💡Okay, so Jake has 3 snakes, and each of them lays 2 eggs. That means there are a total of 3 times 2, which is 6 baby snakes. Now, most of these baby snakes sell for 250 each, but there's one super rare one that sells for 4 times as much. First, I need to figure out how many snakes are selling at the standard price. Since there are 6 baby snakes in total and one is super rare, that leaves 5 snakes that sell for 250 each. So, the revenue from the 5 standard snakes would be 5 multiplied by 250. Let's do that calculation: 5 times 250 is 1250. So, Jake makes 1250 from the standard snakes.Next, the super rare snake sells for 4 times the standard price. The standard price is 250, so 4 times that would be 4 times 250, which is 1000. So, the super rare snake sells for 1000.Now, to find the total revenue Jake gets from selling all the snakes, I need to add the revenue from the standard snakes and the super rare snake together. That would be 1250 plus 1000, which equals 2250.So, Jake makes a total of 2250 from selling all the baby snakes.

💡First, I recognize that the combination formula ( C_{n}^{m} ) represents the number of ways to choose ( m ) elements from a set of ( n ) elements without considering the order.I recall the identity that states:[C_{n}^{m} + C_{n}^{m-1} = C_{n+1}^{m}]This identity allows me to combine the two combinations into a single expression.Applying this identity to the given problem:[C_{10}^{9} + C_{10}^{8} = C_{11}^{9}]Next, I use the symmetry property of combinations:[C_{n}^{m} = C_{n}^{n-m}]Applying this to ( C_{11}^{9} ):[C_{11}^{9} = C_{11}^{2}]Now, I calculate ( C_{11}^{2} ) using the combination formula:[C_{11}^{2} = frac{11 times 10}{2!} = frac{110}{2} = 55]Therefore, the sum of ( C_{10}^{9} + C_{10}^{8} ) is 55.

💡Alright, so I've got these two math problems here, and I need to figure them out step by step. Let me start with the first one.**Problem 1:** The sum of a number and (-frac{2}{11}) equals (-frac{1}{3}). What is the number?Okay, so I need to find a number that when added to (-frac{2}{11}) gives me (-frac{1}{3}). Let's denote this unknown number as (x). So, the equation would be:[ x + left(-frac{2}{11}right) = -frac{1}{3} ]Simplifying that, it's:[ x - frac{2}{11} = -frac{1}{3} ]Now, to find (x), I need to isolate it on one side of the equation. That means I should get rid of the (-frac{2}{11}) that's being subtracted from (x). To do that, I can add (frac{2}{11}) to both sides of the equation:[ x - frac{2}{11} + frac{2}{11} = -frac{1}{3} + frac{2}{11} ]On the left side, (-frac{2}{11} + frac{2}{11}) cancels out, leaving just (x). So now I have:[ x = -frac{1}{3} + frac{2}{11} ]Alright, now I need to add these two fractions. To add fractions, they need a common denominator. The denominators here are 3 and 11. The least common multiple of 3 and 11 is 33, so I'll convert both fractions to have 33 as the denominator.For (-frac{1}{3}):[ -frac{1}{3} = -frac{1 times 11}{3 times 11} = -frac{11}{33} ]And for (frac{2}{11}):[ frac{2}{11} = frac{2 times 3}{11 times 3} = frac{6}{33} ]Now, adding them together:[ -frac{11}{33} + frac{6}{33} = frac{-11 + 6}{33} = frac{-5}{33} ]So, (x = -frac{5}{33}). That seems straightforward.**Problem 2:** What is the difference when subtracting the sum of (-frac{1}{3}) and (frac{1}{2}) from (-2)?Alright, let's break this down. First, I need to find the sum of (-frac{1}{3}) and (frac{1}{2}), and then subtract that sum from (-2).Let's start with the sum:[ -frac{1}{3} + frac{1}{2} ]Again, to add these fractions, I need a common denominator. The denominators are 3 and 2, so the least common multiple is 6.Converting both fractions:For (-frac{1}{3}):[ -frac{1}{3} = -frac{2}{6} ]And for (frac{1}{2}):[ frac{1}{2} = frac{3}{6} ]Now, adding them together:[ -frac{2}{6} + frac{3}{6} = frac{-2 + 3}{6} = frac{1}{6} ]So, the sum of (-frac{1}{3}) and (frac{1}{2}) is (frac{1}{6}).Now, I need to subtract this sum from (-2):[ -2 - frac{1}{6} ]To subtract these, it's helpful to express (-2) as a fraction with the same denominator as (frac{1}{6}). So, (-2) can be written as (-frac{12}{6}).Now, subtracting:[ -frac{12}{6} - frac{1}{6} = frac{-12 - 1}{6} = frac{-13}{6} ]So, the difference is (-frac{13}{6}).Wait, let me double-check that. If I have (-2) and I subtract (frac{1}{6}), that should be more negative, right? Because subtracting a positive number from a negative number makes it more negative. So, (-2) minus (frac{1}{6}) is indeed (-2frac{1}{6}), which is the same as (-frac{13}{6}). Yeah, that makes sense.So, summarizing:1. The unknown number is (-frac{5}{33}).2. The difference is (-frac{13}{6}).I think that's it. I don't see any mistakes in my calculations, but let me just go through them again quickly.For the first problem:[ x - frac{2}{11} = -frac{1}{3} ][ x = -frac{1}{3} + frac{2}{11} ]Convert to common denominator 33:[ -frac{11}{33} + frac{6}{33} = -frac{5}{33} ]Looks good.For the second problem:Sum of (-frac{1}{3}) and (frac{1}{2}):[ -frac{2}{6} + frac{3}{6} = frac{1}{6} ]Subtract from (-2):[ -2 - frac{1}{6} = -frac{12}{6} - frac{1}{6} = -frac{13}{6} ]Yep, that's correct.Alright, I'm confident with these answers.

💡To rewrite the expression ( y^2 + 10y + 33 ) as a combination of a square of a binomial and an integer, I'll start by completing the square. First, I'll focus on the quadratic and linear terms: ( y^2 + 10y ). To complete the square, I need to find a value that, when added and subtracted, forms a perfect square trinomial. The formula for completing the square involves taking half of the coefficient of ( y ), which is ( 10 ), resulting in ( 5 ). Squaring this value gives ( 25 ). So, I can rewrite the expression as:[y^2 + 10y + 25 - 25 + 33]This simplifies to:[(y + 5)^2 + 8]Here, ( (y + 5)^2 ) is the square of the binomial, and ( 8 ) is the integer that completes the expression.

💡First, I need to determine the birth rate per certain number of people, given the death rate, net growth rate, and initial population.I'll start by calculating the net population increase using the net growth rate and the initial population. This will help me understand how many additional people are added to the population.Next, I'll use the relationship between births, deaths, and net growth to set up an equation. The net growth is the difference between the number of births and the number of deaths.With the known death rate, I can substitute the values into the equation to solve for the birth rate.Finally, by performing the necessary calculations, I'll find the birth rate per certain number of people.

💡Okay, so I have this problem here: "The coordinates of the fixed point through which the line ( mx - y + 2m + 1 = 0 ) passes are (　　)" with options A: (-2, 1), B: (2, 1), C: (1, -2), D: (1, 2). I need to figure out which one is the correct answer.First, let me understand what the question is asking. It says that there's a line given by the equation ( mx - y + 2m + 1 = 0 ), and this line passes through a fixed point regardless of the value of ( m ). So, no matter what ( m ) is, the line will always go through this specific point. My task is to find the coordinates of that fixed point.Hmm, okay. So, I think this is about finding a point that satisfies the equation for all values of ( m ). That means, if I plug in the coordinates of this fixed point into the equation, it should hold true for any ( m ).Let me write down the equation again: ( mx - y + 2m + 1 = 0 ). Maybe I can rearrange this equation to make it easier to analyze. Let me try grouping the terms with ( m ) together.So, ( mx + 2m - y + 1 = 0 ). I can factor out the ( m ) from the first two terms: ( m(x + 2) - y + 1 = 0 ).Now, this equation is of the form ( m(x + 2) + (-y + 1) = 0 ). Since this equation must hold true for all values of ( m ), the coefficients of ( m ) and the constant term must both be zero. Otherwise, the equation would only hold for specific values of ( m ), not all.So, setting the coefficient of ( m ) to zero: ( x + 2 = 0 ). And setting the constant term to zero: ( -y + 1 = 0 ).Let me solve these equations one by one. Starting with ( x + 2 = 0 ), subtracting 2 from both sides gives ( x = -2 ).Next, solving ( -y + 1 = 0 ). Subtracting 1 from both sides gives ( -y = -1 ), and multiplying both sides by -1 gives ( y = 1 ).So, the fixed point is at ( (-2, 1) ). Let me check the options given: A is (-2, 1), which matches what I found. So, I think A is the correct answer.But wait, just to make sure I didn't make any mistakes, let me plug this point back into the original equation and see if it holds for any ( m ).Plugging ( x = -2 ) and ( y = 1 ) into ( mx - y + 2m + 1 = 0 ):( m(-2) - 1 + 2m + 1 = 0 ).Simplify this: ( -2m - 1 + 2m + 1 = 0 ).Combine like terms: ( (-2m + 2m) + (-1 + 1) = 0 ).Which simplifies to ( 0 + 0 = 0 ), which is true. So, yes, this point satisfies the equation for any ( m ).Just to double-check, maybe I should test another point from the options to ensure that it doesn't also satisfy the equation for all ( m ). Let's try option B: (2, 1).Plugging ( x = 2 ) and ( y = 1 ) into the equation:( m(2) - 1 + 2m + 1 = 0 ).Simplify: ( 2m - 1 + 2m + 1 = 0 ).Combine like terms: ( 4m + 0 = 0 ).So, ( 4m = 0 ) implies ( m = 0 ). But the equation should hold for all ( m ), not just ( m = 0 ). Therefore, option B is not a fixed point for all ( m ).Similarly, let's test option C: (1, -2).Plugging ( x = 1 ) and ( y = -2 ):( m(1) - (-2) + 2m + 1 = 0 ).Simplify: ( m + 2 + 2m + 1 = 0 ).Combine like terms: ( 3m + 3 = 0 ).Which gives ( 3m = -3 ) so ( m = -1 ). Again, this only holds for ( m = -1 ), not all ( m ). So, option C is not the fixed point.Lastly, option D: (1, 2).Plugging ( x = 1 ) and ( y = 2 ):( m(1) - 2 + 2m + 1 = 0 ).Simplify: ( m - 2 + 2m + 1 = 0 ).Combine like terms: ( 3m - 1 = 0 ).Which gives ( 3m = 1 ) so ( m = 1/3 ). Again, only holds for ( m = 1/3 ), not all ( m ). So, D is not the fixed point.Therefore, after testing all options, only option A: (-2, 1) satisfies the equation for all values of ( m ). So, I'm confident that A is the correct answer.I think the key idea here was to recognize that for the equation to hold for all ( m ), the coefficients of ( m ) and the constant term must each be zero. By setting them to zero, I could solve for ( x ) and ( y ) to find the fixed point. It's a common technique in problems involving parameters like ( m ) where you want something to hold true regardless of the parameter's value.Another way to think about it is that the line ( mx - y + 2m + 1 = 0 ) can be rewritten as ( m(x + 2) + (-y + 1) = 0 ). Since ( m ) can be any real number, the only way this equation can be true for all ( m ) is if both ( x + 2 = 0 ) and ( -y + 1 = 0 ). This ensures that the equation doesn't depend on ( m ) anymore and is just ( 0 = 0 ), which is always true.So, yeah, that makes sense. I think I've got it. The fixed point is (-2, 1), which is option A.**Final Answer**boxed{A}