💡Okay, so I have this problem about a cone. The slant height is 6 cm, and when you unfold the lateral surface, the central angle is 120 degrees. I need to find the radius of the base of the cone. Hmm, let me think about how to approach this.First, I remember that when you unfold a cone's lateral surface, it forms a sector of a circle. The radius of this sector is the same as the slant height of the cone. So in this case, the radius of the sector is 6 cm. That makes sense because the slant height is the distance from the base to the tip of the cone, which would correspond to the radius when it's unfolded.Now, the central angle of this sector is 120 degrees. I think that relates to how much of the full circle this sector represents. A full circle is 360 degrees, so 120 degrees is a third of that. Maybe that means the arc length of the sector is a third of the circumference of a full circle with radius 6 cm.Let me write that down. The circumference of a full circle is 2πr, so for radius 6 cm, that would be 2π*6 = 12π cm. If the sector is 120 degrees, which is a third of 360 degrees, then the arc length of the sector should be a third of 12π cm. So, arc length = (120/360)*12π = (1/3)*12π = 4π cm.Wait, but what is this arc length? I think it's the circumference of the base of the cone. Because when you unfold the cone, the arc of the sector becomes the circumference of the base. So, if the arc length is 4π cm, that should be equal to 2πr, where r is the radius of the base of the cone.So, setting them equal: 2πr = 4π. If I divide both sides by π, I get 2r = 4. Then, dividing both sides by 2, I get r = 2 cm.Let me double-check that. The slant height is 6 cm, which is the radius of the sector. The central angle is 120 degrees, so the arc length is (120/360)*2π*6 = (1/3)*12π = 4π. This arc length is the circumference of the base, so 2πr = 4π, leading to r = 2 cm. Yeah, that seems consistent.I don't think I made any mistakes here. The key was realizing that the arc length of the sector corresponds to the circumference of the base of the cone. Once I had that, it was just a matter of setting up the equation and solving for r.So, I think the radius of the base of the cone is 2 cm.

💡Okay, so I have this problem where I need to find the value of 'a' such that the line ax + 2y + 6 = 0 is parallel to but not coincident with the line x + (a - 1)y + (a² - 1) = 0. Hmm, let me think about how to approach this.First, I remember that for two lines to be parallel, their slopes must be equal. So, I need to find the slopes of both lines and set them equal to each other. But I also need to make sure that the lines aren't coinciding, which means their y-intercepts should be different.Let me start by rewriting both equations in slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept.Starting with the first line: ax + 2y + 6 = 0.I'll solve for y:2y = -ax - 6 y = (-a/2)x - 3So, the slope (m₁) of the first line is -a/2, and the y-intercept is -3.Now, moving on to the second line: x + (a - 1)y + (a² - 1) = 0.Again, I'll solve for y:(a - 1)y = -x - (a² - 1) y = (-1/(a - 1))x - (a² - 1)/(a - 1)So, the slope (m₂) of the second line is -1/(a - 1), and the y-intercept is -(a² - 1)/(a - 1).Since the lines are parallel, their slopes must be equal. Therefore, I can set m₁ equal to m₂:-a/2 = -1/(a - 1)Let me solve this equation for 'a'. First, I can multiply both sides by 2(a - 1) to eliminate the denominators:-a/2 * 2(a - 1) = -1/(a - 1) * 2(a - 1)Simplifying both sides:-a(a - 1) = -2Multiplying both sides by -1 to make it positive:a(a - 1) = 2Expanding the left side:a² - a = 2Bringing all terms to one side to form a quadratic equation:a² - a - 2 = 0Now, I need to solve this quadratic equation. Let's factor it:Looking for two numbers that multiply to -2 and add up to -1. Hmm, those numbers would be -2 and 1.So, factoring:(a - 2)(a + 1) = 0Setting each factor equal to zero:a - 2 = 0 ⇒ a = 2 a + 1 = 0 ⇒ a = -1So, the possible values for 'a' are 2 and -1. But I need to check which of these makes the lines parallel but not coincident.Let me check for a = 2 first.Plugging a = 2 into the first line:2x + 2y + 6 = 0 Dividing by 2: x + y + 3 = 0 So, y = -x - 3Now, plugging a = 2 into the second line:x + (2 - 1)y + (2² - 1) = 0 Simplifying: x + y + (4 - 1) = 0 ⇒ x + y + 3 = 0 So, y = -x - 3Wait a minute, both lines become y = -x - 3 when a = 2. That means they are the same line, which contradicts the condition that they are not coincident. So, a = 2 is not a valid solution.Now, let's check a = -1.Plugging a = -1 into the first line:-1x + 2y + 6 = 0 Which is: -x + 2y + 6 = 0 Solving for y: 2y = x - 6 ⇒ y = (1/2)x - 3Now, plugging a = -1 into the second line:x + (-1 - 1)y + ((-1)² - 1) = 0 Simplifying: x - 2y + (1 - 1) = 0 ⇒ x - 2y + 0 = 0 ⇒ x - 2y = 0 Solving for y: -2y = -x ⇒ y = (1/2)xSo, the first line is y = (1/2)x - 3 and the second line is y = (1/2)x. Their slopes are both 1/2, so they are parallel. However, their y-intercepts are different (-3 and 0), so they are not coincident.Therefore, a = -1 is the correct value that satisfies the given conditions.I think that's it. I started by finding the slopes, set them equal, solved for 'a', and then checked both solutions to ensure they don't make the lines coincide. It seems like a = -1 is the only valid solution here.

💡Alright, so I'm trying to figure out this problem about a crank-slider mechanism. It's something I remember from my mechanics class, but I need to brush up on the details. Let me start by understanding what the problem is asking.We have a crank-slider mechanism with two main components: the crank (OA) and the connecting rod (AB). Both OA and AB are 90 cm long. The crank is rotating with a constant angular velocity of 10 rad/s. There's a point L on the connecting rod AB such that AL is one-third of AB. So, AL is 30 cm, and LB is 60 cm.The problem asks for three things: the equations of motion for point L, the equation of its trajectory, and its speed. Okay, let's break this down step by step.First, I need to represent the coordinates of point B, which is the end of the connecting rod. Since OA is rotating, point A is moving in a circular path around the origin. The coordinates of point B can be expressed using trigonometric functions because of the circular motion.Given that OA is 90 cm and it's rotating with an angular velocity of 10 rad/s, the angle θ at any time t is θ = ωt = 10t. So, the coordinates of point B would be:x_B = OA * cos(θ) = 90 * cos(10t)y_B = OA * sin(θ) = 90 * sin(10t)Okay, that makes sense. Now, point L is one-third of the way from A to B on the connecting rod. Since A is at the origin (0,0), point L's coordinates should be a weighted average between A and B.So, if AL is one-third of AB, then L divides AB in a 1:2 ratio. Therefore, the coordinates of L can be found using the section formula. The coordinates of L would be:x_L = (2 * x_A + 1 * x_B) / (1 + 2) = (0 + x_B) / 3 = x_B / 3y_L = (2 * y_A + 1 * y_B) / (1 + 2) = (0 + y_B) / 3 = y_B / 3Substituting the values of x_B and y_B, we get:x_L = (90 * cos(10t)) / 3 = 30 * cos(10t)y_L = (90 * sin(10t)) / 3 = 30 * sin(10t)So, the equations of motion for point L are:x_L = 30 cos(10t)y_L = 30 sin(10t)That seems straightforward. Now, for the trajectory of point L. Since x_L and y_L are expressed in terms of cosine and sine functions with the same argument, the trajectory should be a circle. Let me verify that.If we eliminate the parameter t from the equations:x_L = 30 cos(10t) => cos(10t) = x_L / 30y_L = 30 sin(10t) => sin(10t) = y_L / 30We know that cos²(θ) + sin²(θ) = 1, so:(x_L / 30)² + (y_L / 30)² = 1=> (x_L² + y_L²) / 900 = 1=> x_L² + y_L² = 900So, the trajectory of point L is a circle with radius 30 cm centered at the origin. That makes sense because L is moving in a circular path as the crank rotates.Now, onto the speed of point L. Speed is the magnitude of the velocity vector. To find the velocity, I need to differentiate the position equations with respect to time.Let's compute the derivatives:dx_L/dt = d/dt [30 cos(10t)] = -30 * 10 sin(10t) = -300 sin(10t)dy_L/dt = d/dt [30 sin(10t)] = 30 * 10 cos(10t) = 300 cos(10t)So, the velocity components are:v_x = -300 sin(10t)v_y = 300 cos(10t)The speed is the magnitude of the velocity vector:v = sqrt(v_x² + v_y²) = sqrt[(-300 sin(10t))² + (300 cos(10t))²]= sqrt[90000 sin²(10t) + 90000 cos²(10t)]= sqrt[90000 (sin²(10t) + cos²(10t))]= sqrt[90000 * 1]= 300 cm/sSo, the speed of point L is constant at 300 cm/s. That's interesting because even though the direction of the velocity changes, the magnitude remains the same due to the uniform angular velocity of the crank.Let me recap to make sure I didn't miss anything. The equations of motion for L are x = 30 cos(10t) and y = 30 sin(10t). The trajectory is a circle of radius 30 cm. The speed is a constant 300 cm/s.I think that covers all the parts of the problem. I should double-check my calculations to ensure there are no errors.Checking the derivatives:d/dt [cos(kt)] = -k sin(kt)d/dt [sin(kt)] = k cos(kt)Yes, so multiplying by 30 and 10 gives the correct velocity components. The magnitude calculation also looks right because sin² + cos² = 1, simplifying the expression to 300 cm/s.Everything seems consistent. I don't see any mistakes in my reasoning.**Final Answer**The equations of motion for point ( mathrm{L} ) are ( x = 30 cos(10t) ) and ( y = 30 sin(10t) ), its trajectory is a circle with radius 30 cm, and its speed is ( boxed{300 text{cm/s}} ).

💡First, I recognize that Aziz just celebrated his 36th birthday, which means he was born in the year ( Y - 36 ), where ( Y ) is the current year.His parents had been living in America for 3 years before he was born. Therefore, the year his parents moved to America is ( (Y - 36) - 3 ).Given that his parents moved to America in 1982, I can set up the equation:[(Y - 36) - 3 = 1982]Simplifying the equation:[Y - 39 = 1982]Solving for ( Y ):[Y = 1982 + 39][Y = 2021]Therefore, the current year is 2021.

💡Okay, so I have this problem where Alex picks a favorite point (x, y) in the first quadrant on the unit circle. That means x² + y² = 1, right? And this point is θ radians counterclockwise from the positive x-axis. So, I can express x and y in terms of cosine and sine of θ. That makes sense because on the unit circle, x is cosθ and y is sinθ.Then, Alex computes the inverse cosine of (4x + 3y)/5 and gets θ. So, mathematically, that means θ = arccos[(4x + 3y)/5]. Hmm, interesting. So, if I substitute x and y with cosθ and sinθ, the equation becomes θ = arccos[(4cosθ + 3sinθ)/5].Now, I need to figure out what tanθ is. Let me think about how to approach this. The expression inside the arccos looks like a linear combination of cosine and sine, which reminds me of the cosine of a difference of angles formula. The formula is cos(A - B) = cosA cosB + sinA sinB. So, maybe I can rewrite (4cosθ + 3sinθ)/5 in that form.Let me set up the equation: (4cosθ + 3sinθ)/5 = cos(θ - φ), where φ is some angle. To find φ, I need to match the coefficients. So, comparing to the formula, 4/5 should be cosφ and 3/5 should be sinφ. Let me check if that works.If cosφ = 4/5 and sinφ = 3/5, then φ is an angle in the first quadrant since both sine and cosine are positive. That makes sense because we're dealing with the first quadrant. So, φ is such that tanφ = (3/5)/(4/5) = 3/4. So, φ = arctan(3/4).Now, going back to the equation: (4cosθ + 3sinθ)/5 = cos(θ - φ). So, arccos[(4cosθ + 3sinθ)/5] = arccos[cos(θ - φ)]. Since arccos(cosα) = α when α is in the range [0, π], which it is here because θ is in the first quadrant and φ is also in the first quadrant, so θ - φ is between -π/2 and π/2, but since θ is equal to arccos[cos(θ - φ)], it must be that θ = θ - φ or θ = -(θ - φ). Wait, that doesn't make sense. Let me think again.Actually, arccos(cosα) = |α| if α is in [0, π], but since θ is in the first quadrant and φ is also in the first quadrant, θ - φ could be positive or negative. But since arccos returns a value between 0 and π, and θ is equal to that, we have θ = θ - φ or θ = φ - θ. Wait, that might not be the right way to think about it.Let me consider that arccos(cos(θ - φ)) = |θ - φ| if θ - φ is in [0, π], but since θ and φ are both in the first quadrant, θ - φ could be positive or negative. But since arccos returns a value between 0 and π, and θ is equal to that, we must have θ = |θ - φ|. Hmm, that seems a bit confusing.Wait, maybe another approach. Since θ = arccos[(4cosθ + 3sinθ)/5], and we know that (4cosθ + 3sinθ)/5 = cos(θ - φ), then θ = arccos(cos(θ - φ)). So, arccos(cos(θ - φ)) = |θ - φ| if θ - φ is in [0, π], but since θ and φ are both in the first quadrant, θ - φ could be positive or negative. But since θ is equal to that, we have θ = |θ - φ|.Wait, that can't be right because if θ = |θ - φ|, then either θ = θ - φ or θ = φ - θ. The first case, θ = θ - φ, would imply φ = 0, which isn't true because φ is arctan(3/4). The second case, θ = φ - θ, implies 2θ = φ, so θ = φ/2.Ah, that makes sense. So, θ is half of φ. Since φ = arctan(3/4), then θ = (1/2) arctan(3/4). Now, I need to find tanθ. So, tanθ = tan[(1/2) arctan(3/4)].To find tan(θ), I can use the double-angle formula for tangent. Wait, no, since θ is half of φ, I can use the half-angle formula. The half-angle formula for tangent is tan(θ/2) = (sinθ)/(1 + cosθ). But in this case, θ is half of φ, so θ = φ/2. So, tanθ = tan(φ/2).Given that φ = arctan(3/4), I can find tan(φ/2). Let me recall the half-angle identity for tangent: tan(φ/2) = sinφ/(1 + cosφ). I already know sinφ and cosφ from earlier: sinφ = 3/5 and cosφ = 4/5.So, plugging these into the formula: tan(φ/2) = (3/5)/(1 + 4/5) = (3/5)/(9/5) = (3/5)*(5/9) = 3/9 = 1/3.Therefore, tanθ = 1/3.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from θ = arccos[(4cosθ + 3sinθ)/5], I recognized that (4cosθ + 3sinθ)/5 is equal to cos(θ - φ) where φ is such that cosφ = 4/5 and sinφ = 3/5. Then, since θ = arccos(cos(θ - φ)), it implies θ = |θ - φ|, but since θ is in the first quadrant and φ is also in the first quadrant, θ must be less than φ, so θ = φ - θ, leading to 2θ = φ. Therefore, θ = φ/2.Then, using the half-angle formula for tangent, tan(φ/2) = sinφ/(1 + cosφ) = (3/5)/(1 + 4/5) = (3/5)/(9/5) = 1/3. That seems correct.So, the final answer is tanθ = 1/3.

💡Okay, so I have this problem here: I need to find the interval where the zero of the function ( f(x) = e^x + x - 2 ) is located. The options are A: (-2, -1), B: (-1, 0), C: (0, 1), and D: (1, 2). Hmm, I remember that to find where a function crosses zero, I can use the Intermediate Value Theorem. That theorem says that if a function is continuous on an interval [a, b], and if f(a) and f(b) have opposite signs, then there must be at least one c in (a, b) where f(c) = 0. First, I should check if ( f(x) ) is continuous. Well, ( e^x ) is continuous everywhere, and polynomials like x - 2 are also continuous everywhere. So the sum of two continuous functions is continuous. That means ( f(x) ) is continuous on all real numbers, including all the intervals given in the options. Good, so I can apply the Intermediate Value Theorem here.Now, I need to evaluate ( f(x) ) at the endpoints of each interval and see where the sign changes. Let me start with option A: (-2, -1). I'll plug in x = -2 and x = -1 into the function.For x = -2:( f(-2) = e^{-2} + (-2) - 2 )Calculating ( e^{-2} ), which is approximately 0.135. So,( f(-2) ≈ 0.135 - 2 - 2 = 0.135 - 4 = -3.865 )That's a negative value.For x = -1:( f(-1) = e^{-1} + (-1) - 2 )( e^{-1} ) is approximately 0.368, so,( f(-1) ≈ 0.368 - 1 - 2 = 0.368 - 3 = -2.632 )Still negative. Hmm, so both endpoints give negative values. That means the function doesn't cross zero in this interval because it's negative at both ends. So option A is probably not the correct interval.Moving on to option B: (-1, 0). Let's evaluate at x = -1 and x = 0.Wait, I already calculated ( f(-1) ≈ -2.632 ). Now, for x = 0:( f(0) = e^{0} + 0 - 2 )( e^{0} = 1 ), so,( f(0) = 1 + 0 - 2 = -1 )Still negative. Hmm, both endpoints are negative again. So, does that mean there's no zero in this interval? Wait, but I remember that sometimes functions can have a minimum or maximum in between, so maybe the function dips below zero and then comes back up, but in this case, both endpoints are negative. Wait, but let me double-check my calculations.Wait, at x = 0, ( f(0) = e^0 + 0 - 2 = 1 - 2 = -1 ). That's correct. At x = -1, it's about -2.632. So both are negative. So, maybe I made a mistake here. Let me check another point in between, like x = -0.5.Calculating ( f(-0.5) = e^{-0.5} + (-0.5) - 2 )( e^{-0.5} ≈ 0.6065 )So, ( f(-0.5) ≈ 0.6065 - 0.5 - 2 = 0.6065 - 2.5 = -1.8935 )Still negative. Hmm, so maybe the function is always negative in this interval. So, perhaps option B is not the correct interval either.Wait, but let me check option C: (0, 1). Let's compute f(0) and f(1).We already have f(0) = -1.For x = 1:( f(1) = e^{1} + 1 - 2 )( e ≈ 2.718 ), so,( f(1) ≈ 2.718 + 1 - 2 = 2.718 - 1 = 1.718 )That's positive. So, f(0) is -1 and f(1) is approximately 1.718. Since f(0) is negative and f(1) is positive, by the Intermediate Value Theorem, there must be a zero between 0 and 1. So, that would suggest option C is the correct interval.Wait, but the initial answer suggested option B. Did I make a mistake? Let me go back.Wait, in the initial problem, the options are A: (-2, -1), B: (-1, 0), C: (0, 1), D: (1, 2). So, according to my calculations, f(-2) ≈ -3.865, f(-1) ≈ -2.632, f(0) = -1, f(1) ≈ 1.718, f(2) would be e^2 + 2 - 2 = e^2 ≈ 7.389, which is positive.So, f(x) is negative at x = -2, -1, 0, and positive at x = 1 and 2. So, the function crosses zero between 0 and 1, which is option C.Wait, but the initial answer said option B. Maybe I made a mistake in evaluating f(-1). Let me recalculate f(-1):( f(-1) = e^{-1} + (-1) - 2 = 1/e - 3 ≈ 0.3679 - 3 ≈ -2.632 ). That's correct.Wait, maybe I should check another point in option B, like x = -0.5, which I did earlier and got f(-0.5) ≈ -1.8935, still negative.Wait, so if f(-1) is -2.632 and f(0) is -1, both negative, then the function doesn't cross zero in (-1, 0). So, why did the initial answer say option B? Maybe I'm misunderstanding the problem.Wait, perhaps I should plot the function or think about its behavior. Let's see, as x approaches negative infinity, e^x approaches zero, so f(x) ≈ x - 2, which goes to negative infinity. As x increases, e^x grows exponentially, so f(x) will eventually become positive. So, the function must cross zero somewhere.Wait, but according to my calculations, f(0) is -1, f(1) is positive, so the zero is in (0,1). So, option C.Wait, maybe the initial answer was wrong. Let me double-check.Alternatively, perhaps I should compute f(-1) and f(0) again.f(-1) = e^{-1} + (-1) - 2 ≈ 0.3679 - 3 ≈ -2.632.f(0) = e^0 + 0 - 2 = 1 - 2 = -1.So, both negative. So, no zero in (-1, 0). So, the zero must be in (0,1), which is option C.Wait, but the initial answer said B. Maybe I need to check if I misread the function. The function is f(x) = e^x + x - 2.Yes, that's correct. So, perhaps the initial answer was incorrect. Alternatively, maybe I made a mistake in calculations.Wait, let me compute f(-1) again:e^{-1} is approximately 0.3679, so 0.3679 -1 -2 = 0.3679 -3 = -2.6321.f(0) = 1 -2 = -1.f(1) = e +1 -2 ≈ 2.718 -1 ≈ 1.718.So, f(0) is -1, f(1) is positive. So, the zero is in (0,1). So, option C.Wait, but the initial answer said B. Maybe the initial answer was wrong. Alternatively, perhaps the function is f(x) = e^{x} + x -2, which is what I have.Alternatively, maybe I should compute f(-0.5):f(-0.5) = e^{-0.5} + (-0.5) -2 ≈ 0.6065 -0.5 -2 ≈ 0.6065 -2.5 ≈ -1.8935.Still negative.f(-0.25):e^{-0.25} ≈ 0.7788, so f(-0.25) ≈ 0.7788 -0.25 -2 ≈ 0.7788 -2.25 ≈ -1.4712.Still negative.f(-0.1):e^{-0.1} ≈ 0.9048, so f(-0.1) ≈ 0.9048 -0.1 -2 ≈ 0.9048 -2.1 ≈ -1.1952.Still negative.f(0) = -1.So, from x = -1 to x = 0, f(x) remains negative. So, no zero in (-1,0).Wait, but maybe I should check x = 0.5:f(0.5) = e^{0.5} +0.5 -2 ≈ 1.6487 +0.5 -2 ≈ 2.1487 -2 ≈ 0.1487.Positive. So, f(0.5) is positive.So, f(0) = -1, f(0.5) ≈ 0.1487. So, the function crosses zero between 0 and 0.5.Therefore, the zero is in (0,1), which is option C.Wait, but the initial answer said B. Maybe the initial answer was incorrect. Alternatively, perhaps I made a mistake in interpreting the function.Wait, let me check if the function is f(x) = e^{x} + x -2 or f(x) = e^{x + x -2}. No, the function is e^x + x -2.Alternatively, maybe the initial answer was based on a different function.Alternatively, perhaps I should consider the derivative to see if the function is increasing or decreasing.f'(x) = e^x +1. Since e^x is always positive, f'(x) is always positive. So, the function is strictly increasing.Therefore, since it's strictly increasing, it can have only one zero. And since f(0) = -1 and f(1) ≈ 1.718, the zero is between 0 and 1.So, the correct interval is C: (0,1).Therefore, the initial answer was wrong, and the correct answer is C.