answer:1. Simplify the left-hand side expression: [ (2x - 10x + 24)^2 = (-8x + 24)^2 ] [ (-8x + 24)^2 + 4 ] 2. Analyze the expressions: - Since (-8x + 24)^2 is a square, it is always nonnegative. - Adding 4 to (-8x + 24)^2 results in a value that is at least 4 (since the smallest value (-8x + 24)^2 can take is 0 when x = 3). - The right-hand side, -2|x|, is always non-positive, ranging from 0 (when x = 0) to negative infinity. 3. Conclude that no real number x can satisfy (-8x + 24)^2 + 4 = -2|x| since the left-hand side is always at least 4, and the right-hand side is always less than or equal to 0. [ boxed{0} text{ solutions} ]

answer:1. We need to find the smallest natural number ( n ) in decimal representation that ends in 6, and if we remove the last 6 and prepend a 6 to the remaining digits, we get four times the original number ( n ). This can be formalized as follows: 2. Let ( n ) be written as ( 10a + 6 ), where ( a ) is the integer part of ( n ) without the last digit. 3. Removing the last 6 and prepending a 6 yields ( 6 times 10^m + a ), where ( m ) is one less than the number of digits in ( a ). 4. According to the problem, this new number must be equal to ( 4n ). Thus, we have the equation: [ 4n = 6 times 10^m + a ] 5. Substituting ( n = 10a + 6 ) into the equation: [ 4(10a + 6) = 6 times 10^m + a ] Simplifying this, we get: [ 40a + 24 = 6 times 10^m + a ] [ 39a + 24 = 6 times 10^m ] 6. Solving for ( a ): [ a = frac{6 times 10^m - 24}{39} ] 7. For ( a ) to be an integer, the numerator ( 6 times 10^m - 24 ) must be divisible by 39. Testing the smallest values of ( m ) (starting from 1): - For ( m = 5 ), calculate the numerator: [ 6 times 10^5 - 24 = 600000 - 24 = 599976 ] - Check for divisibility by 39: [ 599976 div 39 = 15384 text{ (an integer)} ] 8. Therefore, with ( m = 5 ): [ n = 10a + 6 = 10 cdot 15384 + 6 = 153846 ] 9. Verification: Removing the last 6 from 153846 and prepending a 6 gives 615384. Checking the condition: [ 4 times 153846 = 615384 ] As expected, the condition holds true. # Conclusion: [ boxed{153846} ]

answer:From the problem statement, we know that the distance d from the center of the circle to the line equals OP cdot sin(30^circ), because ∠POQ is 120°, it forms two 30° angles at P and O (when we draw a radius from the circle center to meet at a 90° angle with the line). As sin(30^circ) = frac{1}{2}, we obtain: d = OP cdot sin(30^circ) = frac {1}{2} Now, calculating the distance from the center O (0,0) to the line x+y=m using the point-to-line distance formula gives us: d = frac{|Ax+By+C|}{sqrt{A^2+B^2}} For the line x+y = m, we have A=1, B=1, text{and} C=-m, thus: d = frac{|1cdot0+1cdot0-m|}{sqrt{1^2+1^2}} = frac{|-m|}{sqrt{2}} = frac{m}{sqrt{2}} By equating the two expressions for d, we get: frac{m}{sqrt{2}} = frac{1}{2} Rightarrow m = frac{sqrt{2}}{2} Since m>0, the only positive solution for m is: boxed{m = frac{sqrt{2}}{2}} Therefore, the correct answer is B. This problem examines the positional relationship between a line and a circle, as well as the point-to-line distance formula. The key to solving this problem is to calculate the distance from the centre of the circle (0,0) to the line.

answer:To determine which of the given statements are correct, we analyze each option in detail: - **Option A: ain {a,b,c}** This statement asserts that the element a is within the set that includes a, b, and c. Since a set includes all its explicitly listed elements, it directly follows that: [a in {a,b,c}] Therefore, option A is indeed correct. - **Option B: varnothing in {0}** This statement suggests that the empty set varnothing is an element of the set containing only the element 0. The empty set varnothing is a subset of every set, which means: [varnothing subseteq {0}] However, varnothing being a subset (subseteq) does not mean it is an element (in) of the set {0}. Therefore, option B is incorrect because it misuses the subset notation to imply membership. - **Option C: {0,1} subset mathbb{N}** Here, we're looking at whether the set containing 0 and 1 is a subset of the natural numbers mathbb{N}. Given that the set of natural numbers mathbb{N} includes 0, 1, and other positive integers, we observe: [{0,1} subset mathbb{N}] Thus, option C is correct because both 0 and 1 are indeed elements of the set of natural numbers, and other elements in mathbb{N} are not in {0,1}, fulfilling the subset condition. - **Option D: sqrt{2} in mathbb{Q}** This statement checks if the square root of 2, sqrt{2}, is a rational number (an element of mathbb{Q}). However, it is a well-known fact that sqrt{2} is an irrational number and cannot be expressed as a ratio of two integers. Therefore: [sqrt{2} notin mathbb{Q}] Hence, option D is incorrect because sqrt{2} is not a rational number. Given these analyses, the correct options that match the criteria are options A and C. Therefore, the correct answer is encapsulated as: [boxed{A text{ and } C}]