / 2 = 30. Since Randy has 5 less than half as many coconut trees, we subtract 5 from 30. 30 - 5 = 25 coconut trees. Now, to find the total number of trees on Randy's farm, we add the number of mango trees to the number of coconut trees. 60 mango trees + 25 coconut trees = boxed{85} trees in total.

answer:To find the sum of the prime factors of 143, we start by testing divisibility by smaller prime numbers. 1. 143 is not divisible by 2 (it is odd). 2. 143 is not divisible by 3 (sum of digits 1 + 4 + 3 = 8 is not divisible by 3). 3. 143 is not divisible by 5 (it does not end in 0 or 5). 4. 143 is not divisible by 7 or 11 based on quick modular arithmetic checks. 5. We test divisibility by 11 and find that 143 does not divide evenly (since frac{143}{11} approx 13 but leaves a remainder when calculated precisely). 6. We test divisibility by 13, and find frac{143}{13} = 11, which is a whole number. Since 11 is also a prime number, we conclude that the prime factors of 143 are 11 and 13. Thus, the sum of the prime factors of 143 is 11 + 13 = boxed{24}.

answer:Given: ( p(x) = (x + a) q(x) ) is a real polynomial of degree ( n ). The largest absolute value of the coefficients of ( p(x) ) is ( h ) and the largest absolute value of the coefficients of ( q(x) ) is ( k ). We need to prove that ( k leq hn ). 1. Write the polynomials ( p(x) ) and ( q(x) ) in expanded form: [ p(x) = p_0 + p_1 x + p_2 x^2 + cdots + p_n x^n ] [ q(x) = q_0 + q_1 x + q_2 x^2 + cdots + q_{n-1} x^{n-1} ] Given ( h = max |p_i| ) and ( k = max |q_i| ). 2. Compute the coefficients of ( p(x) = (x + a) q(x) ): [ p(x) = (x + a)(q_0 + q_1 x + q_2 x^2 + cdots + q_{n-1} x^{n-1}) ] Expanding this, we get: [ p(x) = q_0 x + q_1 x^2 + q_2 x^3 + cdots + q_{n-1} x^n + a q_0 + a q_1 x + a q_2 x^2 + cdots + a q_{n-1} x^{n-1} ] Collecting the terms, we get: [ p_n = q_{n-1} ] [ p_{n-1} = q_{n-2} + a q_{n-1} ] [ p_{n-2} = q_{n-3} + a q_{n-2} ] [ vdots ] [ p_1 = q_0 + a q_1 ] [ p_0 = a q_0 ] 3. Prove for the case when ( |a| geq 1 ): Apply induction on the coefficients of ( q(x) ) to show ( |q_i| leq (i+1)h ). - **Base Case:** For ( i = 0 ), [ q_0 = frac{p_0}{a} ] So, [ |q_0| leq frac{|p_0|}{|a|} leq h quad text{(since ( |a| geq 1 ))} ] - **Induction Hypothesis:** Assume that ( |q_i| leq (i+1)h ) for some ( i ). - **Inductive Step:** Show that ( |q_{i+1}| leq (i+2)h ). [ q_{i+1} = frac{p_{i+1} - q_i}{a} ] Thus, [ |q_{i+1}| = left| frac{p_{i+1} - q_i}{a} right| leq frac{|p_{i+1}| + |q_i|}{|a|} leq h + (i+1)h = (i+2)h quad text{(since ( |a| geq 1 ))} ] Therefore, by induction, ( |q_i| leq (i+1)h ) for all ( i < n ). 4. Prove for the case when ( 0 < |a| < 1 ): Apply induction on the coefficients starting from ( q_{n-1} ) to show ( |q_{n-i}| leq ih ). - **Base Case:** For ( i = 1 ), [ q_{n-1} = p_n ] So, [ |q_{n-1}| leq h ] - **Induction Hypothesis:** Assume that ( |q_{n-i}| leq ih ) for some ( i ). - **Inductive Step:** Show that ( |q_{n-i-1}| leq (i+1)h ). [ q_{n-i-1} = p_{n-i-1} - a q_{n-i} ] Thus, [ |q_{n-i-1}| = |p_{n-i-1} - a q_{n-i}| leq |p_{n-i-1}| + |a| |q_{n-i}| leq h + |a| ih leq h + ih = (i+1)h quad text{(since ( |a| < 1 ))} ] Therefore, by induction, ( |q_{n-i}| leq ih ) for all ( 1 leq i leq n ). 5. Combine both cases ( |a| geq 1 ) and ( 0 < |a| < 1 ), in either case ( k leq max (h, 2h, cdots, nh) = nh ). # Conclusion: [ boxed{k leq hn} ]

answer:1. h(x) = f(x) - g(x) = log_{a}(1 + x) - log_{a}(1 - x). We have 1 + x > 0 and 1 - x > 0. Solving these inequalities, we get -1 < x < 1. The domain of h(x) is (-1, 1). Now, let's check the parity of h(x): h(-x) = log_{a}(frac{1 - x}{1 + x}) = -log_{a}(frac{1 + x}{1 - x}) = -h(x). Therefore, function h(x) is an odd function. 2. Given f(3) = 2, we find a = 2. Since h(x) < 0, We have 1 + x < 1 - x Rightarrow x < 0. And considering the domain of h(x), we have x in (-1, 1). Therefore, x in (-1, 0). 3. h(x) = log_{a}(frac{1 + x}{1 - x}) = log_{a}(-1 - frac{2}{x - 1}). Let phi(x) = -1 - frac{2}{x - 1}. We know that phi(x) = -1 - frac{2}{x - 1} is strictly increasing on [0, frac{1}{2}]. Therefore, when a > 1, h(x) is strictly increasing on [0, frac{1}{2}]. Also, h(0) = 0 and h(frac{1}{2}) = 1, which implies a = 3. When 0 < a < 1, h(x) is strictly decreasing on [0, frac{1}{2}]. But since h(0) = 1 contradicts h(0) = 0, we have no valid value for a in this case. In conclusion: boxed{a = 3}.