answer：First, identify the initial number of quarters Sara has, which is 21.Next, determine how many additional quarters her dad gave her, which is 49.To find the total number of quarters Sara has now, add the initial amount to the additional amount: 21 + 49.Performing the addition gives a total of 70 quarters.Therefore, Sara now has 70 quarters.

answer：Alright, so I've got this problem here about mathematical induction, and I need to figure out for which set of numbers a proposition holds true. The proposition is verified for n=1, and then it's assumed to hold for some arbitrary natural number k (where k is at least 1). From there, it's proven that if it holds for k, it also holds for k+2. The question is asking which of the options A, B, C, or D is correct based on this information.First, let me recall what mathematical induction is. It's a method of proving that a statement holds true for all natural numbers. The process involves two main steps: the base case and the inductive step. In the base case, we verify that the statement is true for the initial value, usually n=1. In the inductive step, we assume that the statement is true for some arbitrary natural number k, and then we prove that it must also be true for k+1.But in this problem, instead of proving it for k+1, it's proven for k+2. That's interesting. So, if the proposition holds for k, it holds for k+2. Hmm, that seems like a variation of the standard induction method. I wonder how that affects the conclusion.Let me think about it step by step. The base case is n=1, which is verified. So, we know the proposition is true for n=1. Now, assuming it's true for k, we prove it's true for k+2. So, starting from k=1, if it's true for 1, then it's true for 1+2=3. Then, assuming it's true for k=3, it's true for 3+2=5. Continuing this way, it seems like the proposition is being proven for all odd numbers: 1, 3, 5, 7, and so on.Wait, but what about the even numbers? If we start at n=1 and keep adding 2 each time, we never reach an even number. So, does that mean the proposition isn't necessarily true for even numbers? The problem doesn't mention anything about even numbers, so I can't assume it holds for them. It only shows that if it's true for k, it's true for k+2, starting from k=1.So, if we only have the base case at n=1 and the inductive step proving k to k+2, we can only conclude that the proposition holds for all positive odd integers. That would be option B. But let me double-check to make sure I'm not missing something.Is there a way that this could also cover even numbers? If we had a base case at n=2, then we could use the same inductive step to cover all even numbers. But since the base case is at n=1, which is odd, and we're adding 2 each time, we stay within the odd numbers. Therefore, the proposition doesn't necessarily hold for even numbers.Option A says "All positive integers." But since we're only covering the odd ones, that's too broad. Option C is "All positive even numbers," which we don't have evidence for because our base case is odd. Option D is "None of the above," but since option B seems to fit, D isn't necessary.Wait, but what if someone argues that since we can start at n=1 and keep adding 2 indefinitely, we cover all odd numbers, but not necessarily all positive integers. So, the proposition is only proven for the odd numbers, not for the evens. Therefore, the correct answer should be B: All positive odd numbers.Yeah, I think that makes sense. The key here is that the inductive step skips every other number, so starting from an odd base case, we only cover the odd numbers. If we had started from an even base case, we'd cover the evens, but since we started at 1, it's the odds.I don't see any flaws in this reasoning. It seems pretty straightforward once you break it down. So, I'm confident that the answer is B.

answer：Okay, so I have this problem: "How many lattice points lie on the graph of the equation ( x^2 - y^2 = 53 )?" Hmm, lattice points mean points where both ( x ) and ( y ) are integers, right? So I need to find all integer solutions ( (x, y) ) that satisfy this equation.First, I remember that ( x^2 - y^2 ) can be factored into ( (x + y)(x - y) ). So, rewriting the equation, it becomes ( (x + y)(x - y) = 53 ). Now, 53 is a prime number, isn't it? That means its only positive divisors are 1 and 53. But since we're dealing with integers, we also have to consider negative divisors, so -1 and -53.So, the possible pairs of integers ( (a, b) ) such that ( a times b = 53 ) are:1. ( (53, 1) )2. ( (-53, -1) )3. ( (1, 53) )4. ( (-1, -53) )Wait, but actually, since multiplication is commutative, some of these might result in the same solutions, but let me check each pair separately to be thorough.Let me denote ( a = x + y ) and ( b = x - y ). So, for each factor pair ( (a, b) ), I can set up a system of equations:1. ( x + y = a )2. ( x - y = b )Then, solving for ( x ) and ( y ) in each case.Starting with the first pair ( (53, 1) ):- ( x + y = 53 )- ( x - y = 1 )If I add these two equations together, I get:( (x + y) + (x - y) = 53 + 1 )( 2x = 54 )So, ( x = 27 ).Subtracting the second equation from the first:( (x + y) - (x - y) = 53 - 1 )( 2y = 52 )So, ( y = 26 ).Alright, that gives me one solution: ( (27, 26) ).Next, the pair ( (-53, -1) ):- ( x + y = -53 )- ( x - y = -1 )Adding the equations:( (x + y) + (x - y) = -53 + (-1) )( 2x = -54 )So, ( x = -27 ).Subtracting the second equation from the first:( (x + y) - (x - y) = -53 - (-1) )( 2y = -52 )So, ( y = -26 ).Another solution: ( (-27, -26) ).Moving on to the pair ( (1, 53) ):- ( x + y = 1 )- ( x - y = 53 )Adding the equations:( (x + y) + (x - y) = 1 + 53 )( 2x = 54 )So, ( x = 27 ).Subtracting the second equation from the first:( (x + y) - (x - y) = 1 - 53 )( 2y = -52 )So, ( y = -26 ).That's another solution: ( (27, -26) ).Lastly, the pair ( (-1, -53) ):- ( x + y = -1 )- ( x - y = -53 )Adding the equations:( (x + y) + (x - y) = -1 + (-53) )( 2x = -54 )So, ( x = -27 ).Subtracting the second equation from the first:( (x + y) - (x - y) = -1 - (-53) )( 2y = 52 )So, ( y = 26 ).And another solution: ( (-27, 26) ).So, in total, I have four solutions:1. ( (27, 26) )2. ( (-27, -26) )3. ( (27, -26) )4. ( (-27, 26) )Let me double-check each of these to make sure they satisfy the original equation ( x^2 - y^2 = 53 ).1. For ( (27, 26) ): ( 27^2 - 26^2 = 729 - 676 = 53 ). Correct.2. For ( (-27, -26) ): ( (-27)^2 - (-26)^2 = 729 - 676 = 53 ). Correct.3. For ( (27, -26) ): ( 27^2 - (-26)^2 = 729 - 676 = 53 ). Correct.4. For ( (-27, 26) ): ( (-27)^2 - 26^2 = 729 - 676 = 53 ). Correct.All four points satisfy the equation. So, are there any other factor pairs I might have missed? Since 53 is prime, the only factor pairs are the ones I considered: ( (53, 1) ), ( (-53, -1) ), ( (1, 53) ), and ( (-1, -53) ). So, I think I've covered all possibilities.Therefore, the number of lattice points on the graph is 4.**Final Answer**The number of lattice points on the graph is boxed{4}.

answer：To find the number halfway between (frac{1}{6}) and (frac{5}{6}), we can use the concept of the average of two numbers. The halfway point between two numbers is their average.1. **Identify the two numbers:** [ frac{1}{6} quad text{and} quad frac{5}{6} ]2. **Calculate the average:** [ text{Average} = frac{frac{1}{6} + frac{5}{6}}{2} ]3. **Add the fractions:** [ frac{1}{6} + frac{5}{6} = frac{6}{6} = 1 ]4. **Divide by 2 to find the average:** [ frac{1}{2} = frac{1}{2} ]Therefore, the number halfway from (frac{1}{6}) to (frac{5}{6}) is (frac{1}{2}).

answer：First, I'll define the initial number of apples as ( A ).The cafeteria used 20 apples for lunch, so the remaining apples are ( A - 20 ).After using the apples, they bought 28 more, making the total number of apples ( A - 20 + 28 ).According to the problem, the total number of apples after these transactions is 46. This gives the equation:[A - 20 + 28 = 46]Simplifying the equation:[A + 8 = 46]Solving for ( A ):[A = 46 - 8][A = 38]Therefore, the cafeteria initially had 38 apples.

answer：Alright, so I have this problem where I need to find the radius of a circle that's inscribed within three other circles. These three circles all touch each other externally, and their radii are given as 5, 10, and 20. The formula provided is a bit complex, but I think I can handle it step by step.First, let me write down the formula again to make sure I have it right:[frac{1}{r} = frac{1}{a} + frac{1}{b} + frac{1}{c} + 2 sqrt{frac{1}{ab} + frac{1}{ac} + frac{1}{bc}}]Okay, so I need to plug in the values of a, b, and c into this formula. The radii are a = 5, b = 10, and c = 20. Let me substitute these values in:[frac{1}{r} = frac{1}{5} + frac{1}{10} + frac{1}{20} + 2 sqrt{frac{1}{5 times 10} + frac{1}{5 times 20} + frac{1}{10 times 20}}]Now, I need to compute each part of this equation step by step. Let's break it down.First, calculate the individual reciprocals:- (frac{1}{5}) is 0.2- (frac{1}{10}) is 0.1- (frac{1}{20}) is 0.05Adding these together: 0.2 + 0.1 + 0.05 = 0.35Next, let's compute the terms inside the square root:- (frac{1}{5 times 10}) is (frac{1}{50}), which is 0.02- (frac{1}{5 times 20}) is (frac{1}{100}), which is 0.01- (frac{1}{10 times 20}) is (frac{1}{200}), which is 0.005Adding these together: 0.02 + 0.01 + 0.005 = 0.035Now, take the square root of 0.035. Hmm, I don't remember the exact value, but I can approximate it. Let's see, the square of 0.18 is 0.0324, and the square of 0.19 is 0.0361. Since 0.035 is between these two, the square root should be approximately 0.187.So, (sqrt{0.035} approx 0.187). Then, multiply this by 2: 2 * 0.187 = 0.374Now, add this result to the earlier sum of reciprocals:0.35 + 0.374 = 0.724So, (frac{1}{r} = 0.724). To find r, I need to take the reciprocal of this value:r = (frac{1}{0.724})Calculating this, 1 divided by 0.724. Let me do this division carefully. 0.724 goes into 1 once, with a remainder. Let me convert this into a more precise calculation.Alternatively, I can use a calculator for better accuracy, but since I'm doing this manually, let's approximate:0.724 * 1.38 = approximately 1.0 (since 0.724 * 1.38 ≈ 1.0). So, r ≈ 1.38.Wait, let me check that multiplication:0.724 * 1.38:First, 0.724 * 1 = 0.7240.724 * 0.3 = 0.21720.724 * 0.08 = 0.05792Adding these together: 0.724 + 0.2172 = 0.9412; 0.9412 + 0.05792 ≈ 0.99912That's very close to 1.0, so 1.38 is a good approximation for r.But just to be more precise, let's see how much more we need to add to 1.38 to get closer to 1.0.0.724 * 1.38 = ~0.99912, which is just 0.00088 less than 1.0.So, to get the exact value, we can set up the equation:0.724 * x = 1.0x = 1.0 / 0.724 ≈ 1.381So, rounding to three decimal places, r ≈ 1.381.Therefore, the radius of the inscribed circle is approximately 1.381 units.Wait, let me double-check all my steps to make sure I didn't make any mistakes.1. Calculated reciprocals correctly: 1/5=0.2, 1/10=0.1, 1/20=0.05. Sum is 0.35. That seems right.2. Calculated the terms inside the square root: 1/(5*10)=0.02, 1/(5*20)=0.01, 1/(10*20)=0.005. Sum is 0.035. Correct.3. Square root of 0.035: estimated between 0.18 and 0.19, approximated as 0.187. Multiplying by 2 gives 0.374. Adding to 0.35 gives 0.724. Correct.4. Reciprocal of 0.724 is approximately 1.381. That seems accurate.So, all steps seem correct. I think my answer is right.**Final Answer**The radius ( r ) is approximately boxed{1.381}.