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question:A coalpowered train is traveling towards a faraway city when the conductor realizes that he forgot to restock the train's coal supply at the last station. The train is traveling on a oneway track, so the conductor has no other option but to continue moving forward towards the city. The train can travel 5 miles for every 2 pounds of coal that it consumes. If the train has 160 pounds of coal remaining, how far can the train travel before it runs out of fuel?

answer:To solve this problem, we start by understanding the efficiency of the train's coal consumption. The train can travel 5 miles for every 2 pounds of coal. This efficiency can be expressed as a rate of miles per pound of coal: [ text{Efficiency} = frac{5 text{ miles}}{2 text{ pounds}} = 2.5 text{ miles per pound} ] Given that the train has 160 pounds of coal remaining, we can calculate the total distance the train can travel by multiplying the amount of coal by the efficiency: [ text{Distance} = 160 text{ pounds} times 2.5 text{ miles per pound} = 400 text{ miles} ] Therefore, the train can travel a total distance of 400 miles before running out of fuel. [ boxed{400} ]

question:An isosceles triangle ABC is inscribed in a circle with angle ACB = 90^o and EF is a chord of the circle such that neither E nor F coincide with C . Lines CE and CF meet AB at D and G respectively. Prove that |CE|cdot |DG| = |EF| cdot |CG| .

answer:1. **Claim 1:** (D, E, F, G) are cyclic. **Proof:** Note that [ angle CEF = angle CAB - angle FCB = 45^circ - angle GCB = angle CGB = angle FGD ] This shows that the quadrilateral (DEFG) is cyclic because opposite angles of a cyclic quadrilateral sum up to (180^circ). Therefore, (angle CEF = angle FGD). (blacksquare) 2. Since (D, E, F, G) are cyclic, we can use the property of cyclic quadrilaterals to establish similarity between triangles. Specifically, we have (triangle CFE sim triangle CDG). 3. To prove the similarity, we note that: [ angle CFE = angle CDG quad text{(since (D, E, F, G) are cyclic)} ] and [ angle ECF = angle GCD quad text{(common angle)} ] 4. By the AA (Angle-Angle) criterion for similarity, (triangle CFE sim triangle CDG). 5. From the similarity of the triangles, we have the ratio of corresponding sides: [ frac{CE}{EF} = frac{CG}{DG} ] 6. Cross-multiplying the above equation, we get: [ CE cdot DG = EF cdot CG ] Thus, we have proved that ( |CE| cdot |DG| = |EF| cdot |CG| ). The final answer is ( boxed{ |CE| cdot |DG| = |EF| cdot |CG| } ).

question:Calculate the leading coefficient in the polynomial 5(x^4 2x^3 + 3x) 9(x^4 x^2 + 1) + 3(3x^4 + x^3 2) after it is simplified.

answer:The leading coefficient in a polynomial is the coefficient associated with the highest power of x. Here, it is (x^4). 1. Identify and extract the coefficients of (x^4) in each term: - From 5(x^4 - 2x^3 + 3x), the coefficient of (x^4) is (5). - From -9(x^4 - x^2 + 1), the coefficient of (x^4) is (-9). - From 3(3x^4 + x^3 - 2), the coefficient of (x^4) is (3 times 3 = 9). 2. Combine these coefficients: - The coefficient of (x^4) is (5 - 9 + 9 = 5). Thus, the leading coefficient of the simplified polynomial is (boxed{5}).

question:a triangle with integer side lengths, one side is twice as long as a second side, and the length of the third side is 20. What is the greatest possible perimeter of the triangle?

answer:Let the sides of the triangle have lengths ( y ), ( 2y ), and 20. The Triangle Inequality implies the following: 1. ( y + 2y > 20 ) (i.e., ( 3y > 20 )) 2. ( y + 20 > 2y ) (i.e., ( y < 20 )) 3. ( 2y + 20 > y ) (always true since ( y > 0 )) From the inequality ( 3y > 20 ), solving for ( y ) gives ( y > frac{20}{3} approx 6.67 ). Thus, since ( y ) needs to be an integer and must also satisfy ( y < 20 ), the possible values for ( y ) are from 7 to 19. To find the greatest possible perimeter: - Calculate the perimeter from the maximum value of ( y ), which is 19: [ text{Perimeter} = y + 2y + 20 = 3y + 20 = 3(19) + 20 = 77 ] Thus, the greatest possible perimeter of the triangle is (boxed{77}).

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