answer：First, calculate the unusable seaweed:10% of 600 pounds = 0.10 * 600 = 60 poundsSubtract the unusable portion to find the usable seaweed:600 pounds - 60 pounds = 540 poundsNext, determine the amount of seaweed for each category before weight loss:For starting fires: 40% of 540 pounds = 0.40 * 540 = 216 poundsFor medicinal purposes: 20% of 540 pounds = 0.20 * 540 = 108 poundsFor human consumption and livestock feed: 40% of 540 pounds = 0.40 * 540 = 216 poundsDivide the 216 pounds into human consumption and livestock feed:For human consumption: 30% of 216 pounds = 0.30 * 216 = 64.8 poundsFor livestock feed: 70% of 216 pounds = 0.70 * 216 = 151.2 poundsCalculate the weight loss during preparation for livestock feed:5% weight loss of 151.2 pounds = 0.05 * 151.2 = 7.56 poundsSubtract the weight loss to find the final amount:151.2 pounds - 7.56 pounds = 143.64 pounds

answer：To determine the greatest integer ( x ) for which ( frac{8}{11} > frac{x}{17} ), I will start by eliminating the fractions to simplify the inequality.First, I multiply both sides of the inequality by 17:[17 times frac{8}{11} > x]This simplifies to:[frac{136}{11} > x]Calculating the division:[12.36 > x]Since ( x ) must be an integer, the largest integer less than 12.36 is 12. Therefore, the greatest integer ( x ) that satisfies the inequality is 12.

answer：Okay, so I need to figure out how to divide a square into smaller squares for any number n greater than 5, and also show that it's impossible to do this for n equals 2 and 3. Hmm, let's start by understanding what it means to divide a square into smaller squares. Each smaller square must fit perfectly within the larger square without overlapping and without leaving any gaps. First, let's think about the simplest cases. If n equals 1, it's just the square itself, so that's trivial. For n equals 4, I can easily divide the square into four equal smaller squares by drawing lines through the midpoints of the sides. Similarly, for n equals 9, I can divide it into a 3x3 grid of smaller squares. But the problem is asking about any n greater than 5, which includes numbers like 6, 7, 8, and so on. I remember hearing about something called the "squaring the square" problem, where the goal is to tile a square with smaller squares, none of which overlap and all of which are different sizes. But in this case, the smaller squares don't have to be different sizes, just smaller than the original. That might make it easier.Maybe I can use induction here. If I can show that it's possible for some base cases and then show that if it's possible for n, it's also possible for n+3, then I can cover all numbers greater than 5. Let's try that.First, let's establish the base cases. I know that n equals 4 is possible, as I mentioned before. What about n equals 6? If I take the original square and divide it into four equal smaller squares, then I can further divide one of those smaller squares into four even smaller squares. That would give me a total of 4 + 3 = 7 squares. Wait, that's n equals 7. Hmm, maybe I need a different approach.Alternatively, maybe I can divide the original square into a 2x2 grid, giving me four squares, and then further divide one of those into a 3x3 grid, which would add five more squares, making a total of nine squares. But that's n equals 9. I need to get to n equals 6, 7, and 8 as base cases.Let me try to visualize dividing the square into six smaller squares. If I divide the original square into a 2x2 grid, that's four squares. Then, if I divide one of those into four smaller squares, that would give me seven squares. But I need six. Maybe I can divide one of the smaller squares into three smaller squares instead of four. But can I divide a square into three smaller squares? I don't think that's possible without overlapping or leaving gaps.Wait, maybe I need to use a different initial division. Instead of dividing into four equal squares, maybe I can divide the original square into unequal smaller squares. For example, if I divide the original square into one large square and five smaller squares around it. But I'm not sure if that would work without overlapping or leaving gaps.I think I need to look up some examples or diagrams to get a better idea. Maybe there's a standard way to divide a square into six, seven, or eight smaller squares. Once I have those base cases, I can use induction to show that for any n greater than 5, it's possible.Now, moving on to the second part of the problem: showing that it's impossible to divide a square into two or three smaller squares. For n equals 2, if I try to divide the original square into two smaller squares, each smaller square would have to occupy half of the original square. But a square can't be divided into two smaller squares without overlapping or leaving gaps because the sides wouldn't align properly. Similarly, for n equals 3, trying to divide the original square into three smaller squares would require each smaller square to occupy a third of the original square. Again, this isn't possible because the sides wouldn't align properly, and there would either be overlaps or gaps. I think the key here is that dividing a square into smaller squares requires that the sides of the smaller squares align perfectly with the sides of the original square and with each other. For n equals 2 and 3, this alignment isn't possible without violating the conditions of the problem.Overall, I think the strategy is to establish base cases for n equals 6, 7, and 8, and then use induction to show that for any n greater than 5, it's possible to divide a square into n smaller squares. For n equals 2 and 3, the division isn't possible due to geometric constraints.

answer：To solve the problem, we need to determine the relationship between the coordinates of the two points given that they are symmetric with respect to the origin. 1. **Understanding Symmetry with Respect to the Origin**: - If a point ((x, y)) is symmetric to another point with respect to the origin, the second point must be ((-x, -y)).2. **Setting Up Equations**: - Given the points ((a + 2, 2)) and ((4, -b)), we can write the equations based on their symmetry: [ a + 2 = -4 quad text{and} quad 2 = -(-b) ]3. **Solving for (a) and (b)**: - From the first equation: [ a + 2 = -4 implies a = -6 ] - From the second equation: [ 2 = b implies b = 2 ]4. **Calculating the Product (ab)**: - Multiply the values of (a) and (b): [ ab = (-6) times 2 = -12 ]Therefore, the value of (ab) is (-12).

answer：First, I recognize that we need to find the value of (cos 2alpha) given that (cos alpha = dfrac{2sqrt{5}}{5}). To do this, I'll use the double angle identity for cosine, which is:[cos 2alpha = 2cos^2alpha - 1]Next, I'll substitute the given value of (cos alpha) into the identity:[cos 2alpha = 2left(dfrac{2sqrt{5}}{5}right)^2 - 1]Now, I'll calculate the square of (dfrac{2sqrt{5}}{5}):[left(dfrac{2sqrt{5}}{5}right)^2 = dfrac{4 cdot 5}{25} = dfrac{20}{25} = dfrac{4}{5}]Substituting this back into the equation for (cos 2alpha):[cos 2alpha = 2left(dfrac{4}{5}right) - 1 = dfrac{8}{5} - 1 = dfrac{8}{5} - dfrac{5}{5} = dfrac{3}{5}]Therefore, the value of (cos 2alpha) is (dfrac{3}{5}).

answer：First, I need to identify the missing number in the equation 10010 - 12 * some number * 2 = 9938. Let's call the missing number "x".Next, I'll simplify the equation by combining the constants that multiply x. This gives me 10010 - 24x = 9938.To isolate x, I'll subtract 10010 from both sides of the equation, resulting in -24x = -72.Finally, I'll divide both sides by -24 to solve for x, which gives me x = 3.