answer:Mike had 16 video games, but 8 of them weren't working. So, the number of working games is: 16 (total games) - 8 (non-working games) = 8 working games If he sells each working game for 7, the total amount of money he could earn is: 8 (working games) x 7 (price per game) = 56 So, Mike could earn boxed{56} by selling his working video games.

answer:We are given that (AD) is the altitude of the right-angled triangle (triangle ABC) from vertex (A) to side (BC). Point (O) is an arbitrary point on (AD). Lines (BO) and (CO) are extended to intersect (AC) at (F) and (AB) at (E) respectively. We are to prove that (angle EDO = angle FDO). 1. Let (alpha = angle FDO) and (beta = angle EDO). 2. Consider the right triangles (triangle ADC) and (triangle ADB): [ tan angle DAC = frac{AD}{DC} ] [ tan angle DAB = frac{AD}{DB} ] 3. From the given points and definitions, we can use the relationships derived from the tan of the angles: [ frac{tan beta}{tan angle DAC} = frac{EA}{EC} quad text{(Using right triangle properties)} ] Similarly, [ frac{tan alpha}{tan angle DAB} = frac{FA}{FB} ] 4. We divide the two equations: [ frac{frac{tan beta}{tan angle DAC}}{frac{tan alpha}{tan angle DAB}} = frac{frac{EA}{EC}}{frac{FA}{FB}} ] 5. Simplifying the above equation, using that (frac{tan angle DAC}{tan angle DAB} = frac{frac{AD}{DC}}{frac{AD}{DB}} = frac{DC}{DB}): [ frac{tan beta}{tan alpha} = frac{EA}{EC} times frac{FB}{FA} times frac{DC}{DB} ] 6. By Ceva's theorem in trigonometric form, for concurrent cevians (BO), (CO), and (AD) in triangle (ABC), we have: [ frac{EA}{EC} times frac{FB}{FA} times frac{DC}{DB} = 1 ] 7. Substituting this into the equation obtained: [ frac{tan beta}{tan alpha} = 1 Rightarrow tan beta = tan alpha ] 8. Since (alpha) and (beta) are acute angles, we have: [ alpha = beta ] Hence, we conclude that: [ angle EDO = angle FDO ] [ boxed{} ]

answer:To solve this problem, let's break down the information provided and the calculations step by step: 1. **Total Matches Played in the Group**: Each team plays against every other team once in a round-robin format. With 4 teams, the total number of matches played is calculated by the formula for combinations, which is frac{n(n-1)}{2} where n is the number of teams. Thus, the total matches are frac{4 times 3}{2} = 6 matches. 2. **Points Distribution**: - **Japan**: With 6 points and knowing that a win gives 3 points, Japan must have won 2 matches (2 times 3 = 6 points) and lost 1 match (since each team plays 3 matches in total). - **Spain and Germany**: Both have 4 points. Since a win gives 3 points and a draw gives 1 point, each of them must have won 1 match (3 points), drawn 1 match (1 point), and lost 1 match to make up their 3 matches. 3. **Determining Costa Rica's Points**: - Given there's only one draw in the group and knowing Spain and Germany's points, it's clear they were involved in the draw together. This leaves no room for Costa Rica to have drawn any match. - Therefore, Costa Rica's outcomes must involve wins and losses only. Since Japan, Spain, and Germany account for 4 wins total (2 by Japan, 1 each by Spain and Germany), and there are 6 matches in total, Costa Rica must have won the remaining match (since 6 matches - 4 wins = 2 matches, and Costa Rica cannot draw). - Thus, Costa Rica won 1 match (earning 3 points) and lost 2 matches (earning 0 points for each loss). Therefore, Costa Rica's points are calculated as 3 times 1 + 0 times 2 = 3 points. Hence, the correct answer is boxed{text{D: 3}}.

answer:Given the problem: In △ABC, if sin^2 A + sin^2 B + sin^2 C = 2, then the maximum value of (cos A + cos B + 2cos C) is ( ). Step-by-step: 1. We start with the given equation: sin^2 A + sin^2 B + sin^2 C = 2. 2. Recognize that: sin^2 theta = 1 - cos^2 theta, text{ for any angle } theta. 3. Substitute (sin^2 theta = 1 - cos^2 theta) into the original equation: [ begin{aligned} sin^2 A + sin^2 B + sin^2 C &= 2, (1 - cos^2 A) + (1 - cos^2 B) + (1 - cos^2 C) &= 2, 3 - (cos^2 A + cos^2 B + cos^2 C) &= 2, cos^2 A + cos^2 B + cos^2 C &= 1. end{aligned} ] 4. Utilizing the identity: cos 2theta = 2cos^2 theta - 1, we rewrite the expression in terms of (2cos^2 theta - 1). 5. Consider the expression we need to maximize: cos A + cos B + 2cos C. 6. From trigonometric identities and properties, we can deduce that for the given values of (cos) functions: - Use (cos^2 A + cos^2 B + cos^2 C = 1). - Let's consider if one of (cos A, cos B, cos C = 0), then: - If ( cos A = 0 ), then (angle A = frac{pi}{2},) and thus: [ begin{aligned} angle B + angle C &= frac{pi}{2}, cos A + cos B + 2 cos C &= 0 + cos B + 2cos C. end{aligned} ] 7. Let’s analyze the segment ( cos B + 2 cos C ): - Given (angle B + angle C = frac{pi}{2},) where (cos C = sin B). 8. Thus, the equation becomes: [ cos B + 2sin B. ] 9. To maximize (cos B + 2 sin B,) we differentiate (f(B) = cos B + 2 sin B): - The derivative: [ f'(B) = -sin B + 2cos B. ] 10. Set (f'(B) = 0) and solve: [ -sin B + 2 cos B = 0, sin B = 2 cos B, tan B = 2, B = tan^{-1}(2). ] 11. Substitute (B) back: [ f(B) = cos(tan^{-1}(2)) + 2 sin(tan^{-1}(2)), Let quad tan B = 2 Rightarrow cos B = frac{1}{sqrt{5}}, quad sin B = frac{2}{sqrt{5}}, f(B) = frac{1}{sqrt{5}} + 2 cdot frac{2}{sqrt{5}}, f(B) = frac{1 + 4}{sqrt{5}} = sqrt{5}. ] 12. After examining other conditions for ( cos B = 0) or ( cos C = 0), the highest value remains ( sqrt{5}.) Conclusion: [ boxed{sqrt{5}} ]