answer：First, I need to determine the combined weight of Christine's two cats. The first cat weighs 7 pounds and the second cat weighs 10 pounds.Next, I add the weights of the two cats together to find the total weight:7 pounds + 10 pounds = 17 pounds.The problem states that the dog's weight is a multiple of the combined weight of the two cats, which is 34 pounds. To find the ratio of the dog's weight to the combined weight of the cats, I divide the dog's weight by the combined weight of the cats:34 pounds / 17 pounds = 2.Therefore, the ratio of the dog's weight to the combined weight of the two cats is 2 to 1.

answer：First, I need to rewrite the quadratic expression (6x^2 + 36x + 216) in the form (a(x + b)^2 + c).I'll start by factoring out the coefficient of (x^2), which is 6:[6x^2 + 36x + 216 = 6(x^2 + 6x) + 216]Next, I'll complete the square for the expression inside the parentheses. To do this, I take half of the coefficient of (x), which is 6, divide it by 2 to get 3, and then square it to get 9.I'll add and subtract this square inside the parentheses:[6(x^2 + 6x + 9 - 9) + 216 = 6((x + 3)^2 - 9) + 216]Now, I'll distribute the 6 and simplify:[6(x + 3)^2 - 54 + 216 = 6(x + 3)^2 + 162]So, the quadratic expression in the desired form is (6(x + 3)^2 + 162), where (a = 6), (b = 3), and (c = 162).Finally, I'll find the sum of (a), (b), and (c):[a + b + c = 6 + 3 + 162 = 171]

answer：Okay, so I need to figure out the probability that the sum of two fair six-sided dice is less than 11. Hmm, let's see. First, I remember that when you roll two dice, each die has numbers from 1 to 6. So, the smallest sum you can get is 2 (when both dice show 1), and the largest sum is 12 (when both dice show 6). I think the total number of possible outcomes when rolling two dice is 6 times 6, which is 36. Yeah, that makes sense because for each number on the first die, there are 6 possible numbers on the second die. So, 6 multiplied by 6 equals 36 total outcomes.Now, the problem is asking for the probability that the sum is less than 11. So, I need to find all the possible outcomes where the sum of the two dice is 2, 3, 4, 5, 6, 7, 8, 9, or 10. That seems like a lot of sums to consider. Maybe there's a smarter way to do this instead of listing all the possible combinations.Wait, I remember that sometimes it's easier to calculate the probability of the complement event and then subtract it from 1. The complement of getting a sum less than 11 is getting a sum of 11 or 12. So, if I can find the number of outcomes that result in a sum of 11 or 12, I can subtract that from the total number of outcomes to find the number of favorable outcomes for sums less than 11.Let me try that approach. So, first, how many ways can I get a sum of 11? Let's think: - If the first die is 5, the second die needs to be 6 to make 11.- If the first die is 6, the second die needs to be 5 to make 11.So, there are 2 outcomes that result in a sum of 11.Next, how many ways can I get a sum of 12? Well, both dice need to show 6. So, there's only 1 outcome for that.Therefore, the total number of outcomes that result in a sum of 11 or 12 is 2 plus 1, which equals 3.Now, since there are 36 total outcomes, the probability of getting a sum of 11 or 12 is 3 out of 36, which simplifies to 1 out of 12.But we want the probability of getting a sum less than 11, which is the complement of getting 11 or 12. So, we subtract the probability of getting 11 or 12 from 1.So, 1 minus 1/12 equals 11/12.Therefore, the probability that the sum of the two dice is less than 11 is 11/12.Wait, let me double-check that. If there are 36 total outcomes and 3 outcomes result in sums of 11 or 12, then the number of outcomes for sums less than 11 is 36 minus 3, which is 33. So, 33 out of 36 is indeed 11/12 when simplified.Yeah, that seems right. So, the probability is 11/12.

answer：First, list all possible combinations of three sticks from the four available: (2, 3, 4), (2, 3, 6), (2, 4, 6), and (3, 4, 6).Next, check each combination to see if it satisfies the triangle inequality theorem, which states that the sum of the lengths of any two sides must be greater than the length of the third side.For the combination (2, 3, 4):- 2 + 3 > 4 (5 > 4) ✔️- 2 + 4 > 3 (6 > 3) ✔️- 3 + 4 > 2 (7 > 2) ✔️For the combination (2, 3, 6):- 2 + 3 > 6 (5 > 6) ❌For the combination (2, 4, 6):- 2 + 4 > 6 (6 > 6) ❌For the combination (3, 4, 6):- 3 + 4 > 6 (7 > 6) ✔️- 3 + 6 > 4 (9 > 4) ✔️- 4 + 6 > 3 (10 > 3) ✔️Only two combinations, (2, 3, 4) and (3, 4, 6), satisfy the triangle inequality theorem. Therefore, the number of different triangles that can be formed is 2.

answer：Okay, so I'm trying to figure out this game problem where there are initially m red sprites and n blue sprites. The rules are that if you click on two sprites of the same color, they merge into a red sprite. If you click on two different colors, they merge into a blue sprite. The game ends when only one sprite is left, and I need to determine what color that last sprite will be. The options are that it depends only on the parity of m, only on the parity of n, both m and n, or neither. Alright, let me start by understanding the problem step by step. So, we have two types of sprites: red and blue. When two of the same color are clicked, they merge into a red sprite. If two different colors are clicked, they merge into a blue sprite. The game continues until there's only one sprite left. I need to figure out what determines the color of that last sprite.First, let me think about what happens when we merge two sprites. There are three possible cases:1. Two reds merge into one red.2. Two blues merge into one red.3. A red and a blue merge into one blue.Each time we merge two sprites, the total number of sprites decreases by one. So, starting with m + n sprites, we'll have m + n - 1 merges until we're left with one sprite.Now, I need to see how the number of red and blue sprites changes with each merge. Let's consider each case:1. **Two reds merge into one red:** This reduces the number of red sprites by one (since two are merged into one). The number of blue sprites remains the same.2. **Two blues merge into one red:** This reduces the number of blue sprites by two (since two are merged into one) and increases the number of red sprites by one.3. **A red and a blue merge into one blue:** This reduces the number of red sprites by one and keeps the number of blue sprites the same.So, in each case, the number of red sprites can decrease by one or increase by one, depending on the merge. The number of blue sprites either stays the same or decreases by two.Hmm, interesting. So, the number of blue sprites can only change by zero or two. That means the parity (whether it's odd or even) of the number of blue sprites doesn't change. Because if you start with an even number of blue sprites, subtracting two keeps it even, and if you start with an odd number, subtracting two keeps it odd. Similarly, if you don't change the number of blue sprites, their parity remains the same.On the other hand, the number of red sprites can change by plus one or minus one, so the parity of red sprites can change with each merge.So, the key insight here is that the parity of the number of blue sprites remains constant throughout the game. That is, if you start with an even number of blue sprites, you'll always have an even number (or zero) until the end. If you start with an odd number, you'll always have an odd number until the end.Now, since the game ends when there's only one sprite left, let's think about what that implies. If the number of blue sprites is odd, then it's possible that the last sprite is blue. If it's even, then the last sprite must be red because you can't have an even number of blue sprites when there's only one sprite left.Wait, let me clarify that. If you start with an odd number of blue sprites, you can't reduce it to zero because you can only subtract two each time. So, you'll always end up with one blue sprite left. If you start with an even number of blue sprites, you can reduce it to zero, and then you'll be left with red sprites. But since the game ends when there's only one sprite, if you have zero blue sprites, you must have one red sprite left.So, the color of the last sprite depends solely on whether the initial number of blue sprites is odd or even. If n is odd, the last sprite is blue; if n is even, the last sprite is red.Let me test this with some examples to make sure I'm not making a mistake.**Example 1:**m = 1, n = 1So, one red and one blue. If I merge them, they become a blue sprite. So, the last sprite is blue. Since n was odd (1), the result is blue. That matches my conclusion.**Example 2:**m = 2, n = 1Two reds and one blue. Let's see:- Merge two reds: they become one red. Now we have one red and one blue.- Merge red and blue: they become one blue. So, last sprite is blue. n was 1 (odd), result is blue.Alternatively, could I have merged differently?- Merge red and blue first: becomes blue. Now we have one red and one blue.- Merge red and blue: becomes blue. Still, last sprite is blue.Same result.**Example 3:**m = 1, n = 2One red and two blues.- Merge two blues: becomes one red. Now we have two reds.- Merge two reds: becomes one red. Last sprite is red. n was 2 (even), result is red.Alternatively:- Merge red and blue: becomes blue. Now we have one blue and one blue.- Merge two blues: becomes one red. Last sprite is red.Same result.**Example 4:**m = 3, n = 2Three reds and two blues.- Merge two reds: becomes one red. Now we have two reds and two blues.- Merge two reds: becomes one red. Now we have one red and two blues.- Merge two blues: becomes one red. Now we have two reds.- Merge two reds: becomes one red. Last sprite is red. n was 2 (even), result is red.Alternatively, different merging order:- Merge red and blue: becomes blue. Now we have two reds and one blue.- Merge two reds: becomes one red. Now we have one red and one blue.- Merge red and blue: becomes blue. Last sprite is blue. Wait, that contradicts my previous conclusion.Wait, hold on. In this case, I started with n = 2 (even), but ended up with a blue sprite. That shouldn't happen according to my earlier reasoning. Did I make a mistake in the merging steps?Let me check:Starting with m=3, n=2.First merge: red and blue → blue. Now m=2, n=1.Second merge: two reds → red. Now m=1, n=1.Third merge: red and blue → blue. So, last sprite is blue.But n was initially 2 (even), so according to my earlier conclusion, the last sprite should be red. But in this case, it's blue. That means my reasoning was flawed.Hmm, so what's wrong here?Wait, in this case, n started at 2 (even), but during the merging process, n became 1 (odd). But according to my earlier logic, the parity of n should remain constant. But in this case, n went from even to odd. That contradicts my earlier assertion.Wait, let me think again. When I merged red and blue, n didn't change. It was 2, then after merging red and blue, n became 1. Wait, no, actually, when you merge red and blue, you lose one red and one blue, and gain one blue. So, the number of blue sprites remains the same: n was 2, after merging red and blue, you have n=2 again? Wait, no.Wait, hold on. Let me clarify.When you merge two sprites:- If they are both red: m decreases by 1, n remains the same.- If they are both blue: m increases by 1, n decreases by 2.- If they are red and blue: m decreases by 1, n remains the same.So, in the case where I merged red and blue, n remains the same. So, if I started with n=2, after merging red and blue, n should still be 2.Wait, but in my previous example, I thought that merging red and blue would decrease n by 1, but according to the rules, it doesn't. It just replaces one red and one blue with one blue, so n remains the same.Wait, so in the example where I had m=3, n=2:First merge: red and blue → blue. So, m becomes 2, n remains 2.Second merge: two reds → red. So, m becomes 1, n remains 2.Third merge: red and blue → blue. So, m becomes 0, n remains 2.Wait, but then we have two blue sprites left. Then we need to merge them.Fourth merge: two blues → red. So, m becomes 1, n becomes 0.Fifth merge: two reds → red. So, last sprite is red.Wait, so in this case, starting with n=2 (even), we end up with red. So, my initial mistake was thinking that merging red and blue would decrease n, but it doesn't. It just keeps n the same.So, in my earlier incorrect example, I mistakenly thought that merging red and blue would decrease n, but actually, it doesn't. Therefore, the parity of n remains constant.So, in the case where n is even, you can never end up with an odd number of blue sprites. Therefore, the last sprite must be red if n is even, and blue if n is odd.Let me test this again with the same example:m=3, n=2.First merge: red and blue → blue. Now m=2, n=2.Second merge: two reds → red. Now m=1, n=2.Third merge: red and blue → blue. Now m=0, n=2.Fourth merge: two blues → red. Now m=1, n=0.Fifth merge: two reds → red. Last sprite is red.So, n was even, result is red. That's consistent.Another example:m=1, n=3.First merge: red and blue → blue. Now m=0, n=3.Second merge: two blues → red. Now m=1, n=1.Third merge: red and blue → blue. Last sprite is blue.n was odd, result is blue.Another example:m=2, n=3.First merge: two reds → red. Now m=1, n=3.Second merge: two blues → red. Now m=2, n=1.Third merge: red and blue → blue. Now m=1, n=1.Fourth merge: red and blue → blue. Last sprite is blue.n was odd, result is blue.Wait, but in this case, starting with n=3, we ended with blue. That's consistent.Wait, but what if I merge differently:m=2, n=3.First merge: two blues → red. Now m=3, n=1.Second merge: two reds → red. Now m=2, n=1.Third merge: two reds → red. Now m=1, n=1.Fourth merge: red and blue → blue. Last sprite is blue.Same result.So, it seems that regardless of the merging order, the parity of n determines the last sprite.Therefore, the color of the remaining sprite is only related to the parity of n.So, the answer should be B.

answer：Let's solve the problem step by step.**Given:**- **Store A** offers: 1. 20% off the sticker price. 2. A 120 rebate. 3. An additional 5% discount on the already reduced price.- **Store B** offers 15% off the sticker price.- Parvati saves 30 by purchasing from Store A instead of Store B.**Let** ( x ) **be the sticker price of the laptop.****Calculating the final price at Store A:**1. After a 20% discount: ( 0.8x )2. After a 120 rebate: ( 0.8x - 120 )3. After an additional 5% discount: ( 0.95(0.8x - 120) = 0.76x - 114 )**Calculating the final price at Store B:**- After a 15% discount: ( 0.85x )**Setting up the equation based on the savings:**[0.76x - 114 = 0.85x - 30]**Solving for ( x ):**[0.76x - 0.85x = -30 + 114][-0.09x = 84][x = frac{84}{0.09} = 933.33]**Conclusion:**The sticker price of the laptop is ( 933.33 ), which corresponds to option (C).