answer:First, we need to express 260 as a sum of at least three distinct powers of 2 using its binary representation: 1. The largest power of 2 less than 260 is 2^8 = 256. 2. Subtracting 256 from 260 gives us 4, which is 2^2. 3. We have 260 = 2^8 + 2^2. To include at least one more distinct power of 2, we need to break down one of these further. Let's choose to breakdown 2^2 = 2^1 + 2^1, but since they aren't distinct, let’s adjust: 4. Split 256 as 2^7 + 2^7 (two identical powers). We want distinct terms, so we'll not use this split. Instead, consider 260 = 2^8 + 2^2 + 2^0 = 2^8 + 2^2 + 1. 5. Here, the exponents are 8, 2, 0. The sum of these exponents is 8 + 2 + 0 = boxed{10}.

answer:1. Let p and q be prime numbers, with p > q. 2. Consider the difference of their fourth powers: [ p^4 - q^4 ] 3. Notice that the difference of two fourth powers can be factored using the difference of squares formula: [ p^4 - q^4 = (p^2)^2 - (q^2)^2 = (p^2 - q^2)(p^2 + q^2) ] 4. Further factorize ( p^2 - q^2 ) using the difference of squares again: [ p^2 - q^2 = (p - q)(p + q) ] 5. Thus, the expression: [ p^4 - q^4 = (p - q)(p + q)(p^2 + q^2) ] 6. Each of the factors p - q, p + q, and p^2 + q^2 are natural numbers: - ( p - q geq 2 ) (since p and q are distinct primes and p > q), - ( p + q > 1 ) (since p and q are both at least 2), - ( p^2 + q^2 > 1 ) (since ( p^2 ) and ( q^2 ) are positive). 7. Since each factor in ((p - q)(p + q)(p^2 + q^2)) is greater than 1, the product ( p^4 - q^4 ) must be a composite number. # Conclusion: The difference of fourth powers of two prime numbers cannot be a prime number. [boxed{text{No, it cannot.}}]

answer:First, let's simplify the integral expression given: ∫[(1x^2)/20 + (3x^2)/10] dx This simplifies to: ∫[(1/20)x^2 + (3/10)x^2] dx Combine the terms: ∫[(1/20 + 3/10)x^2] dx Since 3/10 is equivalent to 6/20, we can add the fractions: ∫[(1/20 + 6/20)x^2] dx ∫[(7/20)x^2] dx Now, let's find the integral of this expression on the interval [a, b]: ∫[a to b] (7/20)x^2 dx The antiderivative of x^2 is (1/3)x^3, so we have: (7/20) * (1/3) * [x^3] evaluated from a to b = (7/60) * [b^3 - a^3] Now, let's find the integral of x^2 on the same interval [a, b]: ∫[a to b] x^2 dx The antiderivative of x^2 is (1/3)x^3, so we have: (1/3) * [x^3] evaluated from a to b = (1/3) * [b^3 - a^3] Now, to express the first integral as a percent of the second integral, we divide the first by the second and multiply by 100: [(7/60) * [b^3 - a^3]] / [(1/3) * [b^3 - a^3]] * 100 The [b^3 - a^3] terms cancel out, and we are left with: (7/60) / (1/3) * 100 To divide fractions, we multiply by the reciprocal of the second fraction: (7/60) * (3/1) * 100 = (7/20) * 100 = 7 * 5 = 35 So, the value of the first integral is boxed{35%} of the value of the second integral on the interval [a, b].

answer:From the definition, we have lfloor x rfloor leq x < lfloor x rfloor + 1 Therefore, f(x) = x - lfloor x rfloor geq 0, and f(x) < 1 Thus, conclusions ① and ② are correct. Since f(x + 1) = x + 1 - lfloor x + 1 rfloor = x + 1 - (lfloor x rfloor + 1) = x - lfloor x rfloor = f(x), f(x) is a periodic function. Considering f(-0.1) = -0.1 - lfloor -0.1 rfloor = -0.1 - (-1) = 0.9, and f(0.1) = 0.1 - lfloor 0.1 rfloor = 0.1 - 0 = 0.1 neq f(-0.1), f(x) is not an even function. Therefore, the correct choice is boxed{C}.