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question:Dorothy sells glass frames at half the price that Jemma sells them. Jemma sells the glass frames at a certain price, selling twice as many frames as Dorothy does. If Jemma sold 400 frames, and they made 2500 dollars together in total from the sale of the glass frames, how much does Jemma sell each frame for?
answer:Let's denote the price at which Jemma sells each frame as ( P ) dollars. Since Dorothy sells glass frames at half the price that Jemma sells them, Dorothy sells each frame for ( frac{P}{2} ) dollars. Jemma sells twice as many frames as Dorothy, so if Jemma sold 400 frames, Dorothy sold ( frac{400}{2} = 200 ) frames. The total revenue from Jemma's sales is ( 400 times P ), and the total revenue from Dorothy's sales is ( 200 times frac{P}{2} ). Together, they made 2500, so we can write the equation: [ 400P + 200 times frac{P}{2} = 2500 ] Simplifying the equation: [ 400P + 100P = 2500 ] [ 500P = 2500 ] Now, divide both sides by 500 to find the price at which Jemma sells each frame: [ P = frac{2500}{500} ] [ P = 5 ] Therefore, Jemma sells each frame for boxed{5} .
question:Consider the matrix A,Bin mathcal l{M}_3(mathbb{C}) with A=^tA and B=^tB . Prove that if the polinomial function defined by
[ f(x)=det(A+xB)]
has a multiple root, then det(A+B)=det B .
answer:1. **Define the polynomial function and its properties:** Given the polynomial function ( f(x) = det(A + xB) ), we need to analyze its roots and properties. Since ( A ) is a skew-symmetric matrix (( A = -A^T )) and ( B ) is a symmetric matrix (( B = B^T )), we can use these properties to simplify our analysis. 2. **Express the polynomial function:** The polynomial ( f(x) ) can be expanded as: [ f(x) = det(A + xB) = det(B)x^3 + alpha x^2 + beta x + det(A) ] Since ( A ) is a skew-symmetric matrix of odd order (3x3), it is known that the determinant of a skew-symmetric matrix of odd order is zero. Therefore, ( det(A) = 0 ). 3. **Analyze the symmetry of the polynomial:** We know that for a skew-symmetric matrix ( A ), ( A^T = -A ). This implies that: [ f(-x) = det(A - xB) ] Since ( det(A - xB) = det(-(A + xB)) = (-1)^3 det(A + xB) = -det(A + xB) = -f(x) ), we have: [ f(-x) = -f(x) ] This implies that ( f(x) ) is an odd function. Therefore, the coefficient of ( x^2 ) must be zero, i.e., ( alpha = 0 ). 4. **Simplify the polynomial function:** Given ( alpha = 0 ) and ( det(A) = 0 ), the polynomial simplifies to: [ f(x) = det(B)x^3 + beta x ] 5. **Condition for multiple roots:** For ( f(x) ) to have a multiple root, the discriminant of the polynomial must be zero. The polynomial ( det(B)x^3 + beta x ) can be factored as: [ f(x) = x(det(B)x^2 + beta) ] For ( det(B)x^2 + beta ) to have a multiple root, it must be a perfect square, which implies ( beta = 0 ). 6. **Conclusion:** If ( beta = 0 ), then: [ f(x) = det(B)x^3 ] Evaluating ( f(x) ) at ( x = 1 ), we get: [ f(1) = det(A + B) = det(B) ] Therefore, if the polynomial ( f(x) = det(A + xB) ) has a multiple root, then ( det(A + B) = det(B) ). (blacksquare) The final answer is ( boxed{ det(A + B) = det(B) } )
question:For homework, Juan's teacher asked everyone in the class, to write down the different types of transportation (cars, trucks, bicycles, skateboards etc) they saw on their way home that afternoon. After school, Juan walked home and saw the following: 15 cars, 3 bicycles, 8 pickup trucks and 1 tricycle. How many tires in total were there on the vehicles Juan saw?
answer:To calculate the total number of tires, we need to consider the number of tires on each type of vehicle: - Cars typically have 4 tires each. - Bicycles have 2 tires each. - Pickup trucks, like cars, usually have 4 tires each. - A tricycle has 3 tires. Now, let's calculate the total number of tires for each type of vehicle: - Cars: 15 cars x 4 tires/car = 60 tires - Bicycles: 3 bicycles x 2 tires/bicycle = 6 tires - Pickup trucks: 8 trucks x 4 tires/truck = 32 tires - Tricycle: 1 tricycle x 3 tires/tricycle = 3 tires Finally, we add up all the tires: 60 (cars) + 6 (bicycles) + 32 (trucks) + 3 (tricycle) = 101 tires So, Juan saw a total of boxed{101} tires on his way home.
question:Anthony writes the (n+1)^2 distinct positive integer divisors of 10^n , each once, on a whiteboard. On a move, he may choose any two distinct numbers a and b on the board, erase them both, and write gcd(a, b) twice. Anthony keeps making moves until all of the numbers on the board are the same. Find the minimum possible number of moves Anthony could have made. *Proposed by Andrew Wen*
answer:1. **Understanding the Problem:** - We have (n+1)^2 distinct positive integer divisors of 10^n. - We can choose any two distinct numbers a and b, erase them, and write gcd(a, b) twice. - The goal is to make all numbers on the board the same with the minimum number of moves. 2. **Initial Observations:** - The divisors of 10^n are of the form 2^a cdot 5^b where 0 leq a, b leq n. - There are (n+1)^2 such divisors. - The final state must be all numbers equal to 1 since 1 is a divisor of 10^n and gcd(1, x) = 1 for any x. 3. **Greedy Algorithm:** - We aim to reduce all numbers to 1 as efficiently as possible. - We can pair numbers of the form 2^k and 5^k for 1 leq k leq n. 4. **Step-by-Step Reduction:** - **First Phase:** - Pair 2^k and 5^k for 1 leq k leq n. - Each pair (2^k, 5^k) will result in gcd(2^k, 5^k) = 1. - This operation reduces two numbers to 1 in each move. - We can perform this operation n times, reducing 2n numbers to 1. - After this phase, we have (n+1)^2 - 2n = n^2 + 2n + 1 - 2n = n^2 + 1 numbers left. - **Second Phase:** - Now, we have n^2 + 1 numbers left, including 1. - We can pair 1 with any other number to reduce it to 1. - Each move reduces one non-1 number to 1. - We need n^2 moves to reduce the remaining n^2 numbers to 1. 5. **Total Moves:** - First phase: n moves. - Second phase: n^2 moves. - Total moves: n + n^2 = n^2 + n. Therefore, the minimum possible number of moves Anthony could have made is boxed{n^2 + n}.