answer：First, I need to evaluate the expressions in each inequality to determine their numerical values.For option A:- Calculate (-0.6) squared: (-0.6)² = 0.36- Calculate (-1) cubed: (-1)³ = -1- Now, compare the values: -2 < 0.36 < -1This comparison is not true because 0.36 is greater than -1.For option B:- Calculate (-0.6) squared: (-0.6)² = 0.36- Calculate (-1) cubed: (-1)³ = -1- Now, compare the values: -2 < -1 < 0.36This comparison is true because -2 is less than -1, and -1 is less than 0.36.For option C:- Calculate (-0.6) squared: (-0.6)² = 0.36- Calculate (-1) cubed: (-1)³ = -1- Now, compare the values: 0.36 < -2 < -1This comparison is not true because 0.36 is greater than -2.For option D:- Calculate (-0.6) squared: (-0.6)² = 0.36- Calculate (-1) cubed: (-1)³ = -1- Now, compare the values: -1 < -2 < 0.36This comparison is not true because -1 is greater than -2.After evaluating all options, only option B provides a valid inequality.

answer：First, I need to find the least common multiple (LCM) of the given divisors: 12, 16, 18, 21, and 28. To do this, I'll start by determining the prime factorization of each number.For 12, the prime factors are (2^2 times 3). For 16, it's (2^4). For 18, the prime factors are (2 times 3^2). For 21, it's (3 times 7), and for 28, the prime factors are (2^2 times 7).Next, I'll identify the highest powers of all prime factors across these numbers. The highest power of 2 is (2^4) from 16, the highest power of 3 is (3^2) from 18, and the highest power of 7 is (7^1) from both 21 and 28.Now, I'll calculate the LCM by multiplying these highest powers together: (2^4 times 3^2 times 7 = 16 times 9 times 7 = 1008).The problem states that the smallest number, when diminished by 7, should be divisible by 1008. This means the number we're looking for is 7 more than a multiple of 1008. The smallest such number is (1008 + 7 = 1015).Therefore, the smallest number which, when diminished by 7, is divisible by 12, 16, 18, 21, and 28 is 1015.

answer：First, I recognize that γ is directly proportional to the square of δ, which means γ = cδ², where c is a constant.Next, I use the given values to find the constant c. When γ = 25 and δ = 5, I substitute these into the equation: 25 = c * 5².Simplifying this, I get 25 = 25c. Dividing both sides by 25, I find that c = 1.Now that I have the constant, I can use the equation γ = δ² to find γ when δ = 8.Substituting δ = 8 into the equation, I calculate γ = 8² = 64.Therefore, the value of γ when δ = 8 is 64.

answer：Okay, so I have this problem about a shopping mall buying two types of charging piles for electric vehicles. There are two parts to the problem. Let me try to understand and solve them step by step.**Part 1: Finding the Unit Prices of Charging Piles A and B**Alright, the problem says that the shopping mall bought two types of charging piles, A and B, at costs of 50,000 and 80,000 respectively. It also mentions that the unit price of A is 0.1 million yuan higher than B, and the quantity of B purchased is twice the quantity of A.First, I need to figure out what "unit price" means here. I think it refers to the price per charging pile. So, if A costs 50,000 and B costs 80,000, but A is more expensive by 0.1 million yuan, which is 100,000 yuan. Wait, that doesn't make sense because 50,000 is less than 80,000. Maybe I'm misunderstanding.Wait, the problem says the unit price of A is 0.1 million yuan higher than B. So, if B is x million yuan, then A is x + 0.1 million yuan. But the total costs for A and B are 50,000 and 80,000 respectively. Hmm, but 50,000 and 80,000 are in yuan, while the unit prices are in million yuan. So, I need to convert the total costs to million yuan.50,000 yuan is 0.05 million yuan, and 80,000 yuan is 0.08 million yuan. Okay, that makes more sense.So, let me denote the unit price of B as x million yuan. Then, the unit price of A is x + 0.1 million yuan.The quantity of A purchased would be total cost divided by unit price, so for A, it's 0.05 / (x + 0.1). For B, it's 0.08 / x.According to the problem, the quantity of B is twice the quantity of A. So:0.08 / x = 2 * (0.05 / (x + 0.1))Let me write that equation down:0.08 / x = 2 * (0.05 / (x + 0.1))Simplify the right side:2 * 0.05 = 0.1, so:0.08 / x = 0.1 / (x + 0.1)Now, cross-multiplying:0.08 * (x + 0.1) = 0.1 * xExpanding the left side:0.08x + 0.008 = 0.1xSubtract 0.08x from both sides:0.008 = 0.02xDivide both sides by 0.02:x = 0.008 / 0.02 = 0.4So, the unit price of B is 0.4 million yuan, and the unit price of A is 0.4 + 0.1 = 0.5 million yuan.Let me check if this makes sense. Quantity of A is 0.05 / 0.5 = 0.1 million yuan worth, which is 50,000 yuan. Quantity of B is 0.08 / 0.4 = 0.2 million yuan worth, which is 80,000 yuan. And indeed, the quantity of B is twice the quantity of A (0.2 is twice 0.1). So, that seems correct.**Part 2: Maximizing the Number of Charging Piles of Type A**Now, the shopping mall wants to buy a total of 20 charging piles, a mix of A and B, with a total cost not exceeding 9.58 million yuan. We need to find the maximum number of A piles they can buy.Let me denote the number of A piles as m, and the number of B piles as n. So, m + n = 20.The total cost is 0.5m + 0.4n ≤ 9.58.Since n = 20 - m, substitute into the cost equation:0.5m + 0.4(20 - m) ≤ 9.58Let me compute 0.4 * 20 = 8, so:0.5m + 8 - 0.4m ≤ 9.58Combine like terms:(0.5 - 0.4)m + 8 ≤ 9.580.1m + 8 ≤ 9.58Subtract 8 from both sides:0.1m ≤ 1.58Divide both sides by 0.1:m ≤ 15.8Since m has to be an integer, the maximum number of A piles is 15.Let me verify:If m = 15, then n = 5.Total cost = 0.5*15 + 0.4*5 = 7.5 + 2 = 9.5 million yuan, which is within the 9.58 million limit.If m = 16, then n = 4.Total cost = 0.5*16 + 0.4*4 = 8 + 1.6 = 9.6 million yuan, which exceeds 9.58 million.So, 15 is indeed the maximum number of A piles they can buy.**Final Answer**1. The unit price of charging pile A is boxed{0.5} million yuan, and the unit price of charging pile B is boxed{0.4} million yuan.2. The maximum number of charging piles of type A that can be purchased is boxed{15}.

answer：Okay, so I have this problem here: Given that ( k ) is a real number, and the equation ( frac{y^2}{k} - frac{x^2}{k-2} = 1 ) represents a hyperbola with foci on the ( y )-axis. I need to find the range of the semi-focal distance. Hmm, okay, let me try to break this down step by step.First, I remember that the general form of a hyperbola can be written in two ways, depending on whether it opens vertically or horizontally. Since the foci are on the ( y )-axis, this hyperbola must open upwards and downwards. So, the standard form should be ( frac{y^2}{a^2} - frac{x^2}{b^2} = 1 ), where ( a ) and ( b ) are positive real numbers.Comparing this with the given equation ( frac{y^2}{k} - frac{x^2}{k-2} = 1 ), I can see that ( a^2 = k ) and ( b^2 = k - 2 ). For the equation to represent a hyperbola, both ( a^2 ) and ( b^2 ) must be positive. So, I can write:1. ( k > 0 ) (since ( a^2 = k ) must be positive)2. ( k - 2 > 0 ) (since ( b^2 = k - 2 ) must be positive)From the second inequality, ( k - 2 > 0 ) implies ( k > 2 ). So, combining both conditions, ( k ) must be greater than 2. That gives me the range for ( k ), but I need the range for the semi-focal distance.I recall that for a hyperbola, the distance from the center to each focus (the semi-focal distance, denoted as ( c )) is given by ( c = sqrt{a^2 + b^2} ). Plugging in the values from above, ( a^2 = k ) and ( b^2 = k - 2 ), so:( c = sqrt{k + (k - 2)} = sqrt{2k - 2} )Now, I need to find the range of ( c ) given that ( k > 2 ). Let me express ( c ) in terms of ( k ):( c = sqrt{2k - 2} )Since ( k > 2 ), let's see what happens when ( k ) approaches 2 from the right and as ( k ) increases without bound.1. When ( k ) approaches 2 from the right, ( 2k - 2 ) approaches ( 4 - 2 = 2 ), so ( c ) approaches ( sqrt{2} ).2. As ( k ) increases, ( 2k - 2 ) increases without bound, so ( c ) also increases without bound.Therefore, the semi-focal distance ( c ) must be greater than ( sqrt{2} ) and can be any real number larger than that. So, the range of ( c ) is ( (sqrt{2}, +infty) ).Wait, let me double-check. If ( k > 2 ), then ( 2k - 2 ) is greater than ( 2*2 - 2 = 2 ), so ( c = sqrt{2k - 2} ) is greater than ( sqrt{2} ). That makes sense. And as ( k ) becomes very large, ( c ) becomes very large as well. So, yes, the range is all real numbers greater than ( sqrt{2} ).I think that's it. I don't see any mistakes in my reasoning. So, the range of the semi-focal distance is ( (sqrt{2}, +infty) ).**Final Answer**The range of the semi-focal distance is boxed{(sqrt{2}, +infty)}.

answer：First, I need to understand the program's logic. The program takes an input value `t` and calculates the value of `C` based on whether `t` is less than or equal to 3 or greater than 3.Given that the input `t` is 5, which is greater than 3, I will use the second part of the conditional statement to calculate `C`.The formula for `C` when `t > 3` is `C = 0.2 + 0.1 * (t - 3)`.Substituting `t = 5` into the formula:[C = 0.2 + 0.1 times (5 - 3) = 0.2 + 0.2 = 0.4]Therefore, the output result is 0.4.