answer：Alright, so I need to figure out how many non-congruent triangles with integer side lengths have a perimeter of 18 units, and none of the sides are shorter than 3 units. Hmm, okay, let's break this down step by step.First, I remember that for a set of three sides to form a triangle, they must satisfy the triangle inequality theorem. That means the sum of any two sides must be greater than the third side. Also, since we're dealing with integer side lengths and a fixed perimeter, I need to find all possible combinations of three integers that add up to 18 and meet the triangle inequality.Additionally, each side must be at least 3 units long. So, I can't have any side that's 1 or 2 units. That might limit the number of possible combinations.Let me denote the sides of the triangle as a, b, and c, where a ≤ b ≤ c. This way, I can systematically list the combinations without repeating the same sets in different orders, which would be congruent triangles.Given that a + b + c = 18 and a ≥ 3, I can start by setting a to the minimum value, which is 3, and then find possible values for b and c.So, let's start with a = 3.Then, b must be at least 3 as well, but since a ≤ b ≤ c, b can range from 3 up to some maximum value where a + b > c. But since a + b + c = 18, c = 18 - a - b = 18 - 3 - b = 15 - b.So, the triangle inequality tells us that a + b > c. Substituting c, we get 3 + b > 15 - b. Simplifying that:3 + b > 15 - b 3 + b + b > 15 3 + 2b > 15 2b > 12 b > 6Since b must be an integer, b ≥ 7.But also, since b ≤ c, and c = 15 - b, we have b ≤ (15 - b). So:b ≤ (15 - b) 2b ≤ 15 b ≤ 7.5Since b is an integer, b ≤ 7.Wait, so from the triangle inequality, b > 6, and from b ≤ c, b ≤ 7. So b must be exactly 7.Therefore, when a = 3, b = 7, and c = 15 - 7 = 8.So, one triangle is (3, 7, 8).Now, let's check if there are any other possibilities when a = 3. If I try b = 8, then c would be 15 - 8 = 7, but that would make c = 7, which is less than b = 8, violating the condition that a ≤ b ≤ c. So, that's not valid.Therefore, with a = 3, the only valid triangle is (3, 7, 8).Next, let's try a = 4.So, a = 4. Then, b must be at least 4, and c = 18 - 4 - b = 14 - b.Again, applying the triangle inequality: a + b > c.4 + b > 14 - b 4 + b + b > 14 4 + 2b > 14 2b > 10 b > 5Since b must be an integer, b ≥ 6.Also, since b ≤ c, and c = 14 - b, we have:b ≤ (14 - b) 2b ≤ 14 b ≤ 7So, b can be 6 or 7.Let's check b = 6:c = 14 - 6 = 8So, the triangle is (4, 6, 8). Let's verify the triangle inequality:4 + 6 > 8: 10 > 8, which is true.4 + 8 > 6: 12 > 6, true.6 + 8 > 4: 14 > 4, true.Good.Now, b = 7:c = 14 - 7 = 7So, the triangle is (4, 7, 7). Checking the triangle inequality:4 + 7 > 7: 11 > 7, true.4 + 7 > 7: same as above.7 + 7 > 4: 14 > 4, true.Good.So, with a = 4, we have two triangles: (4, 6, 8) and (4, 7, 7).Next, let's try a = 5.a = 5. Then, b must be at least 5, and c = 18 - 5 - b = 13 - b.Applying the triangle inequality: 5 + b > c.5 + b > 13 - b 5 + b + b > 13 5 + 2b > 13 2b > 8 b > 4But since b must be at least 5, this condition is automatically satisfied.Also, since b ≤ c, and c = 13 - b, we have:b ≤ (13 - b) 2b ≤ 13 b ≤ 6.5Since b is an integer, b ≤ 6.So, b can be 5 or 6.Let's check b = 5:c = 13 - 5 = 8So, the triangle is (5, 5, 8). Checking the triangle inequality:5 + 5 > 8: 10 > 8, true.5 + 8 > 5: 13 > 5, true.5 + 8 > 5: same as above.Good.Now, b = 6:c = 13 - 6 = 7So, the triangle is (5, 6, 7). Checking the triangle inequality:5 + 6 > 7: 11 > 7, true.5 + 7 > 6: 12 > 6, true.6 + 7 > 5: 13 > 5, true.Good.So, with a = 5, we have two triangles: (5, 5, 8) and (5, 6, 7).Next, let's try a = 6.a = 6. Then, b must be at least 6, and c = 18 - 6 - b = 12 - b.Applying the triangle inequality: 6 + b > c.6 + b > 12 - b 6 + b + b > 12 6 + 2b > 12 2b > 6 b > 3But since b must be at least 6, this condition is automatically satisfied.Also, since b ≤ c, and c = 12 - b, we have:b ≤ (12 - b) 2b ≤ 12 b ≤ 6Since b must be at least 6, b = 6.So, c = 12 - 6 = 6Thus, the triangle is (6, 6, 6). Checking the triangle inequality:6 + 6 > 6: 12 > 6, true.6 + 6 > 6: same as above.6 + 6 > 6: same as above.Good.So, with a = 6, we have one triangle: (6, 6, 6).Now, let's check if a can be greater than 6.If a = 7, then b must be at least 7, and c = 18 - 7 - b = 11 - b.But since b ≥ 7, c = 11 - b would be ≤ 4, which contradicts the condition that each side must be at least 3 units. Wait, c would be 11 - b, and if b =7, c =4, which is acceptable since 4 ≥3. Let me check.Wait, a =7, b=7, c=4. But we have to maintain a ≤ b ≤ c, so c must be at least b, which is 7. But c =4 <7, which violates the condition. Therefore, this is not valid.Similarly, if a =7, b=8, c=3, which is even worse.Therefore, a cannot be greater than 6 because it would result in c being less than a or b, violating the a ≤ b ≤ c condition.So, we've covered all possible values of a from 3 to 6.Now, let's list all the triangles we've found:- (3, 7, 8)- (4, 6, 8)- (4, 7, 7)- (5, 5, 8)- (5, 6, 7)- (6, 6, 6)That's a total of 6 triangles.I think that's all. Let me just double-check to make sure I haven't missed any combinations.Starting with a=3, we only found one triangle.With a=4, two triangles.With a=5, two triangles.With a=6, one triangle.Total: 1+2+2+1=6.Yes, that seems correct.I don't think there are any other combinations because when a increases beyond 6, c becomes too small, and when a is smaller, we've already covered all possibilities.So, the final answer should be 6 non-congruent triangles.

answer：Alright, so I have this probability problem here, and I need to figure out the chance that both 20-sided dice show the same color, and also the 6-sided die shows a number greater than 4. Hmm, okay, let's break this down step by step.First, let's understand the dice we're dealing with. There are two 20-sided dice:1. The first die has: - 3 maroon sides, - 9 teal sides, - 7 cyan sides, - 1 sparkly side.2. The second die has: - 5 maroon sides, - 6 teal sides, - 8 cyan sides, - 1 sparkly side.And then there's a standard 6-sided die, which I assume is numbered from 1 to 6.The problem asks for the probability that both 20-sided dice show the same color, and the 6-sided die shows a number greater than 4. So, essentially, we're looking for two things to happen simultaneously:1. Both 20-sided dice land on the same color.2. The 6-sided die lands on either a 5 or a 6.Since these are independent events (the outcome of one die doesn't affect the others), I can calculate the probabilities separately and then multiply them together to get the final probability.Let's start with the first part: the probability that both 20-sided dice show the same color.To find this, I need to consider each color separately and calculate the probability that both dice show that color. Then, I'll add up these probabilities to get the total probability of both dice showing the same color.**1. Maroon:**- The first die has 3 maroon sides out of 20, so the probability of rolling maroon on the first die is 3/20.- The second die has 5 maroon sides out of 20, so the probability of rolling maroon on the second die is 5/20.- Since these are independent events, I multiply the probabilities: (3/20) * (5/20) = 15/400.**2. Teal:**- The first die has 9 teal sides out of 20, so the probability is 9/20.- The second die has 6 teal sides out of 20, so the probability is 6/20.- Multiplying these gives: (9/20) * (6/20) = 54/400.**3. Cyan:**- The first die has 7 cyan sides out of 20, so the probability is 7/20.- The second die has 8 cyan sides out of 20, so the probability is 8/20.- Multiplying these gives: (7/20) * (8/20) = 56/400.**4. Sparkly:**- Both dice have 1 sparkly side each out of 20, so the probability for each is 1/20.- Multiplying these gives: (1/20) * (1/20) = 1/400.Now, to find the total probability that both dice show the same color, I add up all these individual probabilities:15/400 (maroon) + 54/400 (teal) + 56/400 (cyan) + 1/400 (sparkly) = 126/400.Simplifying this fraction by dividing numerator and denominator by 2 gives 63/200.Okay, so the probability that both 20-sided dice show the same color is 63/200.Now, moving on to the second part: the probability that the 6-sided die shows a number greater than 4.A standard 6-sided die has numbers 1 through 6. The numbers greater than 4 are 5 and 6, so there are 2 favorable outcomes.The probability is therefore 2/6, which simplifies to 1/3.Now, since these two events are independent (the outcome of the 20-sided dice doesn't affect the 6-sided die and vice versa), I can find the combined probability by multiplying the two individual probabilities:(63/200) * (1/3) = 63/600.Simplifying this fraction by dividing numerator and denominator by 3 gives 21/200.So, the probability that both 20-sided dice show the same color and the 6-sided die shows a number greater than 4 is 21/200.Just to make sure I didn't make any mistakes, let me recap:- Calculated the probability for each color on both dice, multiplied them, and added them up to get 63/200.- Calculated the probability for the 6-sided die showing greater than 4, which is 1/3.- Multiplied these two probabilities together to get 21/200.Everything seems to check out. I think that's the correct answer.

answer：Alright, so I've got this inequality problem here, and I need to figure out which of the given options must hold true if x is less than y. Let's take it step by step.First, the problem states that x < y. That means x is to the left of y on the number line, right? So x is smaller than y. Now, I need to check each option one by one to see if they must be true given this condition.Starting with Option A: ax < ay. Hmm, okay. So if I multiply both sides of the inequality x < y by a, does the inequality still hold? Well, I remember that when you multiply or divide both sides of an inequality by a positive number, the direction of the inequality stays the same. But if you multiply or divide by a negative number, the inequality flips. So, if a is positive, then ax < ay would indeed be true. But what if a is negative? Then ax would actually be greater than ay, right? And if a is zero, then both ax and ay would be zero, making the inequality ax < ay not hold because they're equal. So, Option A isn't necessarily always true. It depends on the value of a. So, I can probably rule out Option A.Moving on to Option B: x² < y². Okay, so squaring both sides of the inequality. Hmm, I need to be careful here because squaring can sometimes change the inequality depending on the signs of x and y. Let's think about this. If both x and y are positive, then squaring them would preserve the inequality, so x² < y² would hold. But what if x and y are negative? For example, if x = -3 and y = -2, then x < y because -3 is less than -2. But x² would be 9 and y² would be 4, so x² > y², which means x² < y² is not true in this case. So, Option B doesn't always hold true either.Next is Option C: x/3 > y/3. Okay, so dividing both sides of the inequality by 3. Since 3 is a positive number, dividing both sides by 3 should preserve the inequality direction. So, if x < y, then x/3 should be less than y/3, not greater. Therefore, x/3 > y/3 is actually the opposite of what should be true. So, Option C is incorrect.Finally, Option D: -x + 3 > -y + 3. Let's break this down. Starting with x < y, if I multiply both sides by -1, the inequality flips, so -x > -y. Then, if I add 3 to both sides, the inequality remains the same because I'm adding the same number to both sides. So, -x + 3 > -y + 3. That makes sense. So, Option D holds true.Wait, but let me double-check. If x < y, then multiplying by -1 gives -x > -y. Adding 3 to both sides doesn't change the inequality, so yes, -x + 3 > -y + 3 is indeed true. So, Option D is correct.Just to make sure I didn't miss anything, let's go through each option again quickly:- Option A: Depends on a, so not necessarily true.- Option B: Doesn't hold if x and y are negative.- Option C: Opposite of what it should be.- Option D: Correct after flipping and adding.Yeah, I think I'm confident that Option D is the only one that must hold true.

answer：First, I need to convert the original length of the line from meters to centimeters. Since 1 meter is equal to 100 centimeters, I can multiply the given length by 100 to get the length in centimeters.Next, I will subtract the erased portion of 15.25 centimeters from the converted length to find the new length of the line.Finally, I will present the final answer in a clear and concise manner.

answer：Okay, so I need to figure out whether this resident should install the energy-efficient lamp themselves or use the energy service company. There are two parts to this problem: part (a) is for the next 10 months, and part (b) is for the entire lifespan of the lamp. Let me start with part (a).First, I need to understand the costs involved. The resident currently uses a 60-watt incandescent lamp for 100 hours each month. The electricity tariff is 5 rubles per kilowatt-hour (kWh). So, I should calculate the monthly electricity cost for the current lamp.To find the energy consumption in kilowatt-hours, I can use the formula:[ text{Energy (kWh)} = frac{text{Power (W)} times text{Time (hours)}}{1000} ]So, for the 60-watt lamp:[ text{Energy}_1 = frac{60 times 100}{1000} = 6 , text{kWh} ]Then, the monthly cost is:[ text{Cost}_1 = 6 , text{kWh} times 5 , text{rubles/kWh} = 30 , text{rubles} ]Over 10 months, this would be:[ text{Total Cost}_1 = 30 times 10 = 300 , text{rubles} ]Now, if the resident buys the energy-efficient 12-watt lamp, which costs 120 rubles, I need to calculate the new electricity cost.First, the energy consumption for the 12-watt lamp:[ text{Energy}_2 = frac{12 times 100}{1000} = 1.2 , text{kWh} ]Monthly cost:[ text{Cost}_2 = 1.2 times 5 = 6 , text{rubles} ]Total cost over 10 months, including the initial cost of the lamp:[ text{Total Cost}_2 = 120 + (6 times 10) = 120 + 60 = 180 , text{rubles} ]Alternatively, if the resident uses the energy service company, the company will install the lamp, but the resident has to pay 75% of the electricity cost savings over 10 months.First, let's find the electricity cost savings. The difference in energy consumption between the two lamps is:[ Delta text{Energy} = 6 - 1.2 = 4.8 , text{kWh} ]So, the monthly cost savings:[ Delta text{Cost} = 4.8 times 5 = 24 , text{rubles} ]The resident has to pay 75% of this saving to the company:[ text{Payment to Company} = 0.75 times 24 = 18 , text{rubles} ]Therefore, the new monthly cost for the resident is:[ text{New Cost} = text{Cost}_2 + text{Payment to Company} = 6 + 18 = 24 , text{rubles} ]Total cost over 10 months:[ text{Total Cost with Company} = 24 times 10 = 240 , text{rubles} ]Comparing the two options:- Installing themselves: 180 rubles- Using the company: 240 rublesSo, for the next 10 months, it's cheaper to install the lamp themselves.Now, for part (b), considering the entire lifespan of the lamp. Assuming the lamp lasts, say, 3 years (36 months), let's calculate the costs.If the resident installs the lamp themselves:Total cost over 36 months:[ text{Total Cost}_2 = 120 + (6 times 36) = 120 + 216 = 336 , text{rubles} ]If using the company:For the first 10 months, the cost is 240 rubles. For the remaining 26 months, the resident only pays the electricity cost:[ text{Electricity Cost for Remaining Months} = 6 times 26 = 156 , text{rubles} ]Total cost over 36 months:[ text{Total Cost with Company} = 240 + 156 = 396 , text{rubles} ]Comparing the two:- Installing themselves: 336 rubles- Using the company: 396 rublesAgain, installing themselves is cheaper over the entire lifespan.So, in both scenarios, it's more economical for the resident to install the energy-efficient lamp themselves rather than using the energy service company.

answer：Okay, so I have this problem here about a survey on reading classic literature by different age groups. The data is presented in a table with age intervals, a variable x, and a population y. The table looks like this:| Age Interval | [0,10) | [10,15) | [15,20) | [20,25) | [25,30) ||--------------|--------|---------|---------|---------|---------|| Variable x | 1 | 2 | 3 | 4 | 5 || Population y | 2 | 3 | 7 | 8 | a |And they've given me a linear regression equation: y hat equals 2.1 times x hat minus 0.3. So, the equation is:[hat{y} = 2.1hat{x} - 0.3]I need to find the value of 'a' in the population y column. Alright, let's think about how linear regression works. In linear regression, the equation is used to predict the dependent variable (y) based on the independent variable (x). The equation is derived using the method of least squares, which minimizes the sum of the squared differences between the observed and predicted values.One important property of the least squares regression line is that it always passes through the point (x̄, ȳ), where x̄ is the mean of the x-values and ȳ is the mean of the y-values. So, if I can find the mean of x and the mean of y, I can use the regression equation to solve for 'a'.First, let's find the mean of x. The x-values are 1, 2, 3, 4, and 5. So, the mean x̄ is:[x̄ = frac{1 + 2 + 3 + 4 + 5}{5} = frac{15}{5} = 3]Okay, so the mean of x is 3. Now, let's find the mean of y. The y-values are 2, 3, 7, 8, and a. So, the mean ȳ is:[ȳ = frac{2 + 3 + 7 + 8 + a}{5} = frac{20 + a}{5}]Now, according to the regression equation, when x is equal to its mean (3), y should be equal to its mean (ȳ). So, plugging x̄ into the regression equation:[ȳ = 2.1 times 3 - 0.3]Let's compute that:[ȳ = 6.3 - 0.3 = 6]So, the mean of y is 6. But we also have an expression for ȳ in terms of 'a':[ȳ = frac{20 + a}{5}]Setting these equal to each other:[frac{20 + a}{5} = 6]To solve for 'a', multiply both sides by 5:[20 + a = 30]Subtract 20 from both sides:[a = 10]So, the value of 'a' should be 10. Let me double-check my calculations to make sure I didn't make any mistakes.First, the mean of x is definitely 3 because 1+2+3+4+5 is 15, and 15 divided by 5 is 3. Then, plugging x̄ into the regression equation gives 2.1*3 - 0.3, which is 6.3 - 0.3, so 6. That seems correct.Then, the mean of y is (2 + 3 + 7 + 8 + a)/5. Adding up the known y-values: 2 + 3 is 5, plus 7 is 12, plus 8 is 20. So, 20 + a over 5 equals 6. Multiplying both sides by 5 gives 20 + a = 30, so a = 10. Yep, that all checks out.I think I've got it right. So, the value of 'a' is 10.