answer：First, I recognize that the volume of a cylinder is calculated using the formula ( V = pi r^2 h ), where ( V ) is the volume, ( r ) is the radius, and ( h ) is the height.Since both cylinders have the same height, I can set up a ratio of their volumes in terms of their radii:[frac{V_1}{V_2} = frac{pi r_1^2 h}{pi r_2^2 h}]Given that ( V_1 = 40 , text{cc} ) and ( V_2 = 360 , text{cc} ), I can simplify the equation by canceling out the common factors (( pi ) and ( h )):[frac{40}{360} = frac{r_1^2}{r_2^2}]Next, I simplify the fraction on the left side:[frac{1}{9} = frac{r_1^2}{r_2^2}]To find the ratio of the radii, I take the square root of both sides:[sqrt{frac{1}{9}} = frac{r_1}{r_2}]The square root of ( frac{1}{9} ) is ( frac{1}{3} ), so the ratio of the radii of the two cylinders is:[frac{r_1}{r_2} = frac{1}{3}]Therefore, the ratio of the radii of the two cylinders is ( 1:3 ).

answer：Okay, so I have this problem where I'm given that a and b satisfy the equation a + 2b = 1. I need to find the fixed point that the line ax + 3y + b = 0 must pass through. Hmm, fixed point... I think that means there's a specific point (x, y) that lies on this line no matter what values a and b take, as long as they satisfy a + 2b = 1. Alright, let me start by writing down what I know. The equation of the line is ax + 3y + b = 0. And I know that a + 2b = 1. Maybe I can express a in terms of b or vice versa using this equation. Let me try solving for a. From a + 2b = 1, subtracting 2b from both sides gives a = 1 - 2b. Okay, so a is equal to 1 minus twice b. Maybe I can substitute this into the equation of the line to eliminate a. Let's try that. Substituting a = 1 - 2b into ax + 3y + b = 0, we get (1 - 2b)x + 3y + b = 0. Let me expand that: 1*x - 2b*x + 3y + b = 0. So that's x - 2bx + 3y + b = 0. Hmm, I can factor out the b terms. Let me see: x + 3y + b*(-2x + 1) = 0. So, the equation becomes x + 3y + b*(1 - 2x) = 0. Now, since this equation must hold true for all values of b (as long as a and b satisfy a + 2b = 1), the coefficient of b must be zero. Otherwise, changing b would change the equation, and the point wouldn't be fixed. So, the coefficient of b is (1 - 2x), and that should be zero. Setting 1 - 2x = 0, I can solve for x. Let's do that: 1 - 2x = 0 => 2x = 1 => x = 1/2. Okay, so x is 1/2. Now, plugging x = 1/2 back into the equation to find y. The equation was x + 3y + b*(1 - 2x) = 0. But since 1 - 2x is zero, that term drops out. So, we're left with x + 3y = 0. Substituting x = 1/2 into this, we get 1/2 + 3y = 0. Solving for y: 3y = -1/2 => y = -1/6. So, the fixed point is (1/2, -1/6). Let me just double-check my steps to make sure I didn't make a mistake. 1. I started with a + 2b = 1 and expressed a in terms of b: a = 1 - 2b. 2. Substituted this into the line equation: (1 - 2b)x + 3y + b = 0. 3. Expanded and factored out b: x + 3y + b*(1 - 2x) = 0. 4. Since this must hold for all b, set the coefficient of b to zero: 1 - 2x = 0 => x = 1/2. 5. Plugged x back into the remaining equation: 1/2 + 3y = 0 => y = -1/6. Everything seems to check out. I think the fixed point is indeed (1/2, -1/6). **Final Answer**The coordinates of the fixed point are boxed{left( frac{1}{2}, -frac{1}{6} right)}.

answer：Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem says that 3a cos C + b = 0, and I need to find the maximum value of tan B. Hmm, let me try to figure this out step by step.First, I know that in any triangle, the sum of the angles is 180 degrees, so A + B + C = π radians. Also, from the Law of Sines, I remember that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe that can help me relate the sides and angles.Given the equation 3a cos C + b = 0, I can rewrite this as 3a cos C = -b. Since sides of a triangle are positive, this implies that cos C must be negative because 3a is positive and b is positive. So, cos C is negative, which means angle C is obtuse because cosine is negative in the second quadrant (angles between 90 and 180 degrees). That's an important point.Now, let me try to express cos C in terms of the sides. From the Law of Cosines, I know that cos C = (a² + b² - c²)/(2ab). Maybe substituting this into the equation 3a cos C = -b will help.So, substituting, we get:3a * [(a² + b² - c²)/(2ab)] = -bSimplify this:(3a(a² + b² - c²))/(2ab) = -bMultiply both sides by 2ab to eliminate the denominator:3a(a² + b² - c²) = -2ab²Divide both sides by a (since a ≠ 0):3(a² + b² - c²) = -2b²Expand the left side:3a² + 3b² - 3c² = -2b²Bring all terms to one side:3a² + 3b² - 3c² + 2b² = 0Combine like terms:3a² + 5b² - 3c² = 0Hmm, not sure if this is helpful yet. Maybe I should try another approach.Since I have the Law of Sines, let me express sides a and b in terms of angles. Let me denote 2R as the common ratio, so a = 2R sin A, b = 2R sin B, c = 2R sin C.Substituting into the equation 3a cos C + b = 0:3*(2R sin A)*cos C + 2R sin B = 0Factor out 2R:2R*(3 sin A cos C + sin B) = 0Since 2R ≠ 0, we have:3 sin A cos C + sin B = 0Hmm, interesting. So, 3 sin A cos C = - sin B.But in triangle ABC, angles A, B, C are related by A + B + C = π. So, B = π - A - C. Therefore, sin B = sin(π - A - C) = sin(A + C). So, sin B = sin(A + C).So, substituting back, we have:3 sin A cos C + sin(A + C) = 0Now, let's expand sin(A + C):sin(A + C) = sin A cos C + cos A sin CSo, substituting that into the equation:3 sin A cos C + sin A cos C + cos A sin C = 0Combine like terms:(3 sin A cos C + sin A cos C) + cos A sin C = 0Which simplifies to:4 sin A cos C + cos A sin C = 0Hmm, so 4 sin A cos C = - cos A sin CLet me write this as:4 sin A cos C + cos A sin C = 0Wait, that's the same as above. Let me rearrange it:4 sin A cos C = - cos A sin CDivide both sides by cos A cos C (assuming cos A and cos C are not zero, which they aren't since A and C are angles in a triangle, so less than 180 degrees, and C is obtuse, so cos C is negative, but not zero):4 (sin A / cos A) = - (sin C / cos C)Which is:4 tan A = - tan CSo, tan C = -4 tan ABut since angle C is obtuse, tan C is negative because tangent is negative in the second quadrant. So, tan C = -4 tan A makes sense because tan A is positive (since angle A is acute in a triangle with an obtuse angle C).So, tan C = -4 tan A.Now, I need to find tan B. Since B = π - A - C, tan B = tan(π - A - C) = - tan(A + C). Because tan(π - x) = - tan x.So, tan B = - tan(A + C)Now, tan(A + C) can be expanded as:tan(A + C) = (tan A + tan C)/(1 - tan A tan C)So, tan B = - (tan A + tan C)/(1 - tan A tan C)But we know tan C = -4 tan A, so let's substitute that in:tan B = - (tan A + (-4 tan A))/(1 - tan A*(-4 tan A))Simplify numerator and denominator:Numerator: tan A - 4 tan A = -3 tan ADenominator: 1 + 4 tan² ASo, tan B = - (-3 tan A)/(1 + 4 tan² A) = (3 tan A)/(1 + 4 tan² A)So, tan B = (3 tan A)/(1 + 4 tan² A)Now, I need to find the maximum value of tan B. So, let's denote t = tan A, where t > 0 because angle A is acute.So, tan B = (3 t)/(1 + 4 t²)We need to find the maximum value of f(t) = (3 t)/(1 + 4 t²) for t > 0.To find the maximum, we can take the derivative of f(t) with respect to t and set it equal to zero.Compute f'(t):f'(t) = [3*(1 + 4 t²) - 3 t*(8 t)] / (1 + 4 t²)^2Simplify numerator:3*(1 + 4 t²) - 24 t² = 3 + 12 t² - 24 t² = 3 - 12 t²Set f'(t) = 0:3 - 12 t² = 012 t² = 3t² = 3/12 = 1/4t = 1/2 (since t > 0)So, the critical point is at t = 1/2.Now, check if this is a maximum. Since the function f(t) approaches 0 as t approaches 0 and as t approaches infinity, and there's only one critical point, it must be a maximum.So, the maximum value of tan B is f(1/2):f(1/2) = (3*(1/2))/(1 + 4*(1/2)^2) = (3/2)/(1 + 4*(1/4)) = (3/2)/(1 + 1) = (3/2)/2 = 3/4Therefore, the maximum value of tan B is 3/4.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Started with 3a cos C + b = 0, which led to 3 sin A cos C + sin B = 0.2. Expressed sin B as sin(A + C), expanded it, and got 4 sin A cos C + cos A sin C = 0.3. Divided both sides by cos A cos C to get 4 tan A = - tan C, so tan C = -4 tan A.4. Expressed tan B as - tan(A + C), expanded tan(A + C), substituted tan C = -4 tan A, and simplified to get tan B = (3 tan A)/(1 + 4 tan² A).5. Let t = tan A, so tan B = 3t/(1 + 4t²). Took derivative, found critical point at t = 1/2, and evaluated to get 3/4.Seems solid. I think the maximum value is indeed 3/4.**Final Answer**The maximum value of tan B is boxed{dfrac{3}{4}}.

answer：First, I need to determine the minimum number of teachers required to cover all subjects, considering that each teacher can teach up to two subjects.There are 4 math teachers, 3 physics teachers, and 3 chemistry teachers.Since each teacher can teach two subjects, I can have some teachers teaching both math and physics, some teaching both physics and chemistry, and some teaching both chemistry and math.I'll start by assigning teachers to the subject with the highest demand, which is math with 4 teachers needed. I can have all 4 math teachers also teach one of the other subjects (physics or chemistry) since they can teach two subjects. This would cover all the math classes and either all the physics or all the chemistry classes.Let's say I assign 3 of the math teachers to also teach physics. This covers all the physics classes (3 teachers needed) and leaves me with 1 math teacher who can also teach chemistry.Now I have 1 math teacher left who can teach chemistry, but I still need 2 more chemistry teachers. I can assign the remaining 2 chemistry teachers to teach chemistry only, or one of them could also teach the remaining math class if needed.In summary, the minimum number of teachers required would be:- 3 math teachers teaching math and physics- 1 math teacher teaching math and chemistry- 2 chemistry teachers teaching chemistry onlyThis adds up to 3 + 1 + 2 = 6 teachers in total.So, the minimum number of teachers required to cover all the classes is 6.

answer：Okay, so I have this problem about vectors. It says that vectors a and b satisfy |a| = 2|b| = 2, and |a - 4b| = 2√7. I need to find the angle between a and b. Hmm, let me try to break this down step by step.First, let's parse the given information. The magnitude of vector a is 2, and the magnitude of vector b is 1 because |a| = 2|b| = 2. So, |a| = 2 and |b| = 1. Got that.Next, the magnitude of the vector a - 4b is given as 2√7. I remember that the magnitude of a vector difference can be found using the dot product formula. Specifically, |a - 4b|² = (a - 4b) · (a - 4b). Let me write that down:|a - 4b|² = (a - 4b) · (a - 4b)Expanding this dot product, I get:a · a - 4a · b - 4b · a + 16b · bWait, actually, that's not quite right. Let me correct that. The expansion should be:(a - 4b) · (a - 4b) = a · a - 4a · b - 4b · a + 16b · bBut since the dot product is commutative, a · b = b · a, so this simplifies to:a · a - 8a · b + 16b · bOkay, so that's better. Now, let's express each term in terms of magnitudes and the cosine of the angle between them. I remember that a · b = |a||b|cosθ, where θ is the angle between vectors a and b.So, substituting the known magnitudes:a · a = |a|² = 2² = 4Similarly, b · b = |b|² = 1² = 1And a · b = |a||b|cosθ = 2 * 1 * cosθ = 2cosθPutting it all back into the expanded equation:|a - 4b|² = 4 - 8*(2cosθ) + 16*1Wait, hold on. Let me check that again. The expansion was:a · a - 8a · b + 16b · bSo substituting:4 - 8*(2cosθ) + 16*1Wait, that would be 4 - 16cosθ + 16, right? Because 8 times 2cosθ is 16cosθ.So, 4 + 16 is 20, and then minus 16cosθ. So:|a - 4b|² = 20 - 16cosθBut we know that |a - 4b| is given as 2√7, so |a - 4b|² is (2√7)² = 4*7 = 28.So, setting up the equation:28 = 20 - 16cosθNow, solving for cosθ:28 - 20 = -16cosθ8 = -16cosθDivide both sides by -16:cosθ = -8 / 16 = -1/2So, cosθ = -1/2. Now, I need to find the angle θ whose cosine is -1/2.I remember that cos(60°) = 0.5, so cos(120°) = -0.5 because cosine is negative in the second quadrant. So, θ = 120°, which is 2π/3 radians.Let me just double-check my steps to make sure I didn't make a mistake. Starting from the given magnitudes, expanding the dot product correctly, substituting the values, solving for cosθ, and then finding the angle. Everything seems to add up. So, I think the angle between vectors a and b is 120 degrees or 2π/3 radians.

answer：First, I need to determine how many strawberries Kimberly's brother picked. He picked 3 baskets, each containing 15 strawberries.Next, I'll calculate the total number of strawberries Kimberly picked. She picked 8 times the amount her brother picked.Then, I'll find out how many strawberries Kimberly's parents picked. They picked 93 strawberries less than Kimberly.After that, I'll sum up all the strawberries picked by the family members to get the total number.Finally, I'll divide the total number of strawberries equally among the four family members to determine how many strawberries each person gets.