answer:1. **Calculate daily consumption:** The cat consumes frac{1}{4} of a can in the morning and frac{1}{6} in the evening. The total daily consumption is: [ frac{1}{4} + frac{1}{6} = frac{3}{12} + frac{2}{12} = frac{5}{12} text{ cans per day}. ] 2. **Determine total cans used per day:** Starting with 10 cans, calculate the cumulative consumption for each day: - **Monday:** frac{5}{12} - **Tuesday:** frac{5}{12} times 2 = frac{10}{12} - **Wednesday:** frac{5}{12} times 3 = frac{15}{12} - **Thursday:** frac{5}{12} times 4 = frac{20}{12} - **Friday:** frac{5}{12} times 5 = frac{25}{12} - **Saturday:** frac{5}{12} times 6 = frac{30}{12} = 2.5 text{ cans} - **Sunday:** frac{5}{12} times 7 = frac{35}{12} = 2.916 text{ cans} - **Monday (next week):** frac{5}{12} times 8 = frac{40}{12} = 3.33 text{ cans} - **Tuesday (next week):** frac{5}{12} times 9 = frac{45}{12} = 3.75 text{ cans} - **Wednesday (next week):** frac{5}{12} times 10 = frac{50}{12} = 4.166 text{ cans} - **Thursday (next week):** frac{5}{12} times 11 = frac{55}{12} = 4.58 text{ cans} - **Friday (next week):** frac{5}{12} times 12 = frac{60}{12} = 5.00 text{ cans} - **Saturday (next week):** frac{5}{12} times 13 = frac{65}{12} = 5.416 text{ cans} - **Sunday (next week):** frac{5}{12} times 14 = frac{70}{12} = 5.833 text{ cans} - **Monday (next week):** frac{5}{12} times 15 = frac{75}{12} = 6.25 text{ cans} Here, frac{75}{12} cans exceed the initial 10 cans. Therefore, the cat finishes exactly 10 cans sometime during Monday of the next week. 3. **Conclusion:** The cat finishes all the cat food on Monday of the next week. The correct answer is: [ text{Monday (next week)} ] The final answer is boxed{text{Monday (next week)} textbf{(D)}}

answer:To prove that x^4 + y^4 geq xy(x + y)^2, we can start by proving that 2(x^4 + y^4) geq xy(x + y)^2. This is equivalent to proving that 2(x^4 + y^4) geq x^3y + xy^3 + 2x^2y^2. We will prove the inequality by verifying both x^4 + y^4 geq x^3y + xy^3 and x^4 + y^4 geq 2x^2y^2 separately. First, notice that x^4 + y^4 - 2x^2y^2 = (x^2 - y^2)^2. Since a square is always non-negative, (x^2 - y^2)^2 geq 0, so it follows that x^4 + y^4 geq 2x^2y^2. Next, we want to prove that x^4 + y^4 geq x^3y + xy^3. We can factor the left-hand side as follows: x^4 + y^4 - x^3y - xy^3 = (x - y)(x^3 - y^3). Since the factors x - y and x^3 - y^3 have the same sign (both are positive if x > y and negative if x < y), their product is non-negative, so we have: (x - y)(x^3 - y^3) geq 0. From this, it follows that x^4 + y^4 geq x^3y + xy^3. Combining the two proven inequalities, we conclude that x^4 + y^4 geq xy(x + y)^2 for any real numbers x and y. Therefore, the original statement is true. [ boxed{x^4 + y^4 geq xy(x + y)^2} ]

answer:First, we need to determine the value of 2*3 and 4*2 from the operation table. 1. **Find 2*3 and 4*2 using the table:** - From the table, 2*3 = 1. - From the table, 4*2 = 3. 2. **Now find (2*3)*(4*2) using the results from step 1:** - We need to find 1*3. - From the table, 1*3 = 3. Thus, (2*3)*(4*2) = 1*3 = 3. Conclusion: The new problem's solution is 3. The final answer is The final answer given the choices is boxed{text{C. 3}}

answer:To show that [ int_{0}^{1} int_{0}^{1} |f(x) + f(y)| , dx , dy geq int_{0}^{1} |f(x)| , dx, ] for any continuous real-valued function ( f ) on ([0, 1]), we proceed as follows: 1. **Define Sets Based on ( f ):** - Let ( A ) be the set of ( x in [0, 1] ) such that ( f(x) geq 0 ). - Let ( B ) be the set of ( x in [0, 1] ) such that ( f(x) < 0 ). Let ( a = |A| ) and ( b = |B| ) be the measures (lengths) of the sets ( A ) and ( B ), respectively. 2. **Integrals Over Sets ( A ) and ( B ):** - Define ( I_A = int_A f(x) , dx ) and ( I_B = -int_B f(x) , dx ). Note that ( I_A geq 0 ) and ( I_B geq 0 ). The right-hand side (RHS) of the given inequality is: [ int_{0}^{1} |f(x)| , dx = I_A + I_B. ] 3. **Evaluating the Double Integral:** - Consider the integral over ( A times A ): [ int_{A} int_{A} |f(x) + f(y)| , dy , dx. ] Since ( f(x) geq 0 ) for ( x in A ) and ( f(y) geq 0 ) for ( y in A ), we have: [ |f(x) + f(y)| = f(x) + f(y). ] Thus, [ int_{A} int_{A} (f(x) + f(y)) , dy , dx = int_{A} left( a f(x) + I_A right) , dx = a I_A + a I_A = 2a I_A. ] - Similarly, for the integral over ( B times B ): [ int_{B} int_{B} |f(x) + f(y)| , dy , dx. ] Since ( f(x) < 0 ) for ( x in B ) and ( f(y) < 0 ) for ( y in B ), we have: [ |f(x) + f(y)| = -(f(x) + f(y)). ] Thus, [ int_{B} int_{B} (-(f(x) + f(y))) , dy , dx = int_{B} left( b |f(x)| + I_B right) , dx = b I_B + b I_B = 2b I_B. ] 4. **Evaluating Mixed Integrals:** - Consider the integral over ( A times B ): [ int_{A} int_{B} |f(x) + f(y)| , dy , dx. ] Since ( f(x) geq 0 ) for ( x in A ) and ( f(y) < 0 ) for ( y in B ), we have: [ |f(x) + f(y)| = |f(x) + (-|f(y)|)| = |f(x) - |f(y)||. ] To account for all possible relationships between (|f(x)|) and ( |f(y)| ): [ int_{A} int_{B} |f(x) - |f(y)|| , dy , dx geq int_{A} b f(x) , dx + int_{A} a I_B , dx = a I_B - b I_A. ] 5. **Combining All Integrals:** - Sum the integrals: [ left( int_{A} int_{A} + int_{B} int_{B} + 2 int_{A} int_{B} right) |f(x) + f(y)| , dy , dx geq 2a I_A + 2b I_B + 2(a I_B - b I_A). ] Simplify: [ 2a I_A + 2b I_B + 2a I_B - 2b I_A = 2(a I_A + b I_B + a I_B - b I_A). ] 6. **Final Inequality:** - We need to show that: [ 2(a I_A + b I_B - b I_A + a I_B) geq I_A + I_B. ] - Grouping similar terms and noting their positivity: [ 2a I_A + 2b I_B geq I_A + I_B. ] This inequality holds since (a + b = 1) and all terms are non-negative. Thus, we have shown that [ int_{0}^{1} int_{0}^{1} |f(x) + f(y)| , dx , dy geq int_{0}^{1} |f(x)| , dx. ] (blacksquare)