answer:Given the function ( f(x) = x^2 + ax + b cos x ), we want to find the range of values for ( a + b ) under the condition that the set [ {x mid f(x)=0, x in mathbb{R}} ] is equal to the set [ {x mid f(f(x))=0, x in mathbb{R}} neq varnothing. ] 1. Let ( x_0 in {x mid f(x)=0, x in mathbb{R}} ). This means that ( f(x_0) = 0 ). 2. Since ( f(x_0) = 0 ) and the same ( x_0 ) must satisfy ( f(f(x_0)) = 0 ), then: [ fleft(fleft(x_{0}right)right) = 0. ] Given ( f(x_0) = 0 ), this implies: [ f(0) = 0. ] 3. Substitute ( x = 0 ) into the function ( f(x) ): [ f(0) = 0^2 + acdot0 + bcos 0 = b. ] Hence, [ b = 0. ] 4. With ( b = 0 ), the function simplifies to: [ f(x) = x^2 + ax. ] 5. Now we need ( f(f(x)) = 0 ): [ f(f(x)) = f(x^2 + ax) = (x^2 + ax)^2 + a(x^2 + ax). ] 6. Expand and simplify: [ (x^2 + ax)^2 + a(x^2 + ax) = x^4 + 2ax^3 + a^2x^2 + ax^2 + a^2x. ] The function ( f(f(x)) ) becomes: [ f(f(x)) = x^4 + 2ax^3 + (a^2 + a)x^2 + a^2x. ] 7. To satisfy ( f(f(x)) = 0 ), factors of the equation must include terms that will satisfy: [ x (x + a)(x^2 + ax + a) = 0. ] 8. Analyze the polynomial equation ( x^2 + ax + a = 0 ): We need to ensure that this polynomial does not introduce any new roots besides (0, -a). 9. The discriminant must be less than zero for ( x^2 + ax + a = 0 ) to ensure no real roots: [ a^2 - 4a < 0. ] This gives: [ 0 < a < 4. ] 10. Therefore, the valid range for ( a ) is: [ 0 < a < 4. ] 11. Since ( b = 0 ), the possible values for ( a + b ) correspond to the interval: [ [0, 4). ] # Conclusion: The range of values for ( a + b ) is (boxed{[0, 4)}).

answer:To find the value of ( ab ), we can use the given equations to solve for ( a ) and ( b ). From equation 1: ( a - b = 3 ) ... (1) From equation 2: ( a^2 + b^2 = 29 ) ... (2) From equation 3: ( a + c = 2b ) Since we are not given any information about ( c ), we can express ( c ) in terms of ( a ) and ( b ): ( c = 2b - a ) ... (3) Now, let's solve for ( a ) and ( b ) using equations (1) and (2). From equation (1), we can express ( a ) in terms of ( b ): ( a = b + 3 ) ... (4) Substitute equation (4) into equation (2): ( (b + 3)^2 + b^2 = 29 ) ( b^2 + 6b + 9 + b^2 = 29 ) ( 2b^2 + 6b + 9 = 29 ) ( 2b^2 + 6b - 20 = 0 ) ( b^2 + 3b - 10 = 0 ) This is a quadratic equation, which we can solve by factoring: ( (b + 5)(b - 2) = 0 ) So, ( b ) can be either ( -5 ) or ( 2 ). If ( b = -5 ), then from equation (4): ( a = -5 + 3 = -2 ) If ( b = 2 ), then from equation (4): ( a = 2 + 3 = 5 ) Now we have two possible pairs for ( (a, b) ): ( (-2, -5) ) and ( (5, 2) ). To find the value of ( ab ), we multiply ( a ) and ( b ) for each pair: For ( (-2, -5) ): ( ab = -2 times -5 = 10 ) For ( (5, 2) ): ( ab = 5 times 2 = 10 ) In both cases, the value of ( ab ) is ( boxed{10} ).

answer:To solve the problem of finding the minimum number of "bad" lights in a (2017 times 2017) grid where each light can be either "on" or "off," we proceed as follows: 1. **Grid Coloring** Consider the (2017 times 2017) grid and think of it as if it was a chessboard, coloring it in a black and white pattern. Define the first cell ((1, 1)) as black, subsequently coloring the grid such that no two adjacent cells share the same color. 2. **Defining Lights Based on Color** Define lights on black cells as black lights and lights on white cells as white lights. Each light can only be in one of two states: on (assigned a value of 1) or off (assigned a value of 0). 3. **Condition for a "Bad" Light** A light is termed "bad" if it is adjacent to an odd number of lights that are "on". We need to ensure that no light in the grid can be bad. 4. **Coordinate System and Parity Condition** Use the coordinates ((i, j)) for lights on the grid with (i, j in {1, 2, ..., 2017}). Notice that a light at ((i, j)) is black if (i + j) is even, and white if (i + j) is odd. 5. **Sum Definitions** Let (a_{i,j}) represent the sum of the values of the lights adjacent to the light at ((i, j)). The total sum (S) of all (a_{i,j}) where both (i) and (j) are even must be considered. Each (a_{i,j}) must be odd since each cell is adjacent to an odd number of cells of the opposite color on this parity-based grid. 6. **Contradiction Argument** Since the grid’s size ensures each black light would consequently impact the sum (S) by the values of black lights adjacent to it and multiplied by 2 (since they are counted multiple times via the adjacency), then (S) must be an even number contrary to the previous statement of being odd, leading to a contradiction. This implies that there must be at least one bad light. 7. **Minimal "Bad" Light Configuration** Consider a configuration to establish exactly one bad light minimally: - For black lights at ((i, j)): [ text{Assign value 0 if } max{|i|, |j|} equiv 0, 1 pmod{4} ] [ text{Assign value 1 if } max{|i|, |j|} equiv 2, 3 pmod{4} ] - For white lights at ((i, j)): [ text{Assign value 0 if } max{|i|, |j| - 1} equiv 0, 1 pmod{4} ] [ text{Assign value 1 if } max{|i|, |j| - 1} equiv 2, 3 pmod{4} ] 8. **Detailed Analysis for Borders and minimal conditions**: - For the borders and edges of the grid, check if all edge lights have any neighboring effect having seamless transition according to the defined assignments. - For edge lights' placement and conditions when they minimally are impacted by more than three adjacent lights confirming the minimal consistent parity. 9. **Conclusion**: By ensuring these detailed placement and state definitions, the configuration allows only one "bad" light: the light at the center at coordinates ((0, 0)). Thus, the minimum number of "bad" lights in this (2017 times 2017) grid is: [ boxed{1} ]

answer:Emma initially had 230 in her bank account. She withdrew 60 for shoes, so her new balance was: 230 - 60 = 170 The next week, she deposited twice the amount she withdrew, which is: 2 * 60 = 120 Adding this deposit to her new balance, her account now has: 170 + 120 = 290 So, Emma now has boxed{290} in her bank account.