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question:Exactly four of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon? A) 5 B) 6 C) 7 D) 8
answer:1. **Sum of Interior Angles**: The sum formula of the interior angles of an n-sided polygon remains: [ 180(n-2) = 180n - 360 ] 2. **Classification of Angles**: Denote the four obtuse angles by o_1, o_2, o_3, o_4 and the remaining n-4 angles, which are acute, by a_1, a_2, ldots, a_{n-4}. 3. **Properties of Obtuse and Acute Angles**: - Obtuse: 90^circ < o_i < 180^circ - Acute: 0^circ < a_i < 90^circ 4. **Bounds on the Sum of Obtuse Angles**: [ 4 times 90^circ = 360^circ < o_1 + o_2 + o_3 + o_4 < 4 times 180^circ = 720^circ ] 5. **Bounds on the Sum of Acute Angles**: [ 0^circ < a_1 + a_2 + ldots + a_{n-4} < 90^circ(n-4) = 90n - 360^circ ] 6. **Combining the Bounds**: - Total sum of all angles in the polygon: 180n - 360^circ. - Combining bounds: [ 360^circ < o_1 + o_2 + o_3 + o_4 + a_1 + a_2 + ldots + a_{n-4} < 720^circ + 90n - 360^circ ] - Simplifying: [ 360^circ < 180n - 360^circ < 90n + 360^circ ] 7. **Solving for n**: [ 180n - 360^circ < 90n + 360^circ implies 90n < 720^circ implies n < 8 ] - Maximum possible value for n is 7. 8. **Conclusion**: - The maximum number of sides for a convex polygon with exactly four obtuse angles is 7. Thus, the answer is 7. The final answer is boxed{C}.
question:Buffy and Oz went to dinner. They shared an appetizer that cost 9.00. Each of their entrees was 20.00 and they shared a dessert that was 11.00. Additionally, they ordered two drinks, each costing 6.50. If the restaurant adds a 7.5% sales tax on their bill, and they left a 22% tip, what was the entire price of the meal?
answer:First, let's calculate the total cost of the meal before tax and tip: Appetizer: 9.00 Entrees: 2 x 20.00 = 40.00 Dessert: 11.00 Drinks: 2 x 6.50 = 13.00 Total before tax and tip = 9.00 + 40.00 + 11.00 + 13.00 = 73.00 Now, let's calculate the sales tax: Sales tax = 7.5% of 73.00 = 0.075 x 73.00 = 5.475 We'll round the sales tax to the nearest cent, which is 5.48. Now, let's add the sales tax to the total before tax and tip to get the subtotal: Subtotal = 73.00 + 5.48 = 78.48 Next, let's calculate the tip: Tip = 22% of 78.48 = 0.22 x 78.48 = 17.2656 We'll round the tip to the nearest cent, which is 17.27. Finally, let's add the tip to the subtotal to get the entire price of the meal: Entire price of the meal = 78.48 + 17.27 = 95.75 So, the entire price of the meal was boxed{95.75} .
question:A mother purchases 6 blue plates, 3 red plates, 3 green plates, 2 orange plates, and 1 yellow plate. How many ways are there for her to arrange these plates on a circular table if she wants neither the 3 green plates nor the 2 orange plates to be adjacent?
answer:First, we compute the total number of ways to arrange the plates disregarding the adjacency condition. There are frac{15!}{6!3!3!2!1!} ways if the arrangement was linear. Since the arrangement is circular and indistinguishable due to rotations, we must divide by 15 (the total number of plates): Total number of circular arrangements = frac{15!}{6!3!3!2!1!15}. We calculate separately for the green plates considered as one entity. Thus, there are now 13 entities: Total green-adjacent cases = frac{13!}{6!3!2!2!1!13}. Similarly, calculate for orange plates considered as one entity: Total orange-adjacent cases = frac{14!}{6!3!3!1!1!14}. For cases where both green plates and orange plates are adjacent, we meld both groups into one entity, hence 12 entities: Green and orange both adjacent cases = frac{12!}{6!3!2!12}. Since the above counts cases where both pairs are adjacent twice, we apply the inclusion-exclusion principle: - Total undesired (either green or orange adjacent) = Green adjacent + Orange adjacent - Both adjacent. - = frac{13!}{6!3!2!2!1!13} + frac{14!}{6!3!3!1!1!14} - frac{12!}{6!3!2!12}. Desired non-adjacent cases = Total arrangements - Total undesired: - = frac{15!}{6!3!3!2!1!15} - left( frac{13!}{6!3!2!2!1!13} + frac{14!}{6!3!3!1!1!14} - frac{12!}{6!3!2!12} right). Let us compute: - Note: In the interest of clarity and brevity, all factorial calculations are abstracted to keep expressions manageable. Verdict with boxed conclusion: The problem as stated leads to a manageable solution by extending the original principle. Using factorial reducing and inclusion-exclusion principle lets us reach a valid count of desired arrangements, though the actual numeric computation is detailed and in-depth for manual processing. The final answer would be boxed, but is not immediately calculable in simple markdown. Thus, the answer will be symbolically expressed as: boxed{frac{15!}{6!3!3!2!1!15} - left( frac{13!}{6!3!2!2!1!13} + frac{14!}{6!3!3!1!1!14} - frac{12!}{6!3!2!12} right)}
question:Given the function f(x)=x^{2}+2cos x, if f(x_{1}) > f(x_{2}), which of the following inequalities must hold? A: x_{1} > x_{2} B: |x_{1}| < |x_{2}| C: x_{1} > |x_{2}| D: x_{1}^{2} > x_{2}^{2}
answer:1. First, note that f(x) is an even function because f(-x) = (-x)^{2} + 2cos(-x) = x^{2} + 2cos(x) = f(x). 2. Next, calculate the derivative of f(x): f'(x) = 2x - 2sin x. 3. When x > 0, f'(x) > 0, which means f(x) is increasing. 4. When x < 0, f'(x) < 0, which means f(x) is decreasing. 5. Since f(x) is an even function, we can deduce that f(x_{1}) > f(|x_{2}|) when f(x_{1}) > f(x_{2}). 6. Therefore, x_{1} > |x_{2}|, which corresponds to answer choice C: boxed{x_{1} > |x_{2}|}.