answer:First, let's calculate the total calories in the salad: Lettuce: 50 calories Carrots: 2 * 50 = 100 calories Tomatoes: 30 calories Olives: 60 calories Cucumber: 15 calories Dressing: 210 calories Total salad calories = 50 + 100 + 30 + 60 + 15 + 210 = 465 calories Now, let's calculate the total calories in the pizza: Crust: 600 calories Pepperoni: 1/3 * 600 = 200 calories Mushrooms: 2/5 * 600 = 240 calories Cheese: 400 calories Total pizza calories = 600 + 200 + 240 + 400 = 1440 calories Now, let's calculate the calories in the garlic bread: Garlic bread per slice: 200 calories Jackson eats 1.5 slices, so the total calories for garlic bread = 1.5 * 200 = 300 calories Now, let's calculate the calories Jackson consumes: Salad: 3/8 * 465 = 174.375 calories Pizza: 2/7 * 1440 = 411.42857 calories Garlic bread: 300 calories Total calories consumed = 174.375 + 411.42857 + 300 = 885.80357 calories Rounded to the nearest whole number, Jackson consumes boxed{886} calories.

answer:1. **Substitution and Transformation**: Let us introduce the substitutions ( x = frac{1}{a} ), ( y = frac{1}{b} ), and ( z = frac{1}{c} ). Then, the given condition ( ab + bc + ca = abc ) can be rewritten using these substitutions: [ frac{1}{x}frac{1}{y} + frac{1}{y}frac{1}{z} + frac{1}{z}frac{1}{x} = frac{1}{x}frac{1}{y}frac{1}{z} ] Simplifying, we have: [ frac{1}{xyz} (xy + yz + zx) = frac{1}{xyz} ] This implies: [ xy + yz + zx = 1 ] 2. **Expression to Prove**: We need to show that: [ sum frac{x^4 + y^4}{x^3 + y^3} geq 1 ] 3. **Simplification and Testing**: Let us consider an auxiliary inequality. We aim to find a constant ( a ) such that: [ frac{x^4 + y^4}{x^3 + y^3} geq a x + (1 - a) y ] Due to the symmetry in ( x ) and ( y ), a natural attempt is ( a = frac{1}{2} ). Thus, we seek to prove: [ frac{x^4 + y^4}{x^3 + y^3} geq frac{x + y}{2} ] 4. **Verification**: Assume ( t = frac{x}{y} ), then we need to show: [ frac{t^4 + 1}{t^3 + 1} geq frac{t + 1}{2} ] Multiplying both sides by ( 2(t^3 + 1) ) results in: [ 2(t^4 + 1) geq (t^3 + 1)(t + 1) ] Simplifying, we obtain: [ 2t^4 + 2 geq t^4 + t^3 + t + 1 ] Rearranging terms gives: [ t^4 + t^3 - t^3 + t^2 - t + 1 geq t^2 + 1 ] This simplifies to: [ (t-1)^2(t^2 + t + 1) geq 0 ] Since ( t geq 0 ), this inequality is always true, establishing our earlier assumption. 5. **Conclusion**: Given the preceding steps, we have shown that: [ frac{x^4 + y^4}{x^3 + y^3} geq frac{x + y}{2} ] Summing cyclically over ( (x,y,z) ): [ sum frac{x^4 + y^4}{x^3 + y^3} geq sum frac{x+y}{2} = frac{sum x + sum y}{2} = frac{2}{2} = 1 ] Thus, the statement is proven: [ boxed{1} ]

answer:1. Given the equation: [ 16 sin(pi x) cos(pi x) = 16x + frac{1}{x} ] We will use the trigonometric identity: [ sin(2theta) = 2 sin(theta) cos(theta) ] Here, let (theta = pi x). Thus: [ sin(2 pi x) = 2 sin(pi x) cos(pi x) ] Therefore, the equation can be rewritten as: [ 16 sin(pi x) cos(pi x) = 8 sin(2 pi x) ] The original equation becomes: [ 8 sin(2pi x) = 16x + frac{1}{x} ] 2. Since the sine function (sin(2pi x)) oscillates between -1 and 1, the function (8 sin(2pi x)) will oscillate between -8 and 8: [ -8 leq 8 sin(2pi x) leq 8 ] This means: [ -8 leq 16x + frac{1}{x} leq 8 ] 3. Let's consider the cases when (16x + frac{1}{x} = 8): [ 16x + frac{1}{x} = 8 ] Multiplying through by (x) (assuming (x neq 0)): [ 16x^2 + 1 = 8x ] Rearranging the equation: [ 16x^2 - 8x + 1 = 0 ] We solve this quadratic equation for (x) using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 16), (b = -8), and (c = 1): [ x = frac{8 pm sqrt{(-8)^2 - 4 cdot 16 cdot 1}}{2 cdot 16} ] Simplifying inside the square root: [ x = frac{8 pm sqrt{64 - 64}}{32} ] [ x = frac{8 pm sqrt{0}}{32} ] [ x = frac{8}{32} ] [ x = frac{1}{4} ] 4. Now, consider the cases when (16x + frac{1}{x} = -8): [ 16x + frac{1}{x} = -8 ] Multiplying through by (x): [ 16x^2 + 1 = -8x ] Rearranging: [ 16x^2 + 8x + 1 = 0 ] Solving this quadratic equation: [ x = frac{-8 pm sqrt{8^2 - 4 cdot 16 cdot 1}}{2 cdot 16} ] [ x = frac{-8 pm sqrt{64 - 64}}{32} ] [ x = frac{-8 pm sqrt{0}}{32} ] [ x = frac{-8}{32} ] [ x = -frac{1}{4} ] Conclusion: The solutions to the equation are: [ boxed{left{-frac{1}{4}, frac{1}{4}right}} ]

answer:To solve this problem, we need to evaluate if the series telescopes similarly to the original problem. For the first few terms, the structure can be reviewed similarly: [left(frac{1}{3+1} + frac{2}{3^2+1} + frac{4}{3^4+1}right)] Multiplying together the denominators: [(3+1)(3^2+1)(3^4+1)] To attempt telescoping, multiply by (3-1): [(3-1)(3+1)(3^2+1)(3^4+1) = (3^2-1)(3^2+1)(3^4+1) = (3^4-1)(3^4+1) = 3^8-1] Thus, the pattern looks similar, and we test for telescoping by subtracting consecutive terms: [frac{1}{3 + 1} - frac{2}{3^2 + 1} = frac{1}{4} - frac{2}{10} = frac{5}{20} - frac{4}{20} = frac{1}{20}] Continuing this method: [frac{2}{3^2 + 1} - frac{4}{3^4 + 1}] [frac{4}{3^4 + 1} - frac{8}{3^8 + 1}] [dots] It is clear that the series telescopes, and converging towards: [frac{1}{3-1} = frac{1}{2}] The exact value of the series is boxed{frac{1}{2}}.