💡First, I need to convert 80 grams to kilograms. Since there are 1000 grams in a kilogram, I divide 80 by 1000 to get 0.08 kilograms.Next, I'll convert 165 centimeters to meters. There are 100 centimeters in a meter, so I divide 165 by 100 to obtain 1.65 meters.For the third problem, I have 4 jiao and 9 fen. Knowing that 1 jiao equals 0.1 yuan and 1 fen equals 0.01 yuan, I calculate 4 jiao as 0.4 yuan and 9 fen as 0.09 yuan. Adding these together gives 0.49 yuan.The fourth problem involves 13 yuan and 7 fen. Since 7 fen is 0.07 yuan, I add this to 13 yuan to get a total of 13.07 yuan.Finally, I need to convert 5 tons and 26 kilograms to tons. There are 1000 kilograms in a ton, so 26 kilograms is 0.026 tons. Adding this to 5 tons results in 5.026 tons.

💡Okay, so I have this problem where I need to find the value of ( a + b ) given the two equations:1. ( 3a + 5b = 47 )2. ( 7a + 2b = 52 )Hmm, I remember that when you have two equations with two variables, you can solve for the variables using either substitution or elimination. I think elimination might be easier here because the coefficients are manageable.First, maybe I should try to eliminate one of the variables by making the coefficients of either ( a ) or ( b ) the same in both equations. Let me look at the coefficients:- The coefficients of ( a ) are 3 and 7.- The coefficients of ( b ) are 5 and 2.If I multiply the first equation by 7 and the second equation by 3, then the coefficients of ( a ) will both be 21. That should help me eliminate ( a ) by subtracting the equations.So, multiplying the first equation by 7:( 7 times (3a + 5b) = 7 times 47 )Which gives:( 21a + 35b = 329 )And multiplying the second equation by 3:( 3 times (7a + 2b) = 3 times 52 )Which gives:( 21a + 6b = 156 )Now, I have:1. ( 21a + 35b = 329 )2. ( 21a + 6b = 156 )If I subtract the second equation from the first, the ( 21a ) terms will cancel out:( (21a + 35b) - (21a + 6b) = 329 - 156 )Simplifying:( 21a + 35b - 21a - 6b = 173 )Which simplifies further to:( 29b = 173 )Now, solving for ( b ):( b = frac{173}{29} )Calculating that, ( 29 times 6 = 174 ), which is just 1 more than 173, so ( b = 6 - frac{1}{29} ). Wait, that doesn't seem right. Let me check my calculations again.Wait, no, I think I made a mistake in the subtraction step. Let me redo that part:Subtracting the second equation from the first:( 21a + 35b - 21a - 6b = 329 - 156 )So, ( 21a - 21a = 0 ), and ( 35b - 6b = 29b ). On the right side, ( 329 - 156 = 173 ). So, ( 29b = 173 ).Dividing both sides by 29:( b = frac{173}{29} )Calculating ( 29 times 6 = 174 ), which is 1 more than 173, so ( b = 6 - frac{1}{29} ). Wait, that still doesn't make sense because ( 29 times 6 = 174 ), so ( 173 = 29 times 6 - 1 ). Therefore, ( b = 6 - frac{1}{29} ). But that would make ( b ) a fraction, which seems unusual. Maybe I made a mistake earlier.Let me check the multiplication steps again.First equation multiplied by 7:( 3a times 7 = 21a )( 5b times 7 = 35b )( 47 times 7 = 329 ). That seems correct.Second equation multiplied by 3:( 7a times 3 = 21a )( 2b times 3 = 6b )( 52 times 3 = 156 ). That also seems correct.Subtracting the second equation from the first:( 21a + 35b - 21a - 6b = 329 - 156 )Which simplifies to ( 29b = 173 ). So, ( b = frac{173}{29} ).Calculating ( 29 times 6 = 174 ), so ( 173 = 29 times 6 - 1 ), which means ( b = 6 - frac{1}{29} ). That's approximately 5.9655, which is a bit messy. Maybe I should keep it as a fraction for now.So, ( b = frac{173}{29} ).Now, I need to find ( a ). Let's substitute ( b ) back into one of the original equations. I'll use the first equation:( 3a + 5b = 47 )Substituting ( b = frac{173}{29} ):( 3a + 5 times frac{173}{29} = 47 )Calculating ( 5 times frac{173}{29} = frac{865}{29} )So, ( 3a + frac{865}{29} = 47 )Subtracting ( frac{865}{29} ) from both sides:( 3a = 47 - frac{865}{29} )Convert 47 to a fraction with denominator 29:( 47 = frac{47 times 29}{29} = frac{1363}{29} )So, ( 3a = frac{1363}{29} - frac{865}{29} = frac{1363 - 865}{29} = frac{498}{29} )Therefore, ( a = frac{498}{29 times 3} = frac{498}{87} )Simplifying ( frac{498}{87} ):Divide numerator and denominator by 3:( frac{498 ÷ 3}{87 ÷ 3} = frac{166}{29} )So, ( a = frac{166}{29} )Now, to find ( a + b ):( a + b = frac{166}{29} + frac{173}{29} = frac{166 + 173}{29} = frac{339}{29} )Simplifying ( frac{339}{29} ):29 times 11 is 319, and 339 - 319 = 20, so ( frac{339}{29} = 11 + frac{20}{29} )But since the question asks for ( a + b ), and both ( a ) and ( b ) are fractions, I can leave it as ( frac{339}{29} ), but let me check if that reduces further. 339 divided by 29 is 11.689... which doesn't seem to simplify further.Wait, but earlier when I calculated ( b = frac{173}{29} ), I thought it was approximately 5.9655, and ( a = frac{166}{29} ) is approximately 5.7241. Adding them together gives approximately 11.689, which matches ( frac{339}{29} ).But let me double-check my calculations because fractions can be tricky.Starting again:From the two equations:1. ( 3a + 5b = 47 )2. ( 7a + 2b = 52 )I multiplied the first by 7 and the second by 3 to eliminate ( a ):1. ( 21a + 35b = 329 )2. ( 21a + 6b = 156 )Subtracting the second from the first:( 29b = 173 ) → ( b = frac{173}{29} )Then substituting back:( 3a + 5 times frac{173}{29} = 47 )Calculating ( 5 times 173 = 865 ), so ( frac{865}{29} )Then ( 3a = 47 - frac{865}{29} )Converting 47 to ( frac{1363}{29} ), so ( 3a = frac{1363 - 865}{29} = frac{498}{29} )Thus, ( a = frac{498}{87} = frac{166}{29} )Adding ( a + b = frac{166 + 173}{29} = frac{339}{29} )Yes, that seems correct. So, ( a + b = frac{339}{29} ), which can also be written as ( 11 frac{20}{29} ) if needed, but as an improper fraction, it's ( frac{339}{29} ).I think that's the correct answer. To be thorough, I can plug ( a ) and ( b ) back into the original equations to verify.First equation: ( 3a + 5b = 47 )( 3 times frac{166}{29} + 5 times frac{173}{29} = frac{498}{29} + frac{865}{29} = frac{1363}{29} = 47 ). Correct.Second equation: ( 7a + 2b = 52 )( 7 times frac{166}{29} + 2 times frac{173}{29} = frac{1162}{29} + frac{346}{29} = frac{1508}{29} = 52 ). Correct.So, both equations are satisfied, which means my values for ( a ) and ( b ) are correct, and thus ( a + b = frac{339}{29} ).

💡Okay, so I have this problem here: Let ( m ) be an odd integer with exactly 17 positive divisors. I need to find the number of positive divisors of ( 16m^3 ). Hmm, alright, let me try to work through this step by step.First, I remember that the number of positive divisors of a number can be found using its prime factorization. If a number ( n ) has a prime factorization of ( n = p_1^{a_1} p_2^{a_2} dots p_k^{a_k} ), then the number of positive divisors of ( n ) is ( (a_1 + 1)(a_2 + 1) dots (a_k + 1) ). So, that's the formula I need to use here.Given that ( m ) has exactly 17 positive divisors, I can set up the equation ( (a_1 + 1)(a_2 + 1) dots (a_k + 1) = 17 ). Now, 17 is a prime number, which means it can only be factored as 17 × 1. So, this tells me that ( m ) must be of the form ( p^{16} ), where ( p ) is a prime number. Because if I have ( (16 + 1) = 17 ), that's the only way to get 17 as the product of integers greater than 1.But wait, the problem also says that ( m ) is an odd integer. Since 2 is the only even prime, and ( m ) is odd, ( p ) must be an odd prime. So, ( m = q^{16} ) where ( q ) is an odd prime. That makes sense.Now, I need to find the number of positive divisors of ( 16m^3 ). Let's break this down. First, let's compute ( m^3 ). Since ( m = q^{16} ), then ( m^3 = (q^{16})^3 = q^{48} ). So, ( m^3 = q^{48} ).Next, let's compute ( 16m^3 ). Since 16 is ( 2^4 ), we can write ( 16m^3 = 2^4 times q^{48} ). So, the prime factorization of ( 16m^3 ) is ( 2^4 times q^{48} ).Now, to find the number of positive divisors, I can apply the formula I mentioned earlier. For each prime factor, take the exponent, add 1, and multiply them together. So, for the prime 2, the exponent is 4, so we have ( 4 + 1 = 5 ). For the prime ( q ), the exponent is 48, so we have ( 48 + 1 = 49 ).Therefore, the number of positive divisors of ( 16m^3 ) is ( 5 times 49 ). Let me compute that: 5 multiplied by 49 is 245. So, there are 245 positive divisors of ( 16m^3 ).Wait, let me double-check my steps to make sure I didn't make any mistakes. First, I determined that since ( m ) has 17 divisors, it must be of the form ( p^{16} ) because 17 is prime. Then, since ( m ) is odd, ( p ) must be an odd prime. Then, I calculated ( m^3 ) as ( q^{48} ), which seems correct. Then, I wrote ( 16m^3 ) as ( 2^4 times q^{48} ), which is accurate because 16 is ( 2^4 ). Finally, applying the number of divisors formula, I added 1 to each exponent and multiplied, getting 5 and 49, which multiply to 245. That seems right.I don't think I made any mistakes here. So, I feel confident that the number of positive divisors of ( 16m^3 ) is 245.

💡Okay, so Jane has these big empty cans that she's using as pots for her sunflowers. She's really into gardening, I guess. Anyway, she has 54 sunflower seeds, and there are 9 cans. She wants to put an equal number of seeds in each can. Hmm, okay, so I need to figure out how many seeds go into each can.Alright, let's break this down. She has 54 seeds and 9 cans. So, if she wants to distribute them equally, I think I need to divide the total number of seeds by the number of cans. That makes sense because if you have a certain number of items and you want to split them evenly into groups, division is the way to go.So, mathematically, it's 54 divided by 9. Let me write that out: 54 ÷ 9. Now, what's 54 divided by 9? I remember that 9 times 6 is 54 because 9 times 5 is 45, and then adding another 9 gets you to 54. So, 9 times 6 equals 54. That means 54 divided by 9 is 6.Wait, let me double-check that. If I multiply 9 by 6, do I get 54? Yes, because 9 times 6 is indeed 54. So, that checks out. Therefore, if Jane puts 6 seeds in each can, she'll use all 54 seeds equally across the 9 cans.I think that's it. She'll place 6 seeds in each can. It seems straightforward, but just to make sure I didn't make any mistakes, let me visualize it. Imagine each can is getting 6 seeds. So, the first can gets 6, the second can gets another 6, and so on until the ninth can. If I add up all the seeds in each can, it should total 54.Let's do a quick addition: 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6. That's nine 6s added together. 6 times 9 is 54, so that adds up correctly. Okay, that makes me feel confident that 6 is the right answer.I don't think there's anything more to it. It's a simple division problem, but it's good to go through the steps to ensure accuracy, especially when dealing with equal distribution. So, yeah, Jane will put 6 seeds in each can.

💡First, I need to understand what the problem is asking. It's asking for the result of multiplying ten thousand by ten thousand.I know that ten thousand can be written as 10,000. So, the problem becomes 10,000 multiplied by 10,000.To make this easier, I can express 10,000 in scientific notation. 10,000 is the same as (1 times 10^4).Now, I can rewrite the multiplication using scientific notation:[(1 times 10^4) times (1 times 10^4)]Next, I'll use the properties of exponents to multiply these two numbers. When multiplying numbers with the same base, I add their exponents:[1 times 10^{4+4} = 1 times 10^8]Now, I need to convert (1 times 10^8) back to standard form. (10^8) means 1 followed by 8 zeros, which is 100,000,000.Looking at the answer choices, 100,000,000 is called "one hundred million," which corresponds to option (D).So, the final answer is (D) one hundred million.

💡Alright, so I have this problem here: α is an angle in the second quadrant, and tan α is equal to -1/2. I need to find the value of cos α. Hmm, okay, let's break this down step by step.First, I remember that the tangent of an angle is equal to the sine of the angle divided by the cosine of the angle. So, tan α = sin α / cos α. In this case, tan α is -1/2, which means sin α divided by cos α equals -1/2. So, I can write that as:tan α = sin α / cos α = -1/2From this equation, I can express sin α in terms of cos α. Let's rearrange it:sin α = (-1/2) * cos αOkay, so sin α is equal to negative half of cos α. Now, since α is in the second quadrant, I recall that in the second quadrant, sine is positive and cosine is negative. Wait, that seems conflicting because if sin α is positive and cos α is negative, then tan α, which is sin α over cos α, should be negative, which matches the given tan α = -1/2. So, that makes sense.But wait, in my earlier step, I wrote sin α = (-1/2) * cos α. If cos α is negative, then sin α would be positive because negative times negative is positive. That aligns with the second quadrant where sine is positive and cosine is negative. So, that seems consistent.Now, I need to find cos α. I know that sin² α + cos² α = 1. That's the Pythagorean identity. So, maybe I can use that. Let me substitute sin α from the earlier equation into this identity.So, sin α = (-1/2) * cos α. Therefore, sin² α = [(-1/2) * cos α]^2. Let's compute that:sin² α = (1/4) * cos² αBecause when you square a negative, it becomes positive, and (1/2)^2 is 1/4.Now, substitute sin² α back into the Pythagorean identity:sin² α + cos² α = 1(1/4) * cos² α + cos² α = 1Let me combine these terms. To add them, they need a common denominator. The first term is (1/4) cos² α, and the second term is cos² α, which is the same as (4/4) cos² α. So, adding them together:(1/4) cos² α + (4/4) cos² α = (5/4) cos² αSo, (5/4) cos² α = 1Now, to solve for cos² α, I can multiply both sides by 4/5:cos² α = (4/5) * 1cos² α = 4/5Now, to find cos α, I take the square root of both sides:cos α = ±√(4/5) = ±2/√5But, since α is in the second quadrant, where cosine is negative, we take the negative value:cos α = -2/√5Hmm, but usually, we rationalize the denominator. So, multiplying numerator and denominator by √5:cos α = -2√5 / 5So, that should be the value of cos α.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Started with tan α = -1/2, which is sin α / cos α = -1/2.2. Expressed sin α as (-1/2) cos α.3. Used the Pythagorean identity sin² α + cos² α = 1.4. Substituted sin² α with (1/4) cos² α.5. Combined terms to get (5/4) cos² α = 1.6. Solved for cos² α = 4/5.7. Took the square root to get cos α = ±2/√5.8. Since α is in the second quadrant, cos α is negative, so cos α = -2/√5, which rationalizes to -2√5 / 5.Everything seems to check out. I don't see any errors in the calculations or reasoning.Just to visualize, in the second quadrant, the reference angle would have tan equal to 1/2, but since it's in the second quadrant, the actual angle has a negative tangent. The sides of the triangle would be opposite = 1, adjacent = -2 (since cosine is negative), and hypotenuse would be √(1² + (-2)²) = √5. Therefore, cosine is adjacent over hypotenuse, which is -2/√5, or -2√5/5 after rationalizing.Yeah, that makes sense. So, I think I've got the right answer here.