answer:Let's start by calculating the total cost of the pencils. Since each box of pencils costs 2 and there are 10 boxes in a carton, the cost of one carton of pencils is: 10 boxes/carton * 2/box = 20/carton The school bought 20 cartons of pencils, so the total cost for the pencils is: 20 cartons * 20/carton = 400 Now, we know that the school spent 600 in total, and we have already accounted for 400 spent on pencils. This means that the remaining 200 was spent on markers. Let's denote the number of boxes in a carton of markers as "x". Since each carton of markers costs 4, the cost of one carton of markers is: x boxes/carton * 4/box = 4x/carton The school bought 10 cartons of markers, so the total cost for the markers is: 10 cartons * 4x/carton = 40x We know that the total cost for the markers is 200, so we can set up the equation: 40x = 200 Now, we can solve for "x" to find the number of boxes in a carton of markers: x = 200 / 40 x = 5 Therefore, there are boxed{5} boxes in a carton of markers.

answer:To solve this problem, let's break it down into clear, step-by-step calculations: 1. First, we calculate the total number of questions the mathematician needs to write for both projects combined. This is done by adding the number of questions for each project: [518 text{ questions} + 476 text{ questions} = 994 text{ questions}] 2. Next, we need to find out how many questions the mathematician should aim to complete each day. Since he has one week to complete all the questions and one week consists of 7 days, we divide the total number of questions by 7: [994 text{ questions} div 7 text{ days} = 142 text{ questions/day}] Therefore, the mathematician should aim to complete boxed{142} questions per day.

answer:To solve the system of equations left{{begin{array}{l}{2x-y=3}{7x-3y=20}end{array}}right., we follow these steps: 1. From the first equation, we express y in terms of x: 2x - y = 3 implies y = 2x - 3 quad text{(Equation ③)}. 2. We then substitute Equation ③ into the second equation to find x: begin{align*} 7x - 3(2x - 3) &= 20 7x - 6x + 9 &= 20 x &= 20 - 9 x &= 11. end{align*} 3. With x=11, we substitute this value back into Equation ③ to find y: begin{align*} y &= 2(11) - 3 y &= 22 - 3 y &= 19. end{align*} Therefore, the solution to the system of equations is boxed{left{begin{array}{l}{x=11}{y=19}end{array}right.}.

answer:Given the quadratic polynomial ( f(x) = ax^2 + bx + c ) has exactly one root, and the quadratic polynomial ( g(x) = f(3x + 2) - 2f(2x - 1) ) also has exactly one root, we need to find the root of ( f(x) ). 1. **Normalization of ( a ):** Since scaling all coefficients of ( f(x) ) by ( a ) does not alter the roots, we can assume ( a = 1 ) without loss of generality. This allows us to rewrite ( f(x) ) as: [ f(x) = x^2 + bx + c ] 2. **Condition for a Quadratic Polynomial to Have One Root:** A quadratic polynomial ( f(x) ) has exactly one root if and only if its discriminant is zero. The discriminant ( Delta ) of ( f(x) = x^2 + bx + c ) is given by: [ Delta = b^2 - 4ac ] Since ( a = 1 ), ( Delta = b^2 - 4c ). Setting this to zero: [ b^2 - 4c = 0 implies c = frac{b^2}{4} ] 3. **Rewrite ( f(x) ) in Terms of ( b ):** Using ( c = frac{b^2}{4} ), we have: [ f(x) = x^2 + bx + frac{b^2}{4} ] 4. **Construct ( g(x) ):** Now, we need to find ( g(x) = f(3x + 2) - 2f(2x - 1) ). - Compute ( f(3x+2) ): [ f(3x + 2) = (3x + 2)^2 + b(3x + 2) + frac{b^2}{4} ] Expanding and simplifying: [ (3x + 2)^2 = 9x^2 + 12x + 4 ] [ b(3x + 2) = 3bx + 2b ] Thus, [ f(3x + 2) = 9x^2 + 12x + 4 + 3bx + 2b + frac{b^2}{4} ] - Compute ( f(2x-1) ): [ f(2x - 1) = (2x - 1)^2 + b(2x - 1) + frac{b^2}{4} ] Expanding and simplifying: [ (2x - 1)^2 = 4x^2 - 4x + 1 ] [ b(2x - 1) = 2bx - b ] Thus, [ f(2x - 1) = 4x^2 - 4x + 1 + 2bx - b + frac{b^2}{4} ] - Subtract ( 2f(2x - 1) ) from ( f(3x + 2) ): [ g(x) = f(3x + 2) - 2f(2x - 1) ] [ g(x) = left(9x^2 + 12x + 4 + 3bx + 2b + frac{b^2}{4}right) - 2 left(4x^2 - 4x + 1 + 2bx - b + frac{b^2}{4}right) ] Simplifying: [ g(x) = 9x^2 + 12x + 4 + 3bx + 2b + frac{b^2}{4} - (8x^2 - 8x + 2 + 4bx - 2b + frac{b^2}{2}) ] [ g(x) = (9x^2 - 8x^2) + (12x + 8x) + (3bx - 4bx) + (4 - 2) + (2b - 2b) + left(frac{b^2}{4} - frac{b^2}{2}right) ] Reducing further: [ g(x) = x^2 + (20 - b)x + left( 2 + 4b - frac{b^2}{4} right) ] 5. **Condition for ( g(x) ) to Have One Root:** Since ( g(x) ) also has exactly one root, its discriminant must be zero: [ Delta_{g(x)} = (20 - b)^2 - 4 left(2 + 4b - frac{b^2}{4}right) = 0 ] Simplify the calculation: [ (20 - b)^2 - 4 left( 2 + 4b - frac{b^2}{4} right) = 400 - 40b + b^2 - 8 - 16b + b^2 ] [ 0 = 2b^2 - 56b + 392 ] [ 2(b^2 - 28b + 196) = 2(b - 14)^2 ] Thus, [ 2(b - 14)^2 = 0 implies b - 14 = 0 implies b = 14 ] 6. **Find the Root of ( f(x) ) with Given ( b ):** With ( b = 14 ), [ f(x) = x^2 + 14x + frac{14^2}{4} = x^2 + 14x + 49 ] Factoring the polynomial: [ f(x) = (x + 7)^2 ] Therefore, the root of ( f(x) ) is: [ x = -7 ] Conclusion: [ boxed{-7} ]