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question:Geordie takes a toll road on his drive to work and back every day of his fiveday workweek. The toll is 12.50 per car or 7 per motorcycle. Both his car and his motorcycle get 35 miles per gallon of gas and the commute to work is some miles. Gas costs him 3.75 per gallon. Geordie drives his car to work three times a week and his motorcycle to work twice a week. He spends 118 driving to work and back on the same route in a week. How many miles is his commute to work?
answer:Let's break down the costs for Geordie's weekly commute. Cost of tolls for the car: 12.50 per trip × 3 trips = 37.50 Cost of tolls for the motorcycle: 7 per trip × 2 trips = 14 Total cost of tolls for the week: 37.50 + 14 = 51.50 Now, let's calculate the cost of gas for the week. We know the total cost for the week is 118, so the cost of gas would be the total cost minus the cost of tolls. Cost of gas for the week: 118 - 51.50 = 66.50 Since both the car and the motorcycle get 35 miles per gallon, we can calculate the total number of gallons used for the week. Number of gallons used for the week: 66.50 / 3.75 per gallon = 17.7333 gallons Now, let's calculate the total miles driven for the week. Since Geordie drives to work and back five times a week, the total miles driven for the week would be 10 times the one-way commute distance. Total miles driven for the week: 17.7333 gallons × 35 miles per gallon = 620.6665 miles Since this is the total for 10 commutes (5 days to work and 5 days back), we divide by 10 to find the one-way commute distance. One-way commute distance: 620.6665 miles / 10 = 62.06665 miles Therefore, Geordie's commute to work is approximately boxed{62.07} miles.
question:Let p and q be positive integers such that p is a prime, p divides q1 , and p+q divides p^2+2020q^2 . Find the sum of the possible values of p . *Proposed by Andy Xu*
answer:1. **Given Conditions and Initial Setup:** - ( p ) is a prime number. - ( p ) divides ( q-1 ), i.e., ( q equiv 1 pmod{p} ). - ( p+q ) divides ( p^2 + 2020q^2 ). 2. **Simplifying the Divisibility Condition:** - We start with the condition ( p+q mid p^2 + 2020q^2 ). - Rewrite ( p^2 + 2020q^2 ) as: [ p^2 + 2020q^2 = (p+q)(p-q) + 2021q^2 ] - Therefore, ( p+q mid 2021q^2 ). 3. **Using the GCD Condition:** - Since ( p mid q-1 ), we have ( q = kp + 1 ) for some integer ( k ). - This implies ( gcd(p, q) = 1 ) because ( p ) is a prime and does not divide ( 1 ). 4. **Divisibility by 2021:** - Given ( p+q mid 2021q^2 ) and knowing ( gcd(p, q) = 1 ), we conclude ( p+q mid 2021 ). - The prime factorization of 2021 is ( 2021 = 43 times 47 ). 5. **Possible Values for ( p+q ):** - The divisors of 2021 are ( 1, 43, 47, 2021 ). - We need ( p+q ) to be one of these values. Since ( p ) is a prime and ( q ) is a positive integer, we discard ( 1 ) and ( 2021 ) as they do not provide valid solutions. 6. **Solving for ( p ) and ( q ):** - For ( p+q = 43 ): [ q = 43 - p ] Since ( p mid q-1 ): [ p mid (43 - p - 1) implies p mid 42 ] The prime divisors of 42 are ( 2, 3, 7 ). - For ( p+q = 47 ): [ q = 47 - p ] Since ( p mid q-1 ): [ p mid (47 - p - 1) implies p mid 46 ] The prime divisors of 46 are ( 2, 23 ). 7. **Combining All Possible Values:** - From ( p+q = 43 ), we get ( p = 2, 3, 7 ). - From ( p+q = 47 ), we get ( p = 2, 23 ). 8. **Summing the Unique Values of ( p ):** - The unique values of ( p ) are ( 2, 3, 7, 23 ). - Their sum is: [ 2 + 3 + 7 + 23 = 35 ] The final answer is (boxed{35}).
question:a tetrahedron ( A_{1} A_{2} A_{3} A_{4} ), the area of the face opposite vertex ( A_{i} ) is ( S_{i} ) for ( 1 leq i leq 4 ). The angle between the planes of the faces with areas ( S_{k} ) and ( S_{j} ) is denoted as ( langle k, j rangle ) for ( 1 leq k < j leq 4 ). The midpoint of edge ( A_{k} A_{j} ) is denoted as ( M_{k j} ). The triangle formed by ( M_{k j} ) and the other two vertices (excluding vertices ( A_{k} ) and ( A_{j} )) is called a median face of the tetrahedron. The area of this median face is denoted as ( S_{k j} ) for ( 1 leq k < j leq 4 ). Then, [ S_{k j}^{2} = frac{1}{4} left( S_{k}^{2} + S_{j}^{2} + 2 S_{k} cdot S_{j} cdot cos langle k, j rangle right), ] where ( k ) and ( j ) satisfy ( 1 leq k < j leq 4 ).
answer:To prove the given area relation for the tetrahedron A_{1}A_{2}A_{3}A_{4}, we start by considering the areas of the faces of the tetrahedron and the dihedral angles between them. Step 1: Establish the base relations for faces A_3A_4, A_1A_2, and the angles between the faces. 1. By applying the properties of a tetrahedron (specifically, Property 6), we have: [ S_{3}^{2} + S_{4}^{2} = 2S_{3} cdot S_{4} cdot cos langle 3, 4 rangle + S_{1} cdot S_{3} cdot cos langle 1, 3 rangle + S_{1} cdot S_{4} cdot cos langle 1, 4 rangle + S_{2} cdot S_{3} cdot cos langle 2, 3 rangle + S_{2} cdot S_{4} cdot cos langle 2, 4 rangle ] 2. Similarly, we have: [ S_{1}^{2} + S_{2}^{2} = 2S_{1} cdot S_{2} cdot cos langle 1, 2 rangle + S_{1} cdot S_{3} cdot cos langle 1, 3 rangle + S_{1} cdot S_{4} cdot cos langle 1, 4 rangle + S_{2} cdot S_{3} cdot cos langle 2, 3 rangle + S_{2} cdot S_{4} cdot cos langle 2, 4 rangle ] 3. By combining the above results and simplifying, we obtain: [ S_{1} cdot S_{3} cdot cos langle 1, 3 rangle + S_{1} cdot S_{4} cdot cos langle 1, 4 rangle + S_{2} cdot S_{3} cdot cos langle 2, 3 rangle + S_{2} cdot S_{4} cdot cos langle 2, 4 rangle = S_{1}^{2} + S_{2}^{2} - 2S_{1} cdot S_{2} cdot cos langle 1, 2 rangle ] Step 2: Consider the tetrahedrons A_{1}A_{3}A_{4}M_{12} and A_{2}A_{3}A_{4}M_{12}. 1. Using Property 6 for tetrahedron A_{1}A_{3}A_{4}M_{12}: [ S_{12}^{2} = frac{1}{4}S_{3}^{2} + frac{1}{4}S_{4}^{2} + S_{2}^{2} - 2 cdot frac{1}{2}S_{3} cdot frac{1}{2}S_{4} cdot cos langle 3, 4 rangle - 2 cdot frac{1}{2}S_{3} cdot S_{2} cdot cos langle 2, 3 rangle - 2 cdot frac{1}{2}S_{4} cdot S_{2} cdot cos langle 2, 4 rangle ] 2. Similarly, using the same property for tetrahedron A_{2}A_{3}A_{4}M_{12}: [ S_{12}^{2} = frac{1}{4}S_{3}^{2} + frac{1}{4}S_{4}^{2} + S_{1}^{2} - 2 cdot frac{1}{2}S_{3} cdot frac{1}{2}S_{4} cdot cos langle 3, 4 rangle - 2 cdot frac{1}{2}S_{3} cdot S_{1} cdot cos langle 1, 3 rangle - 2 cdot frac{1}{2}S_{4} cdot S_{1} cdot cos langle 1, 4 rangle ] Step 3: Summing the two expressions for S_{12}^2 and simplifying. 1. Adding the equations obtained in Step 2: [ 2S_{12}^{2} = frac{1}{2}(S_{3}^{2} + S_{4}^{2}) + S_{1}^{2} + S_{2}^{2} - S_{1} cdot S_{3} cdot cos langle 1, 3 rangle - S_{1} cdot S_{4} cdot cos langle 1, 4 rangle - S_{2} cdot S_{3} cdot cos langle 2, 3 rangle - S_{2} cdot S_{4} cdot cos langle 2, 4 rangle - S_{3} cdot S_{4} cdot cos langle 3, 4 rangle ] 2. Expressing using simplified form: [ 2S_{12}^{2} = S_{1}^{2} + S_{2}^{2} - frac{1}{2}(S_{1}^{2} + S_{2}^{2} - 2S_{1} cdot S_{2} cdot cos langle 1, 2 rangle) ] 3. Simplifying further: [ 2S_{12}^{2} = frac{1}{2}(S_{1}^{2} + S_{2}^{2}) + S_{1} cdot S_{2} cdot cos langle 1, 2 rangle ] 4. Finally, we obtain: [ S_{12}^{2} = frac{1}{4}(S_{1}^{2} + S_{2}^{2} + 2S_{1} cdot S_{2} cdot cos langle 1, 2 rangle) ] Using the same reasoning, we can conclude that for any 1 leq k < j leq 4: [ S_{kj}^{2} = frac{1}{4}(S_{k}^{2} + S_{j}^{2} + 2S_{k} cdot S_{j} cdot cos langle k, j rangle) ] Therefore, we use Property 5, summing all such terms: [ sum_{1 leq k < j leq 4} S_{kj}^{2} = frac{1}{4} left[sum_{1 leq k < j leq 4}(S_{k}^{2} + S_{j}^{2}) + 2 sum_{1 leq k < j leq 4} S_{k} cdot S_{j} cdot cos langle k, j rangle right] = frac{1}{4}(3sum_{i=1}^{4} S_{i}^{2} + sum_{i=1}^{4} S_{i}^{2}) = sum_{i=1}^{4} S_{i}^{2} ] [ sum_{1 leq k < j leq 4} S_{kj}^{2} = sum_{i=1}^{4} S_{i}^{2} ] Thus, the proof is concluded. blacksquare
question:What is the perimeter of the figure shown if it is a regular decagon (10sided polygon) with each side measuring 3 units?
answer:Given that the figure is a regular decagon, it has 10 equal sides. Each side of the decagon measures 3 units. The perimeter (P) of a polygon is calculated by multiplying the number of sides (n) by the length of one side (s): [ P = n times s ] Substituting the given values: [ P = 10 times 3 = 30 ] Thus, the perimeter of the decagon is (boxed{30}) units.