💡First, I recognize that the refrigeration compartment has a temperature of ( +5^{circ}mathrm{C} ), indicating it is 5 degrees above zero.Next, the freezer compartment is stated to have a temperature of ( -12^{circ}mathrm{C} ), which means it is 12 degrees below zero.Looking at the provided options:- **A:** ( +12^{circ}mathrm{C} ) represents 12 degrees above zero, which does not match the freezer's temperature.- **B:** ( -7^{circ}mathrm{C} ) represents 7 degrees below zero, which is not the same as the freezer's temperature.- **C:** ( -12^{circ}mathrm{C} ) accurately represents 12 degrees below zero, matching the freezer's temperature.- **D:** ( +17^{circ}mathrm{C} ) represents 17 degrees above zero, which is not the freezer's temperature.Therefore, the correct representation for the freezer compartment's temperature is ( -12^{circ}mathrm{C} ), corresponding to option **C**.

💡Okay, so I have this problem where I need to find the value of ( b ) such that the line ( y = frac{1}{3}x + b ) is tangent to the curve ( y = ln x ) for ( x > 0 ). Hmm, tangents... I remember that a tangent line touches the curve at exactly one point and has the same slope as the curve at that point. So, I think I need to find a point where the line and the curve meet and where their slopes are equal.First, let me recall that the derivative of ( y = ln x ) gives the slope of the tangent to the curve at any point ( x ). The derivative of ( ln x ) is ( frac{1}{x} ). So, if the slope of the tangent line is ( frac{1}{3} ), that must be equal to the derivative at the point of tangency. Let me denote the point of tangency as ( (m, ln m) ). Then, the slope of the tangent line at this point is ( frac{1}{m} ). Since the slope of the given line is ( frac{1}{3} ), I can set up the equation:[frac{1}{m} = frac{1}{3}]Solving for ( m ), I get:[m = 3]So, the point of tangency is ( (3, ln 3) ). Now, since this point lies on the tangent line ( y = frac{1}{3}x + b ), I can substitute ( x = 3 ) and ( y = ln 3 ) into the equation to find ( b ):[ln 3 = frac{1}{3} times 3 + b]Simplifying the right side:[ln 3 = 1 + b]Subtracting 1 from both sides gives:[b = ln 3 - 1]Wait, let me double-check that. If I plug ( m = 3 ) back into the point, it's ( (3, ln 3) ). Plugging into the line equation:[ln 3 = frac{1}{3} times 3 + b implies ln 3 = 1 + b implies b = ln 3 - 1]Yes, that seems correct. So, ( b ) is ( ln 3 - 1 ). I think that's the answer.**Final Answer**The value of ( b ) is boxed{ln 3 - 1}.

💡First, I'll define the two numbers as ( x ) and ( y ), where ( x ) is the larger number. According to the problem, the difference between the two numbers is 1670, so I can write the equation:[x - y = 1670 quad text{(1)}]The problem also states that 7.5% of one number is equal to 12.5% of the other number. This can be expressed as:[0.075x = 0.125y quad text{(2)}]To simplify equation (2), I'll divide both sides by 0.075 to isolate ( x ):[x = frac{0.125y}{0.075} = frac{5}{3}y quad text{(3)}]Next, I'll substitute equation (3) into equation (1):[frac{5}{3}y - y = 1670]To combine the terms, I'll find a common denominator, which is 3:[frac{5y - 3y}{3} = 1670][frac{2y}{3} = 1670]Now, I'll multiply both sides by 3 to solve for ( y ):[2y = 1670 times 3][2y = 5010]Finally, I'll divide both sides by 2 to find ( y ):[y = frac{5010}{2} = 2505]With the value of ( y ), I can find ( x ) using equation (3):[x = frac{5}{3} times 2505 = 4175]Therefore, the larger number ( x ) is (boxed{4175}).

💡First, I need to identify all pairs of single-digit numbers whose product is 12. The possible pairs are 3 and 4, as well as 2 and 6.Next, I'll form the two-digit numbers from these pairs. For the pair 3 and 4, the possible numbers are 34 and 43. For the pair 2 and 6, the possible numbers are 26 and 62.Finally, I'll compare these numbers to find the smallest one. Among 26, 34, 43, and 62, the lowest number is 26.

💡Alright, so I have this problem here that asks me to find the sum of all prime numbers between 1 and 150 that satisfy two conditions: they are 3 less than a multiple of 6 and 2 greater than a multiple of 7. The answer choices are given, and I need to figure out which one is correct.First, let me break down the problem. I need to find prime numbers, so I know they have to be greater than 1 and only divisible by 1 and themselves. The primes are between 1 and 150, so I can list them out if needed, but maybe there's a smarter way.The first condition is that the prime number is 3 less than a multiple of 6. So, mathematically, this means that if I take the prime number and add 3 to it, it should be divisible by 6. In other words, the prime number p satisfies p ≡ 3 mod 6. That means p can be written as 6k + 3 for some integer k.The second condition is that the prime number is 2 greater than a multiple of 7. So, similarly, if I subtract 2 from the prime number, it should be divisible by 7. This means p ≡ 2 mod 7, or p can be written as 7m + 2 for some integer m.So, I need to find primes p such that:1. p = 6k + 32. p = 7m + 2Since both expressions equal p, I can set them equal to each other:6k + 3 = 7m + 2Let me rearrange this equation to find a relationship between k and m:6k - 7m = -1Hmm, this is a linear Diophantine equation. I need to find integer solutions for k and m. I remember that to solve such equations, I can use the Extended Euclidean Algorithm to find particular solutions and then find the general solution.First, let's find the greatest common divisor (gcd) of 6 and 7. Since 6 and 7 are coprime, their gcd is 1. Because 1 divides -1, there are integer solutions to this equation.Now, let me find a particular solution. I can try small integers for k and m to see if they satisfy the equation.Let's try k = 1:6(1) - 7m = -16 - 7m = -1-7m = -7m = 1So, k = 1 and m = 1 is a particular solution.The general solution for such equations is given by:k = k0 + (7/d)tm = m0 + (6/d)tWhere d is the gcd of 6 and 7, which is 1, and t is an integer.So, the general solution is:k = 1 + 7tm = 1 + 6tTherefore, substituting back into p = 6k + 3:p = 6(1 + 7t) + 3 = 6 + 42t + 3 = 9 + 42tSo, p can be written as 42t + 9, where t is an integer.Now, since p must be a prime number between 1 and 150, let's find all such p.Let's start with t = 0:p = 42(0) + 9 = 9But 9 is not a prime number because it's divisible by 3.t = 1:p = 42(1) + 9 = 5151 is also not a prime number; it's divisible by 3 and 17.t = 2:p = 42(2) + 9 = 84 + 9 = 9393 is not a prime number; it's divisible by 3 and 31.t = 3:p = 42(3) + 9 = 126 + 9 = 135135 is not a prime number; it's divisible by 5 and 27.t = 4:p = 42(4) + 9 = 168 + 9 = 177But 177 is greater than 150, so we stop here.Wait, so all the numbers generated by p = 42t + 9 up to 150 are 9, 51, 93, 135, and 177. None of these are prime numbers. That means there are no prime numbers between 1 and 150 that satisfy both conditions.Therefore, the sum of all such prime numbers is 0.But just to make sure I didn't make a mistake, let me double-check my reasoning.I started by translating the conditions into congruences:p ≡ 3 mod 6p ≡ 2 mod 7Then, I set up the equation 6k + 3 = 7m + 2, which simplifies to 6k - 7m = -1. I found a particular solution (k=1, m=1) and then wrote the general solution as k = 1 + 7t and m = 1 + 6t. Substituting back, I got p = 42t + 9.Testing t from 0 to 4 gives p values up to 177, but all are non-prime. So, indeed, there are no primes in the given range that satisfy both conditions.Therefore, the answer should be 0.**Final Answer**boxed{A}

💡Alright, so I need to show that for any positive integers (a) and (b) (with (b) not zero), there exist integers (q) and (r) such that (0 leq r < b) and (a = bq + r). This is basically the Euclidean division algorithm, right? I remember hearing about it in my math classes, but I need to actually prove it from scratch.Okay, let's start by understanding what this means. We have two positive integers, (a) and (b), and we want to divide (a) by (b) to get a quotient (q) and a remainder (r). The remainder should be less than (b), and it should satisfy the equation (a = bq + r). So, essentially, we're breaking down (a) into a multiple of (b) plus some leftover amount, which is the remainder.I think the key here is to find such (q) and (r) for any given (a) and (b). Maybe I can approach this by considering how many times (b) fits into (a). That number would be the quotient (q), and whatever is left after subtracting (b) multiplied by (q) from (a) would be the remainder (r).But how do I formalize this? I guess I need to define (q) as the largest integer such that (bq leq a). Then, (r) would naturally be (a - bq). This makes sense because if (q) is the largest integer where (bq) doesn't exceed (a), then (a - bq) should be less than (b), which is exactly what we need for the remainder.Wait, is this always true? Let me test it with some numbers. Suppose (a = 10) and (b = 3). Then, the largest (q) such that (3q leq 10) is (q = 3) because (3 times 3 = 9 leq 10). Then, (r = 10 - 9 = 1), which is indeed less than (3). So, (10 = 3 times 3 + 1). That works.Another example: (a = 7) and (b = 5). The largest (q) is (1) because (5 times 1 = 5 leq 7). Then, (r = 7 - 5 = 2), which is less than (5). So, (7 = 5 times 1 + 2). Perfect.But what if (a) is less than (b)? Let's say (a = 2) and (b = 5). Then, the largest (q) such that (5q leq 2) is (q = 0), because (5 times 0 = 0 leq 2). Then, (r = 2 - 0 = 2), which is still less than (5). So, (2 = 5 times 0 + 2). That still works.Okay, so this method seems to hold in these cases. But how do I prove it in general? Maybe I can use the Well-Ordering Principle, which states that every non-empty set of positive integers has a least element. That might help me find the appropriate (q).Let me define the set (S = {a - bk mid k in mathbb{Z}, a - bk geq 0}). This set contains all the non-negative differences when we subtract multiples of (b) from (a). Since (a) and (b) are positive integers, and (b) is non-zero, (S) is non-empty because at least (k = 0) gives (a - 0 = a), which is in (S).By the Well-Ordering Principle, (S) has a least element, say (r). So, (r = a - bq) for some integer (q). Since (r) is the least element in (S), subtracting (b) from it would give a negative number, which is not in (S). Therefore, (r - b < 0), which implies (r < b).Thus, we have (a = bq + r) with (0 leq r < b). This satisfies the conditions of the Euclidean division algorithm.Wait, let me make sure I didn't skip any steps. I defined (S) as the set of all (a - bk) that are non-negative. Then, by the Well-Ordering Principle, (S) has a least element (r). Since (r) is the smallest such element, (r - b) would be negative, ensuring that (r < b). Therefore, (r) is indeed the remainder when (a) is divided by (b), and (q) is the quotient.This seems solid. I think I've covered all the necessary parts. I considered examples, defined the set properly, applied the Well-Ordering Principle, and ensured that the remainder is less than (b). I don't see any gaps in the logic here.Just to recap:1. Define (S = {a - bk mid k in mathbb{Z}, a - bk geq 0}).2. (S) is non-empty because (a) is positive.3. By the Well-Ordering Principle, (S) has a least element (r).4. (r = a - bq) for some integer (q).5. Since (r) is the least element, (r - b < 0), so (r < b).6. Therefore, (a = bq + r) with (0 leq r < b).Yes, that makes sense. I think this is a valid proof of the Euclidean division algorithm.