answer:For the quadratic equation (10x^2 + 25x + c = 0) to have two real roots, its discriminant must be positive. Therefore, we need: [25^2 - 4 times 10 times c > 0] [625 - 40c > 0] [c < frac{625}{40}] [c < 15.625] The largest integer smaller than (15.625) is 15. Thus, the positive integer values of (c) are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, and 15. We calculate their product: [1 cdot 2 cdot 3 cdot 4 cdot 5 cdot 6 cdot 7 cdot 8 cdot 9 cdot 10 cdot 11 cdot 12 cdot 13 cdot 14 cdot 15 = 1307674368000] So, the product of all valid (c) values is (boxed{1307674368000}).

answer:Since q(x) is a monic cubic polynomial with real coefficients and has 2-3i as a root, it must also include 2+3i as a root. We start by finding the quadratic factor that includes both 2-3i and 2+3i as its roots: begin{align*} left(x - (2-3i)right) left(x - (2+3i)right) &= left(x - 2 + 3iright)left(x - 2 - 3iright) &= (x - 2)^2 - (3i)^2 &= x^2 - 4x + 4 + 9 &= x^2 - 4x + 13. end{align*} Since q(x) is cubic, it has one more real root, say r. Therefore, we can express q(x) as: q(x) = (x^2 - 4x + 13)(x - r). Given that q(x) is monic, the coefficient of the highest degree term is 1. Plugging in x = 0 yields q(0) = 13r, and we know that q(0) = -30. Solving for r: 13r = -30 Rightarrow r = -frac{30}{13}. Thus the polynomial becomes: begin{align*} q(x) &= (x^2 - 4x + 13) left(x + frac{30}{13}right) &= x^3 - frac{22}{13}x^2 + frac{82}{13}x - 30. end{align*} The final answer, in expanded form, is: boxed{x^3 - frac{22}{13}x^2 + frac{82}{13}x - 30}.

answer:1. **Rewrite the Expression**: Starting with the given problem, [ frac{2^{2020} + 2^{2016}}{2^{2020} - 2^{2016}} ] 2. **Factor out 2^{2016} from Both the Numerator and the Denominator**: [ frac{2^{2016} cdot (2^4 + 1)}{2^{2016} cdot (2^4 - 1)} ] Using the property a^m cdot a^n = a^{m+n}, 2^{2016} is factored out. 3. **Simplify the Expression Inside the Parentheses**: [ 2^4 + 1 = 16 + 1 = 17 quad text{and} quad 2^4 - 1 = 16 - 1 = 15 ] The expression now becomes: [ frac{2^{2016} cdot 17}{2^{2016} cdot 15} ] 4. **Cancel out 2^{2016} from the Numerator and the Denominator**: [ frac{17}{15} ] 5. **Conclude with the Final Answer**: [ frac{17{15}} ] The final answer is boxed{frac{17}{15}}

answer:Let's denote the original number of red apples as R and the original number of green apples as G. According to the information given: G = 32 (originally there were 32 green apples) After the truck arrives, the number of green apples becomes: G + 340 = 32 + 340 = 372 Now, we are told that there are 140 more green apples than red apples after the truck arrives. So we can write: 372 = R + 140 Now, let's solve for R: R = 372 - 140 R = 232 So originally, there were 232 red apples. To find out how many more red apples there were than green apples before the truck arrived, we subtract the original number of green apples from the original number of red apples: More red apples than green apples = R - G More red apples than green apples = 232 - 32 More red apples than green apples = 200 Therefore, there were boxed{200} more red apples than green apples before the truck arrived.