💡First, I need to determine the amount of fat in each type of fish. A herring has 40 ounces of fat, an eel has 20 ounces, and a pike has 10 ounces more than an eel, which means 30 ounces of fat.Next, I'll calculate the total fat contributed by each type of fish when Ellianna cooks and serves 40 fish of each type. For the herring, multiplying 40 fish by 40 ounces per fish gives 1600 ounces. For the eel, 40 fish multiplied by 20 ounces per fish equals 800 ounces. For the pike, 40 fish multiplied by 30 ounces per fish results in 1200 ounces.Finally, I'll add up the total fat from all three types of fish: 1600 ounces from herring, 800 ounces from eel, and 1200 ounces from pike. This sum equals 3600 ounces of fat in total.

💡First, I need to simplify the expression ((a + b)^2 - a(2b + a)). I'll start by expanding each part separately.Expanding ((a + b)^2) gives (a^2 + 2ab + b^2). Then, distributing (a) in the second term (a(2b + a)) results in (2ab + a^2).Now, I'll subtract the second expanded term from the first:[a^2 + 2ab + b^2 - 2ab - a^2]Looking at the terms, I notice that (a^2) and (-a^2) cancel each other out, as do (2ab) and (-2ab). This leaves me with just (b^2).Next, I'll substitute the given value of (b = -frac{1}{3}) into the simplified expression:[b^2 = left(-frac{1}{3}right)^2 = frac{1}{9}]So, the final evaluated value of the expression is (frac{1}{9}).

💡Okay, so I've got this problem about a line l with the equation (2 + m)x + (1 - 2m)y + 4 - 3m = 0. There are two parts: part (I) asks me to prove that no matter what real number m is, the line l always passes through a fixed point. Part (II) is about drawing another line l₁ through a specific point M(-1, -2) such that the segment between the two coordinate axes enclosed by l₁ is bisected by point M. I need to find the equation of line l₁.Starting with part (I). Hmm, so I need to show that regardless of the value of m, the line passes through a particular point. That means there's a point (x, y) that satisfies the equation for any m. So, if I can find such an (x, y) that works for all m, that would be the fixed point.Let me write the equation again: (2 + m)x + (1 - 2m)y + 4 - 3m = 0.I can rearrange this equation to group the terms with m together. Let's see:(2x + y + 4) + m(x - 2y - 3) = 0.So, this equation is of the form A + mB = 0, where A = 2x + y + 4 and B = x - 2y - 3.Since this equation must hold for all values of m, the coefficients of m and the constant term must both be zero. Otherwise, if m changes, the equation wouldn't hold anymore. So, for the equation to be true for all m, both A and B must be zero.Therefore, I can set up the following system of equations:1) 2x + y + 4 = 02) x - 2y - 3 = 0Now, I need to solve this system to find the fixed point (x, y).Let me solve equation 2 for x:x = 2y + 3.Now, substitute this into equation 1:2*(2y + 3) + y + 4 = 04y + 6 + y + 4 = 05y + 10 = 05y = -10y = -2.Now, plug y = -2 back into equation 2:x - 2*(-2) - 3 = 0x + 4 - 3 = 0x + 1 = 0x = -1.So, the fixed point is (-1, -2). That makes sense. So, no matter what m is, the line l will always pass through (-1, -2). Cool, that seems straightforward.Moving on to part (II). I need to find the equation of line l₁ that passes through M(-1, -2) and such that the segment between the two coordinate axes is bisected by M. So, M is the midpoint of the segment connecting the x-intercept and y-intercept of l₁.Let me recall that the equation of a line can be written in slope-intercept form as y = kx + b, but since it passes through M(-1, -2), I can write it in point-slope form. Alternatively, since I know the midpoint, maybe I can use the intercept form.Wait, let's think about intercepts. If the line intersects the x-axis at point A(a, 0) and the y-axis at point B(0, b), then the midpoint of AB is ((a/2), (b/2)). But in this case, the midpoint is given as M(-1, -2). So, that means:(a/2) = -1 => a = -2(b/2) = -2 => b = -4.So, the x-intercept is (-2, 0) and the y-intercept is (0, -4). Therefore, the line passes through (-2, 0) and (0, -4). So, I can find the equation of the line using these two points.First, let's find the slope (k):k = (y₂ - y₁)/(x₂ - x₁) = (-4 - 0)/(0 - (-2)) = (-4)/2 = -2.So, the slope is -2. Now, using point-slope form with point M(-1, -2):y - (-2) = -2(x - (-1))y + 2 = -2(x + 1)y + 2 = -2x - 2y = -2x - 4.Alternatively, using the intercepts, the equation can be written as:x/(-2) + y/(-4) = 1Which simplifies to:(-x/2) + (-y/4) = 1Multiply both sides by 4 to eliminate denominators:-2x - y = 4Which can be rewritten as:2x + y + 4 = 0.Wait, that's the same as the equation I got earlier. So, the equation of line l₁ is 2x + y + 4 = 0.Let me double-check. If I plug in M(-1, -2):2*(-1) + (-2) + 4 = -2 -2 +4 = 0. Yes, that works.Also, checking the intercepts:For x-intercept, set y=0: 2x + 0 +4=0 => 2x = -4 => x = -2. Correct.For y-intercept, set x=0: 0 + y +4=0 => y = -4. Correct.So, the midpoint between (-2,0) and (0,-4) is indeed (-1, -2), which is M. So, that checks out.Alternatively, another way to approach part (II) is to let the equation of l₁ be y = kx + c, passing through M(-1, -2). So, substituting:-2 = k*(-1) + c => c = k - 2.So, the equation is y = kx + (k - 2).Now, find the x-intercept by setting y=0:0 = kx + (k - 2) => kx = 2 - k => x = (2 - k)/k.Similarly, the y-intercept is when x=0: y = 0 + (k - 2) => y = k - 2.So, the intercepts are ((2 - k)/k, 0) and (0, k - 2).The midpoint of these two points is:x-coordinate: [(2 - k)/k + 0]/2 = (2 - k)/(2k)y-coordinate: [0 + (k - 2)]/2 = (k - 2)/2But we know the midpoint is (-1, -2). So:(2 - k)/(2k) = -1and(k - 2)/2 = -2Let's solve the second equation first:(k - 2)/2 = -2Multiply both sides by 2:k - 2 = -4k = -2.Now, plug k = -2 into the first equation:(2 - (-2))/(2*(-2)) = (4)/(-4) = -1. Which matches the x-coordinate of the midpoint. So, that's consistent.Therefore, the equation is y = -2x + (-2 - 2) => y = -2x -4, which is the same as 2x + y + 4 = 0.So, both methods lead to the same result. That gives me more confidence that the equation is correct.In summary, for part (I), the fixed point is (-1, -2), and for part (II), the equation of line l₁ is 2x + y + 4 = 0.**Final Answer**Ⅰ. The fixed point is boxed{(-1, -2)}.Ⅱ. The equation of line ( l_1 ) is boxed{2x + y + 4 = 0}.

💡Okay, so I have these three trigonometry problems here, all about finding the value of 2 sin α + cos α given different conditions. Let me try to work through each one step by step. I'll start with the first problem.**Problem 1:**Given that the terminal side of angle α passes through point P(4, -3), find the value of 2 sin α + cos α.Alright, so I remember that in trigonometry, when we have a point on the terminal side of an angle, we can use that point to find the sine and cosine of that angle. The point given is (4, -3), which is in the fourth quadrant because the x-coordinate is positive and the y-coordinate is negative.First, I need to find the radius, which is the distance from the origin to the point P. That's calculated using the Pythagorean theorem. So, the radius r is sqrt(x² + y²). Plugging in the values:r = sqrt(4² + (-3)²) = sqrt(16 + 9) = sqrt(25) = 5.Okay, so the radius is 5. Now, sine of α is y/r, and cosine of α is x/r. So,sin α = y/r = -3/5,cos α = x/r = 4/5.Now, the expression we need to find is 2 sin α + cos α. Let's plug in the values:2 sin α + cos α = 2*(-3/5) + (4/5) = (-6/5) + (4/5) = (-6 + 4)/5 = (-2)/5.So, the value is -2/5. That seems straightforward. Let me just double-check my calculations to make sure I didn't make a mistake. The radius is definitely 5, and the sine and cosine values are correct based on the coordinates. Multiplying and adding them gives -2/5. Yep, that looks right.**Problem 2:**Given that the terminal side of angle α passes through point P(4a, -3a) (a ≠ 0), find the value of 2 sin α + cos α.Hmm, this is similar to the first problem, but now the coordinates are given in terms of a parameter 'a'. So, the point is (4a, -3a). Since a ≠ 0, we don't have to worry about the origin. First, let's find the radius again. Using the Pythagorean theorem:r = sqrt((4a)² + (-3a)²) = sqrt(16a² + 9a²) = sqrt(25a²) = 5|a|.Wait, so the radius is 5|a|. That means depending on whether 'a' is positive or negative, the radius will be positive or negative? No, actually, the radius is always positive because it's a distance. So, regardless of the sign of 'a', r is 5|a|.But when calculating sine and cosine, the signs will depend on the quadrant where the point lies. Since the coordinates are (4a, -3a), the x-coordinate is 4a and the y-coordinate is -3a. So, the sign of x and y depends on the sign of 'a'.Let me consider two cases:1. When a > 0: - x = 4a is positive, - y = -3a is negative. So, the point is in the fourth quadrant.2. When a < 0: - x = 4a is negative, - y = -3a is positive. So, the point is in the second quadrant.Therefore, depending on the sign of 'a', the angle α is in different quadrants, which affects the signs of sine and cosine.Let's compute sine and cosine for both cases.**Case 1: a > 0**- sin α = y/r = (-3a)/(5|a|). Since a > 0, |a| = a. So, sin α = (-3a)/(5a) = -3/5.- cos α = x/r = (4a)/(5a) = 4/5.So, 2 sin α + cos α = 2*(-3/5) + 4/5 = (-6/5) + (4/5) = (-2)/5.**Case 2: a < 0**- sin α = y/r = (-3a)/(5|a|). Since a < 0, |a| = -a. So, sin α = (-3a)/(5*(-a)) = (-3a)/(-5a) = 3/5.- cos α = x/r = (4a)/(5|a|) = (4a)/(5*(-a)) = -4/5.So, 2 sin α + cos α = 2*(3/5) + (-4/5) = (6/5) - (4/5) = 2/5.Therefore, depending on whether 'a' is positive or negative, the value of 2 sin α + cos α is either -2/5 or 2/5. So, the answer is ±2/5.Wait, but the question just says "find the value", not specifying 'a'. So, since 'a' can be positive or negative, the expression can take both values. So, the answer is ±2/5. That makes sense.Let me just verify if I considered all cases correctly. If a is positive, the point is in the fourth quadrant, giving negative sine and positive cosine. If a is negative, the point is in the second quadrant, giving positive sine and negative cosine. Plugging into the expression, we get the two different results. Yep, that seems correct.**Problem 3:**Given that the ratio of the distance from a point P on the terminal side of angle α to the x-axis and to the y-axis is 3:4, find the value of 2 sin α + cos α.Okay, this one is a bit different. It's talking about the ratio of distances from point P to the x-axis and y-axis. So, the distance from P to the x-axis is the absolute value of the y-coordinate, and the distance from P to the y-axis is the absolute value of the x-coordinate.So, if the ratio is 3:4, that means |y| / |x| = 3/4. So, |y| = (3/4)|x|. Therefore, the coordinates of P can be represented as (±4k, ±3k) where k ≠ 0. The signs depend on the quadrant where the point lies.So, point P can be in any of the four quadrants, depending on the signs of x and y. Therefore, we need to consider each quadrant and find the corresponding value of 2 sin α + cos α.Let me denote the coordinates as (x, y) = (±4k, ±3k). Then, the radius r is sqrt(x² + y²) = sqrt((4k)² + (3k)²) = sqrt(16k² + 9k²) = sqrt(25k²) = 5|k|. Since k ≠ 0, r = 5|k|.Now, let's compute sin α and cos α for each quadrant.**First Quadrant:**x = 4k, y = 3k, k > 0- sin α = y/r = 3k / 5k = 3/5- cos α = x/r = 4k / 5k = 4/5- 2 sin α + cos α = 2*(3/5) + 4/5 = 6/5 + 4/5 = 10/5 = 2**Second Quadrant:**x = -4k, y = 3k, k > 0- sin α = y/r = 3k / 5k = 3/5- cos α = x/r = -4k / 5k = -4/5- 2 sin α + cos α = 2*(3/5) + (-4/5) = 6/5 - 4/5 = 2/5**Third Quadrant:**x = -4k, y = -3k, k > 0- sin α = y/r = -3k / 5k = -3/5- cos α = x/r = -4k / 5k = -4/5- 2 sin α + cos α = 2*(-3/5) + (-4/5) = -6/5 - 4/5 = -10/5 = -2**Fourth Quadrant:**x = 4k, y = -3k, k > 0- sin α = y/r = -3k / 5k = -3/5- cos α = x/r = 4k / 5k = 4/5- 2 sin α + cos α = 2*(-3/5) + 4/5 = -6/5 + 4/5 = -2/5So, depending on the quadrant, the value of 2 sin α + cos α can be 2, 2/5, -2, or -2/5.Wait, let me double-check each case.First Quadrant: Positive x and y, so both sine and cosine are positive. Calculations seem correct.Second Quadrant: Negative x, positive y. Sine is positive, cosine is negative. The expression gives 2*(3/5) - 4/5 = 6/5 - 4/5 = 2/5. Correct.Third Quadrant: Negative x and y. Both sine and cosine are negative. 2*(-3/5) + (-4/5) = -6/5 -4/5 = -10/5 = -2. Correct.Fourth Quadrant: Positive x, negative y. Sine is negative, cosine is positive. 2*(-3/5) + 4/5 = -6/5 + 4/5 = -2/5. Correct.So, all four cases are covered, and the possible values are 2, 2/5, -2, and -2/5.But the question says "find the value of 2 sin α + cos α." It doesn't specify the quadrant, so we need to consider all possibilities. So, the answer is that it can be 2, 2/5, -2, or -2/5 depending on the quadrant where the terminal side lies.Wait, but in the problem statement, it just says "the ratio of the distance from a point P on the terminal side of angle α to the x-axis and to the y-axis is 3:4." So, it doesn't specify the quadrant, so we have to consider all four quadrants. Therefore, the expression can take four different values.Is there a way to express this without listing all four? Or is it acceptable to list all four possible values? Since the problem doesn't specify, I think it's appropriate to provide all possible values.Alternatively, if the problem expects a single answer, maybe it's expecting the absolute value or something, but I don't think so because the expression can be positive or negative depending on the quadrant.So, to be thorough, I should present all four possible values.Wait, but in the first problem, the point was given in the fourth quadrant, so the value was -2/5. In the second problem, depending on 'a', it was ±2/5. In the third problem, depending on the quadrant, it can be 2, 2/5, -2, or -2/5.Therefore, in the third problem, the answer is four possible values.I think that's correct.Let me just recap:1. First problem: specific point, specific quadrant, specific value: -2/5.2. Second problem: point with parameter 'a', two possible quadrants, two possible values: ±2/5.3. Third problem: ratio given, four possible quadrants, four possible values: 2, 2/5, -2, -2/5.Yes, that all makes sense.I think I've worked through each problem carefully, considered all possible cases, and double-checked my calculations. I feel confident about these answers.**Final Answer**1. boxed{-dfrac{2}{5}}2. boxed{pm dfrac{2}{5}}3. boxed{2}, boxed{dfrac{2}{5}}, boxed{-2}, and boxed{-dfrac{2}{5}}

💡Okay, so I have this problem about triangle ABC where AB is 4 units long, angle B is π/3 radians, and angle A is between π/6 and π/2 radians. I need to find the range of the dot product of vectors AB and AC. Hmm, okay, let me try to figure this out step by step.First, I remember that the dot product of two vectors can be calculated using the formula: [overrightarrow{AB} cdot overrightarrow{AC} = |AB||AC|costheta]where θ is the angle between vectors AB and AC. In this case, θ would be angle A of the triangle because vectors AB and AC originate from point A. So, I can write:[overrightarrow{AB} cdot overrightarrow{AC} = |AB||AC|cos A]I know that |AB| is 4, but I don't know |AC|. Maybe I can find |AC| using the Law of Sines or the Law of Cosines. Let me recall the Law of Sines:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}]In triangle ABC, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. So, in this case, side AB is opposite angle C, side BC is opposite angle A, and side AC is opposite angle B.Wait, let me clarify:- AB is opposite angle C.- BC is opposite angle A.- AC is opposite angle B.Given that AB = 4, angle B = π/3, and angle A is between π/6 and π/2.So, using the Law of Sines:[frac{AB}{sin C} = frac{AC}{sin B}]I need to find AC, so let me rearrange the formula:[AC = frac{AB cdot sin B}{sin C}]But I don't know angle C yet. However, I know that the sum of angles in a triangle is π, so:[A + B + C = π]Therefore,[C = π - A - B = π - A - frac{π}{3} = frac{2π}{3} - A]So, angle C is (2π/3 - A). Now, I can write AC as:[AC = frac{4 cdot sin frac{π}{3}}{sin (frac{2π}{3} - A)}]Simplify sin(2π/3 - A):Using the sine subtraction formula:[sin(frac{2π}{3} - A) = sin frac{2π}{3} cos A - cos frac{2π}{3} sin A]I know that:[sin frac{2π}{3} = frac{sqrt{3}}{2}, quad cos frac{2π}{3} = -frac{1}{2}]So,[sin(frac{2π}{3} - A) = frac{sqrt{3}}{2} cos A - (-frac{1}{2}) sin A = frac{sqrt{3}}{2} cos A + frac{1}{2} sin A]Therefore, AC becomes:[AC = frac{4 cdot frac{sqrt{3}}{2}}{frac{sqrt{3}}{2} cos A + frac{1}{2} sin A} = frac{2sqrt{3}}{frac{sqrt{3}}{2} cos A + frac{1}{2} sin A}]Simplify the denominator:Factor out 1/2:[frac{sqrt{3}}{2} cos A + frac{1}{2} sin A = frac{1}{2} (sqrt{3} cos A + sin A)]So,[AC = frac{2sqrt{3}}{frac{1}{2} (sqrt{3} cos A + sin A)} = frac{4sqrt{3}}{sqrt{3} cos A + sin A}]Now, going back to the dot product:[overrightarrow{AB} cdot overrightarrow{AC} = |AB||AC|cos A = 4 cdot frac{4sqrt{3}}{sqrt{3} cos A + sin A} cdot cos A]Simplify this expression:[overrightarrow{AB} cdot overrightarrow{AC} = frac{16sqrt{3} cos A}{sqrt{3} cos A + sin A}]Hmm, this looks a bit complicated. Maybe I can simplify it further. Let me factor out cos A from the denominator:[sqrt{3} cos A + sin A = cos A (sqrt{3} + tan A)]So, substituting back:[overrightarrow{AB} cdot overrightarrow{AC} = frac{16sqrt{3} cos A}{cos A (sqrt{3} + tan A)} = frac{16sqrt{3}}{sqrt{3} + tan A}]That's better. So now, the dot product is expressed in terms of tan A:[overrightarrow{AB} cdot overrightarrow{AC} = frac{16sqrt{3}}{sqrt{3} + tan A}]Now, I need to find the range of this expression as angle A varies between π/6 and π/2.First, let's note the behavior of tan A in this interval.When A = π/6:[tan frac{π}{6} = frac{1}{sqrt{3}} approx 0.577]When A approaches π/2:[tan A rightarrow infty]So, tan A increases from 1/√3 to infinity as A increases from π/6 to π/2.Therefore, the denominator √3 + tan A increases from √3 + 1/√3 to √3 + infinity.Let me compute √3 + 1/√3:[sqrt{3} + frac{1}{sqrt{3}} = frac{3 + 1}{sqrt{3}} = frac{4}{sqrt{3}} approx 2.309]So, the denominator ranges from approximately 2.309 to infinity.Therefore, the entire expression for the dot product:[frac{16sqrt{3}}{sqrt{3} + tan A}]ranges from:When denominator is smallest (2.309):[frac{16sqrt{3}}{4/sqrt{3}} = frac{16sqrt{3} cdot sqrt{3}}{4} = frac{16 cdot 3}{4} = 12]When denominator approaches infinity:[frac{16sqrt{3}}{infty} rightarrow 0]Therefore, the dot product ranges from 0 to 12.But wait, when A approaches π/2, tan A approaches infinity, so the dot product approaches 0. When A is π/6, the dot product is 12. So, the range is (0, 12).But let me double-check my steps to make sure I didn't make a mistake.1. Used Law of Sines correctly to find AC in terms of angle A.2. Expressed AC correctly and simplified.3. Calculated the dot product correctly as |AB||AC|cos A.4. Simplified the expression step by step, factoring out cos A.5. Expressed the dot product in terms of tan A.6. Analyzed the behavior of tan A in the given interval.7. Calculated the range based on the behavior of tan A.Everything seems to check out. So, the range of the dot product is from 0 to 12, not including the endpoints because angle A is strictly between π/6 and π/2, not including π/2 where the dot product would be 0.Therefore, the range is (0, 12).

💡First, I need to determine the original height ( h ) from which the ball is dropped.Each time the ball rebounds, it reaches 75% of the previous height. This means after the first bounce, it reaches ( 0.75h ), after the second bounce ( (0.75)^2h ), and so on.The total distance traveled by the ball when it touches the floor for the fifth time includes the initial drop, the first four ascents, and the first four descents. The fifth descent is not doubled because the ball does not ascend again after the fifth touch.I can express the total distance as:[h + 2(0.75h) + 2(0.75)^2h + 2(0.75)^3h + 2(0.75)^4h + (0.75)^4h = 305]Simplifying this equation:[h + 1.5h + 1.5(0.75)h + 1.5(0.75)^2h + 1.5(0.75)^3h + (0.75)^4h = 305]Next, I calculate the sum of the series in the parentheses:[1 + 1.5 + 1.125 + 0.84375 + 0.6328125 + 0.31640625 = 5.41896875]Now, I can solve for ( h ):[h times 5.41896875 = 305][h = frac{305}{5.41896875} approx 56.3 text{ cm}]Therefore, the original height from which the ball was dropped is approximately 56.3 cm.