💡To determine the largest prime factor of ( N ), where ( N ) is the sum of the divisors of 300, follow these steps:1. **Factorize 300**: - The prime factorization of 300 is ( 2^2 cdot 3 cdot 5^2 ).2. **Calculate the sum of divisors using the formula**: - The formula for the sum of divisors of a number ( n = p^a cdot q^b cdot r^c ) is ( (1 + p + p^2 + ldots + p^a)(1 + q + q^2 + ldots + q^b)(1 + r + r^2 + ldots + r^c) ). - Applying this to 300: [ N = (1 + 2 + 2^2)(1 + 3)(1 + 5 + 5^2) ]3. **Compute each term individually**: - For ( 2 ): ( 1 + 2 + 4 = 7 ) - For ( 3 ): ( 1 + 3 = 4 ) - For ( 5 ): ( 1 + 5 + 25 = 31 )4. **Multiply these to find ( N )**: [ N = 7 cdot 4 cdot 31 = 868 ]5. **Prime factorize 868**: - ( 868 = 2^2 cdot 7 cdot 31 )6. **Identify the largest prime factor**: - The prime factors of 868 are 2, 7, and 31. - The largest prime factor is 31.Therefore, the largest prime factor of ( N ) is (boxed{31}).

💡Okay, so I have this problem where the Russian alphabet is divided into five groups, and I need to figure out the criterion used for this division. The groups are:1) А, Д, М, П, Т, Ш 2) В, Е, З, К, С, Э, Ю 3) И 4) Ж, Н, О, Ф, Х 5) Б, Г, Л, Р, У, Ц, Ч, Щ, Я First, I need to understand what each group has in common. Maybe it's something related to the shape of the letters, their pronunciation, or perhaps their position in the alphabet. Let me think about each possibility.Starting with the idea of pronunciation, I know that the Russian alphabet has different consonants and vowels. Maybe the groups are based on vowels and consonants? Let's see:Group 1: А, Д, М, П, Т, Ш Group 2: В, Е, З, К, С, Э, Ю Group 3: И Group 4: Ж, Н, О, Ф, Х Group 5: Б, Г, Л, Р, У, Ц, Ч, Щ, Я Looking at this, Group 3 only has И, which is a vowel. Group 1 has А, which is also a vowel, but the rest are consonants. So it's not just vowels and consonants.Maybe it's something about the shape of the letters? Like symmetry. I remember that some letters are symmetrical vertically, horizontally, or both. Let me check:Group 1: А, Д, М, П, Т, Ш - А is symmetrical vertically. - Д is symmetrical vertically. - М is symmetrical vertically. - П is symmetrical vertically. - Т is symmetrical vertically. - Ш is symmetrical vertically. So Group 1 seems to have letters that are symmetrical vertically.Group 2: В, Е, З, К, С, Э, Ю - В is symmetrical horizontally. - Е is symmetrical horizontally. - З is symmetrical horizontally. - К is symmetrical horizontally. - С is symmetrical horizontally. - Э is symmetrical horizontally. - Ю is symmetrical horizontally. So Group 2 has letters that are symmetrical horizontally.Group 3: И - И is symmetrical both vertically and horizontally, so it has central symmetry.Group 4: Ж, Н, О, Ф, Х - Ж is symmetrical vertically and horizontally. - Н is symmetrical vertically and horizontally. - О is symmetrical vertically and horizontally. - Ф is symmetrical vertically and horizontally. - Х is symmetrical vertically and horizontally. So Group 4 has letters that are symmetrical both vertically and horizontally, meaning they have central symmetry.Group 5: Б, Г, Л, Р, У, Ц, Ч, Щ, Я - These letters don't seem to have any symmetry. For example, Б is not symmetrical vertically or horizontally. The same goes for Г, Л, R, У, Ц, Ч, Щ, and Я.So, putting it all together, the criterion seems to be based on the type of symmetry each letter has:1. Vertical symmetry 2. Horizontal symmetry 3. Central symmetry (both vertical and horizontal) 4. All three types of symmetry 5. No symmetry This makes sense because each group corresponds to a specific type of symmetry, and Group 3 is unique because it only has one letter with central symmetry. Group 4 includes letters with all types of symmetry, which might be a bit confusing since central symmetry already includes both vertical and horizontal, but in this case, it's treated as a separate category.I think this is the correct criterion because it logically divides the letters based on their visual properties, and it matches the groups provided. I don't see any other obvious patterns related to pronunciation or position in the alphabet that would explain this division as clearly as symmetry does.

💡Okay, so I have this problem where I need to find the value of g(3) for the function g(x) = 5x³ - 6x² - 3x + 5. Alright, let me think about how to approach this step by step.First, I remember that to find g(3), I just need to substitute x with 3 in the given function. That means wherever I see an x in the equation, I'll replace it with the number 3. So, let me write that out:g(3) = 5*(3)³ - 6*(3)² - 3*(3) + 5.Now, I need to calculate each term one by one to avoid confusion. Let's start with the first term, which is 5*(3)³. I know that the exponent comes before multiplication, so I should calculate 3³ first. 3³ is 3 multiplied by itself three times: 3*3*3. Let me do that: 3*3 is 9, and then 9*3 is 27. So, 3³ is 27. Now, multiply that by 5: 5*27. Hmm, 5*20 is 100, and 5*7 is 35, so adding those together gives me 135. So, the first term is 135.Next, the second term is -6*(3)². Again, I need to handle the exponent first. 3² is 3*3, which is 9. Now, multiply that by -6: -6*9. Let me calculate that: 6*9 is 54, so with the negative sign, it's -54. So, the second term is -54.Moving on to the third term, which is -3*(3). That's straightforward: -3 multiplied by 3. 3*3 is 9, so with the negative sign, it's -9. So, the third term is -9.The last term is just 5, so that's simple. There's no x there, so it remains as 5.Now, I have all the terms calculated:- First term: 135- Second term: -54- Third term: -9- Fourth term: 5I need to add all these together to find g(3). Let me write that out:135 - 54 - 9 + 5.Let me do this step by step to avoid mistakes. Starting with 135, subtract 54: 135 - 54. Hmm, 135 minus 50 is 85, and then minus 4 more is 81. So, 135 - 54 equals 81.Now, take that result and subtract 9: 81 - 9. That's straightforward; 81 minus 9 is 72.Finally, add 5 to 72: 72 + 5 equals 77.So, putting it all together, g(3) equals 77.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting with 5*(3)³: 3³ is 27, 5*27 is 135. Correct. Then -6*(3)²: 3² is 9, -6*9 is -54. Correct. Then -3*(3) is -9. Correct. And the last term is +5. So, adding them up: 135 - 54 is 81, 81 - 9 is 72, 72 + 5 is 77. Yep, that seems right.I think I did everything correctly. Substituted x with 3, calculated each term step by step, and added them up. So, I'm confident that g(3) is 77.

💡First, I need to determine the total number of marks in the exam and the passing marks. Let's denote the total marks as ( T ) and the passing marks as ( P ).From the problem, a candidate who scores 30% of the marks fails by 60 marks. This means:[0.30T = P - 60 quad text{(Equation 1)}]Another candidate who scores 45% of the marks gets 30 marks more than necessary for passing. This means:[0.45T = P + 30 quad text{(Equation 2)}]To find the value of ( T ), I'll subtract Equation 1 from Equation 2 to eliminate ( P ):[0.45T - 0.30T = (P + 30) - (P - 60)][0.15T = 90][T = frac{90}{0.15} = 600]Now that I have the total marks ( T = 600 ), I can substitute this back into Equation 1 to find ( P ):[0.30 times 600 = P - 60][180 = P - 60][P = 180 + 60 = 240]Therefore, the number of marks required to pass is 240.

💡First, I need to calculate the product of three and the given value of ( x ), which is 4.Next, I add 36 to this product to find the certain number.Finally, by performing these calculations, I determine that the certain number is 48.

💡Okay, so I have this problem about calculating the cost of using whiteboards for one day at Oakland High. Let me try to understand what's being asked here. There are 5 classes in a building block, and each class uses 2 whiteboards. Each whiteboard needs about 20 ml of ink per day, and the ink costs 50 cents per ml. I need to find out how much it would cost in total to use all these whiteboards for one day.Alright, let's break this down step by step. First, I need to figure out how much ink is used by all the whiteboards in one day. Since each whiteboard uses 20 ml of ink, and there are 2 whiteboards per class, that means each class uses 2 times 20 ml of ink. So, 2 times 20 is 40 ml per class.Now, there are 5 classes in total. So, I need to multiply the ink used per class by the number of classes. That would be 40 ml per class times 5 classes. 40 times 5 is 200 ml. So, altogether, all the whiteboards use 200 ml of ink in one day.Next, I need to calculate the cost of this ink. The ink costs 50 cents per ml. So, if I have 200 ml of ink needed, and each ml costs 50 cents, I can find the total cost by multiplying the amount of ink by the cost per ml. That would be 200 ml times 0.50 per ml.Wait, let me make sure I'm doing this right. 200 ml times 0.50 per ml. So, 200 times 0.50 is... 200 times 0.5 is 100. So, the total cost would be 100.00.Hmm, that seems straightforward, but let me double-check to make sure I didn't miss anything. So, starting from the beginning: 5 classes, each with 2 whiteboards, each whiteboard needing 20 ml of ink. So, 5 classes times 2 whiteboards is 10 whiteboards in total. Each whiteboard uses 20 ml, so 10 whiteboards times 20 ml is 200 ml total ink needed.Then, the cost is 50 cents per ml, so 200 ml times 0.50 is indeed 100.00. Okay, that seems consistent.I guess another way to look at it is to calculate the cost per whiteboard first. Each whiteboard uses 20 ml, and ink costs 0.50 per ml. So, the cost per whiteboard is 20 times 0.50, which is 10.00 per whiteboard per day.Since there are 2 whiteboards per class, the cost per class would be 2 times 10.00, which is 20.00 per class per day.With 5 classes, the total cost would be 5 times 20.00, which again is 100.00.Okay, so both methods give me the same result, which is reassuring. It looks like the total cost to use all the whiteboards for one day is 100.00.I wonder if there's any other way to approach this problem. Maybe by calculating the cost per ml first and then scaling it up. But I think the way I did it is pretty straightforward.Just to be thorough, let's consider if there are any hidden costs or factors I might have missed. For example, is the ink cost per ml inclusive of any taxes or additional fees? The problem doesn't specify, so I think it's safe to assume that 0.50 per ml is the total cost.Also, are there any discounts for buying ink in larger quantities? Again, the problem doesn't mention any, so I'll proceed with the given rate.Another thing to consider is whether the ink usage is exactly 20 ml per whiteboard per day, or if there's any variation. But since the problem states "about 20 ml," I'll take that as an average and use it for the calculation.In summary, by calculating the total ink needed and then multiplying by the cost per ml, or by calculating the cost per whiteboard and scaling up to the total number of whiteboards, I arrive at the same total cost of 100.00 for one day's use of all the whiteboards in the building block at Oakland High.