💡First, I need to determine the total amount of flour Jim has available to bake 2 loaves of bread. Each loaf requires 200g of flour, so for 2 loaves, Jim needs a total of 400g of flour.Next, I'll add up the flour Jim already has. He has 200g in the cupboard and 100g on the kitchen counter, which sums up to 300g.To find out how much flour is in the pantry, I'll subtract the flour he already has from the total amount needed. 400g (total needed) - 300g (cupboard + counter) = 100g.Therefore, Jim has 100g of flour in the pantry.

💡Alright, so I have this problem about a cosine function and some properties related to it. Let me try to understand and solve it step by step.First, the function given is ( f(x) = cos(omega x + frac{pi}{3}) ) where ( omega > 0 ). It says that the distance between any two adjacent axes of symmetry on its graph is ( frac{pi}{2} ). Hmm, okay, so I need to figure out what ( omega ) is because that will help me determine the rest of the properties.I remember that for a cosine function ( cos(kx + phi) ), the period is ( frac{2pi}{k} ). The axes of symmetry for a cosine function are the vertical lines passing through the maxima and minima. The distance between two adjacent axes of symmetry should be half the period because the period is the distance between two corresponding points on the graph, like two maxima or two minima. So, if the distance between two adjacent axes is ( frac{pi}{2} ), then half the period is ( frac{pi}{2} ), which means the full period is ( pi ).So, the period ( T ) is ( frac{2pi}{omega} ), and we know ( T = pi ). Therefore, ( frac{2pi}{omega} = pi ), which simplifies to ( omega = 2 ). So, the function becomes ( f(x) = cos(2x + frac{pi}{3}) ).Now, part (1) asks for the intervals of monotonicity and the center of symmetry. Let's tackle monotonicity first. The function ( cos(2x + frac{pi}{3}) ) is a cosine function with a phase shift and a frequency change. The standard cosine function ( cos(theta) ) is increasing on intervals where its derivative is positive and decreasing where the derivative is negative.Let me compute the derivative of ( f(x) ). The derivative of ( cos(2x + frac{pi}{3}) ) with respect to ( x ) is ( -2sin(2x + frac{pi}{3}) ). So, ( f'(x) = -2sin(2x + frac{pi}{3}) ).To find where the function is increasing or decreasing, we need to find where the derivative is positive or negative. Let's set ( f'(x) > 0 ):( -2sin(2x + frac{pi}{3}) > 0 )( sin(2x + frac{pi}{3}) < 0 )Similarly, for ( f'(x) < 0 ):( -2sin(2x + frac{pi}{3}) < 0 )( sin(2x + frac{pi}{3}) > 0 )So, the function ( f(x) ) is increasing when ( sin(2x + frac{pi}{3}) < 0 ) and decreasing when ( sin(2x + frac{pi}{3}) > 0 ).Let me solve for ( x ) in both cases.First, when is ( sin(2x + frac{pi}{3}) < 0 )?The sine function is negative in the intervals ( (2kpi - pi, 2kpi) ) for any integer ( k ). So, we have:( 2kpi - pi < 2x + frac{pi}{3} < 2kpi )Subtract ( frac{pi}{3} ) from all parts:( 2kpi - pi - frac{pi}{3} < 2x < 2kpi - frac{pi}{3} )Simplify the left side:( 2kpi - frac{4pi}{3} < 2x < 2kpi - frac{pi}{3} )Divide by 2:( kpi - frac{2pi}{3} < x < kpi - frac{pi}{6} )So, the function is increasing on intervals ( (kpi - frac{2pi}{3}, kpi - frac{pi}{6}) ) for any integer ( k ).Similarly, when is ( sin(2x + frac{pi}{3}) > 0 )?The sine function is positive in the intervals ( (2kpi, 2kpi + pi) ). So:( 2kpi < 2x + frac{pi}{3} < 2kpi + pi )Subtract ( frac{pi}{3} ):( 2kpi - frac{pi}{3} < 2x < 2kpi + frac{2pi}{3} )Divide by 2:( kpi - frac{pi}{6} < x < kpi + frac{pi}{3} )So, the function is decreasing on intervals ( (kpi - frac{pi}{6}, kpi + frac{pi}{3}) ) for any integer ( k ).Alright, so that's the monotonicity part. Now, the center of symmetry. The function ( f(x) = cos(2x + frac{pi}{3}) ) is a cosine function, which is symmetric about its maxima and minima. The axes of symmetry are the vertical lines passing through these points.For a standard cosine function ( cos(kx + phi) ), the axes of symmetry occur at the maxima and minima, which are at ( x = frac{2npi - phi}{k} ) for maxima and ( x = frac{(2n + 1)pi - phi}{k} ) for minima, where ( n ) is an integer.In our case, ( k = 2 ) and ( phi = frac{pi}{3} ). So, the maxima occur at:( x = frac{2npi - frac{pi}{3}}{2} = npi - frac{pi}{6} )And the minima occur at:( x = frac{(2n + 1)pi - frac{pi}{3}}{2} = frac{(2n + 1)pi}{2} - frac{pi}{6} = npi + frac{pi}{2} - frac{pi}{6} = npi + frac{pi}{3} )Wait, but the problem mentions the "center of symmetry." I think this refers to the points about which the function is symmetric. For a cosine function, each axis of symmetry is a vertical line through a maximum or minimum. So, the centers of symmetry would be these points.But actually, in the context of symmetry, the center of symmetry is a point, not a line. So, perhaps it's referring to the points where the function has rotational symmetry. For a standard cosine function, it's symmetric about the points where it crosses the x-axis, but in this case, with a phase shift, it might be different.Wait, let me think again. The function ( cos(2x + frac{pi}{3}) ) can be rewritten as ( cos[2(x + frac{pi}{6})] ). So, it's a cosine function with a horizontal compression by a factor of 2 and a phase shift to the left by ( frac{pi}{6} ).The standard cosine function ( cos(2x) ) has centers of symmetry at points where it crosses the x-axis, which are at ( x = frac{pi}{4} + frac{kpi}{2} ) for integer ( k ). But with the phase shift, these centers of symmetry would shift accordingly.Wait, maybe I'm overcomplicating. Let me recall that for a function to have a center of symmetry at a point ( (a, b) ), it must satisfy ( f(a + h) + f(a - h) = 2b ) for all ( h ). For the cosine function, which is symmetric about its maxima and minima, the centers of symmetry would be the points where the function crosses the x-axis, i.e., the points where ( f(x) = 0 ).So, let's find the points where ( cos(2x + frac{pi}{3}) = 0 ). That occurs when ( 2x + frac{pi}{3} = frac{pi}{2} + kpi ), where ( k ) is an integer.Solving for ( x ):( 2x = frac{pi}{2} + kpi - frac{pi}{3} )( 2x = frac{3pi}{6} + frac{6kpi}{6} - frac{2pi}{6} )( 2x = frac{(3 + 6k - 2)pi}{6} )( 2x = frac{(1 + 6k)pi}{6} )( x = frac{(1 + 6k)pi}{12} )( x = frac{pi}{12} + frac{kpi}{2} )So, the centers of symmetry are at ( x = frac{pi}{12} + frac{kpi}{2} ) for integer ( k ). Each of these points is a center of symmetry, meaning the function is symmetric about these points.Therefore, the centers of symmetry are at ( left( frac{pi}{12} + frac{kpi}{2}, 0 right) ) for any integer ( k ).Alright, that seems to make sense. So, summarizing part (1):- The function ( f(x) = cos(2x + frac{pi}{3}) ) is increasing on intervals ( (kpi - frac{2pi}{3}, kpi - frac{pi}{6}) ) and decreasing on intervals ( (kpi - frac{pi}{6}, kpi + frac{pi}{3}) ) for any integer ( k ).- The centers of symmetry are at points ( left( frac{pi}{12} + frac{kpi}{2}, 0 right) ) for any integer ( k ).Moving on to part (2): The equation ( 2cos^2x + mcos x + 2 = 0 ) has real solutions for ( x in (0, frac{pi}{2}) ). We need to find the range of the real number ( m ).First, let me note that ( x ) is in the interval ( (0, frac{pi}{2}) ), so ( cos x ) is positive and decreasing from 1 to 0. Therefore, ( cos x ) takes values in ( (0, 1) ).Let me set ( t = cos x ). Then, ( t in (0, 1) ). The equation becomes:( 2t^2 + mt + 2 = 0 )We need this quadratic equation in ( t ) to have real solutions in ( t in (0, 1) ).First, for the quadratic equation ( 2t^2 + mt + 2 = 0 ) to have real solutions, the discriminant must be non-negative.Discriminant ( D = m^2 - 16 geq 0 )So, ( m^2 geq 16 )Which implies ( m leq -4 ) or ( m geq 4 ).But since ( t in (0, 1) ), we need to ensure that at least one of the roots lies in ( (0, 1) ).Let me denote the quadratic function as ( f(t) = 2t^2 + mt + 2 ).We need ( f(t) = 0 ) to have at least one root in ( (0, 1) ).Given that ( f(t) ) is a quadratic opening upwards (since the coefficient of ( t^2 ) is positive), the function will have a minimum between its roots.But since the leading coefficient is positive, if both roots are real, the function will be negative between the roots and positive outside.Given that ( f(0) = 2(0)^2 + m(0) + 2 = 2 > 0 )And ( f(1) = 2(1)^2 + m(1) + 2 = 2 + m + 2 = m + 4 )So, for the quadratic to cross the t-axis between 0 and 1, we need ( f(1) < 0 ), because ( f(0) > 0 ). If ( f(1) < 0 ), then by the Intermediate Value Theorem, there is a root between 0 and 1.So, ( f(1) = m + 4 < 0 )Thus, ( m < -4 )But we also have the discriminant condition ( m^2 geq 16 ), which gives ( m leq -4 ) or ( m geq 4 ). However, since ( m < -4 ) is more restrictive, the range of ( m ) is ( m < -4 ).Wait, let me verify this.If ( m < -4 ), then ( f(1) = m + 4 < 0 ), and since ( f(0) = 2 > 0 ), there must be a root in ( (0, 1) ). Also, the discriminant is positive because ( m^2 > 16 ), so there are two real roots. Since the quadratic opens upwards, one root is between 0 and 1, and the other is less than 0 or greater than 1. But since ( t in (0, 1) ), only the root in ( (0, 1) ) is relevant.Alternatively, if ( m = -4 ), then ( f(1) = 0 ), so ( t = 1 ) is a root. But ( t = 1 ) corresponds to ( x = 0 ), which is not in the open interval ( (0, frac{pi}{2}) ). Therefore, ( m = -4 ) is excluded.Similarly, if ( m geq 4 ), then ( f(1) = m + 4 geq 8 > 0 ), so both ( f(0) ) and ( f(1) ) are positive. Since the quadratic opens upwards, if both roots are real, they must both be less than 0 or both greater than 1. But since ( t in (0, 1) ), there are no roots in ( (0, 1) ) in this case.Therefore, the only possibility for the equation to have a solution in ( t in (0, 1) ) is when ( m < -4 ).So, the range of ( m ) is ( m < -4 ).I think that's it. Let me just recap:- For part (1), we found ( omega = 2 ), determined the intervals where the function is increasing and decreasing, and identified the centers of symmetry.- For part (2), by substituting ( t = cos x ), we transformed the equation into a quadratic in ( t ), analyzed the conditions for real roots in ( (0, 1) ), and concluded that ( m ) must be less than -4.Yeah, that seems solid. I don't think I missed anything here.

💡Okay, so I have this problem where I need to find two integers, a and b, such that the difference between their squares is 144. And I need to figure out the largest possible value of d, which is a + b. Hmm, let me think about how to approach this.First, I remember that the difference of squares can be factored. So, a² - b² equals (a - b)(a + b). That's a useful identity. So, in this case, (a - b)(a + b) = 144. That means that the product of (a - b) and (a + b) is 144.Now, since a and b are integers, both (a - b) and (a + b) must also be integers. Moreover, (a + b) is the value we're trying to maximize, which is d. So, I need to find pairs of integers whose product is 144 and then see which pair gives me the largest possible sum, a + b.Let me list out all the factor pairs of 144. Starting from the smallest:1 and 144,2 and 72,3 and 48,4 and 36,6 and 24,8 and 18,9 and 16,12 and 12.Wait, but since (a - b) and (a + b) are both integers, and (a + b) is larger than (a - b) because a and b are positive integers (I assume they are positive; the problem doesn't specify, but let's go with that for now), so I can consider each of these factor pairs where the first number is (a - b) and the second is (a + b).So, for each pair, I can set up a system of equations:a - b = x,a + b = y,where x and y are factors of 144, and y > x.Then, solving for a and b:Adding the two equations: 2a = x + y => a = (x + y)/2,Subtracting the two equations: 2b = y - x => b = (y - x)/2.Since a and b must be integers, both (x + y) and (y - x) must be even numbers. That means that x and y must be both even or both odd. But looking at the factor pairs of 144, all the factors are even except for 1 and 3, etc. Wait, 144 is even, so all its factors are even or odd? Let me check.144 is 12 squared, which is 2^4 * 3^2. So, all its factors are combinations of 2s and 3s. So, factors can be even or odd. For example, 1 is odd, 3 is odd, 9 is odd, etc. But when paired with another factor, if one is odd, the other must be even because 144 is even. So, actually, in all factor pairs, one will be even and the other will be odd or both even.Wait, but if one is even and the other is odd, then x + y and y - x will be odd, meaning a and b would not be integers. So, to have a and b as integers, both x and y must be even. Because if both are even, then x + y and y - x will be even, so a and b will be integers.So, I need to consider only the factor pairs where both factors are even. Let me go back to the list:1 and 144: 1 is odd, 144 is even. So, discard this pair because one is odd, the other is even.2 and 72: both even. Good.3 and 48: 3 is odd, 48 is even. Discard.4 and 36: both even. Good.6 and 24: both even. Good.8 and 18: both even. Good.9 and 16: 9 is odd, 16 is even. Discard.12 and 12: both even. Good.So, the valid factor pairs where both factors are even are:(2, 72), (4, 36), (6, 24), (8, 18), (12, 12).Now, for each of these pairs, I can compute a and b and see which gives me the largest a + b.Let's start with the first pair: (2, 72).So, a - b = 2,a + b = 72.Adding these equations: 2a = 74 => a = 37.Subtracting: 2b = 70 => b = 35.So, a = 37, b = 35. Then, a + b = 72.Next pair: (4, 36).a - b = 4,a + b = 36.Adding: 2a = 40 => a = 20.Subtracting: 2b = 32 => b = 16.So, a = 20, b = 16. Then, a + b = 36.Hmm, that's smaller than 72.Next pair: (6, 24).a - b = 6,a + b = 24.Adding: 2a = 30 => a = 15.Subtracting: 2b = 18 => b = 9.So, a = 15, b = 9. Then, a + b = 24.Even smaller.Next pair: (8, 18).a - b = 8,a + b = 18.Adding: 2a = 26 => a = 13.Subtracting: 2b = 10 => b = 5.So, a = 13, b = 5. Then, a + b = 18.Still smaller.Last pair: (12, 12).a - b = 12,a + b = 12.Adding: 2a = 24 => a = 12.Subtracting: 2b = 0 => b = 0.Wait, b = 0. Is 0 considered an integer? Yes, it is. So, a = 12, b = 0. Then, a + b = 12.That's the smallest.So, among all these, the largest a + b is 72, from the first valid pair (2, 72).Wait, but let me double-check. Is there any other factor pair that I might have missed where both factors are even and could give a larger a + b?Looking back, I considered all factor pairs where both factors are even. The next possible pair after (2, 72) would be (72, 2), but that's the same as (2, 72) because multiplication is commutative. So, no, I don't think I missed any.But just to be thorough, let me consider if negative integers are allowed. The problem says integers, so a and b could be negative as well.If a and b are negative, then a + b could be negative, but we're looking for the largest possible value of d = a + b, so the maximum would still be 72, because if a and b are negative, their sum would be negative, which is smaller than 72.Wait, unless a is positive and b is negative, or vice versa. Let me think about that.If a is positive and b is negative, then a + b could be larger than 72? Let me check.Suppose a is positive and b is negative. Let me denote b as -c, where c is positive. Then, a² - b² = a² - c² = 144.So, same as before, (a - c)(a + c) = 144.But in this case, a and c are positive integers, so a + c is positive, and a - c is positive as well because a > c (since a² > c²).So, same as before, factor pairs where both factors are even.So, same result, a + c = 72, a - c = 2, leading to a = 37, c = 35, so b = -35.Thus, a + b = 37 + (-35) = 2, which is smaller than 72.Alternatively, if a is negative and b is positive, similar result.So, regardless, the maximum a + b is 72.Wait, but let me think again. If a is positive and b is negative, could a + b be larger than 72?Suppose a is very large and b is a large negative number such that a² - b² = 144.But actually, since a² - b² = (a - b)(a + b) = 144, and if a is positive and b is negative, then a - b is a + |b|, which is larger than a + b.Wait, let me clarify.If b is negative, say b = -c where c > 0, then a² - b² = a² - c² = 144.So, (a - c)(a + c) = 144.In this case, a + c is still positive, and a - c is positive as well because a > c.So, same as before, the factor pairs are the same.But when we solve for a and c, we get a = (x + y)/2 and c = (y - x)/2, where x and y are the factors.Then, a + b = a + (-c) = a - c.Which is equal to x, which is one of the factors.So, in the first case, when we took x = 2 and y = 72, a + b = 72, which is the maximum.But if we consider x = 72 and y = 2, then a + b = 2, which is smaller.So, regardless of how we arrange the factors, the maximum a + b is 72.Therefore, I think 72 is indeed the largest possible value of d = a + b.But just to make sure, let me see if there are any other factor pairs where both factors are even and could result in a larger a + b.Wait, 144 can also be factored as ( -2, -72 ), but that would give a negative sum, which is not larger than 72.Similarly, other negative factor pairs would result in negative sums, which are smaller than 72.So, no, I don't think there's a way to get a larger sum than 72.Therefore, the largest possible value of d = a + b is 72.

💡Okay, so I have this function g(x) which is equal to (x - 2) divided by (x - 5). The question is asking me to find the value of x where the inverse of g, which is g⁻¹(x), is undefined. Hmm, okay, let's break this down step by step.First off, I remember that for a function and its inverse, the domain of the original function becomes the range of the inverse function, and vice versa. So, if I can figure out where g(x) is undefined, that might help me understand where g⁻¹(x) is undefined. But wait, the question is specifically about g⁻¹(x), so maybe I should focus on finding the inverse function first and then determine where it's undefined.Alright, let's try to find the inverse of g(x). To find the inverse, I need to switch the roles of x and y and then solve for y. So, starting with the equation:y = (x - 2)/(x - 5)To find the inverse, I'll swap x and y:x = (y - 2)/(y - 5)Now, I need to solve this equation for y. Let's do that step by step. Multiply both sides of the equation by (y - 5) to get rid of the denominator:x(y - 5) = y - 2Now, distribute x on the left side:xy - 5x = y - 2Hmm, now I have terms with y on both sides. Let me get all the y terms on one side and the constants on the other. So, subtract y from both sides and add 5x to both sides:xy - y = 5x - 2Factor out y from the left side:y(x - 1) = 5x - 2Now, solve for y by dividing both sides by (x - 1):y = (5x - 2)/(x - 1)So, the inverse function g⁻¹(x) is (5x - 2)/(x - 1). Now, I need to find where this inverse function is undefined. A function is undefined where its denominator is zero because division by zero is not allowed. So, let's set the denominator equal to zero and solve for x:x - 1 = 0Adding 1 to both sides:x = 1So, g⁻¹(x) is undefined at x = 1. Therefore, the value of x where g⁻¹(x) is undefined is 1.Wait a second, let me double-check my steps to make sure I didn't make a mistake. Starting from the inverse function:g⁻¹(x) = (5x - 2)/(x - 1)The denominator is x - 1, so setting that equal to zero gives x = 1. That seems straightforward. Let me also consider the original function g(x). Its denominator is x - 5, so g(x) is undefined at x = 5. But since we're dealing with the inverse function, the undefined point should correspond to the original function's range. The range of g(x) is all real numbers except for the value that makes the inverse function undefined. Let me see if that aligns.The original function g(x) = (x - 2)/(x - 5). To find its range, I can set y = (x - 2)/(x - 5) and solve for x in terms of y, which is essentially what I did when finding the inverse. The inverse function is g⁻¹(y) = (5y - 2)/(y - 1). So, the range of g(x) is all real numbers except y = 1 because that's where the inverse function is undefined. Therefore, the inverse function g⁻¹(x) is undefined at x = 1, which matches what I found earlier.Just to be thorough, let me plug x = 1 into the inverse function:g⁻¹(1) = (5*1 - 2)/(1 - 1) = (5 - 2)/0 = 3/0Which is undefined, as expected. So, everything checks out. I think I've got the right answer here.Another way to think about it is that the inverse function essentially "undoes" what the original function does. So, if the original function has a vertical asymptote at x = 5, the inverse function should have a horizontal asymptote corresponding to that. Wait, actually, the horizontal asymptote of the original function would become the vertical asymptote of the inverse function. Let me recall, for rational functions, the horizontal asymptote depends on the degrees of the numerator and denominator. Both the numerator and denominator of g(x) are degree 1, so the horizontal asymptote is the ratio of the leading coefficients, which is 1/1 = 1. Therefore, the inverse function should have a vertical asymptote at x = 1, which is exactly where it's undefined. That makes sense.So, putting it all together, after finding the inverse function and analyzing its undefined points, as well as considering the asymptotes and the relationship between a function and its inverse, I'm confident that the value of x where g⁻¹(x) is undefined is indeed 1.**Final Answer**The value of ( x ) where ( g^{-1}(x) ) is undefined is boxed{1}.

💡Okay, so I have this function ( g(x) = frac{x - 5}{x - 6} ), and I need to find for what value of ( x ) its inverse function ( g^{-1}(x) ) is undefined. Hmm, let me think about how to approach this.First, I remember that to find the inverse of a function, I usually switch the roles of ( x ) and ( y ) and then solve for ( y ). So, let me start by writing ( y = frac{x - 5}{x - 6} ). Now, to find the inverse, I need to solve this equation for ( x ) in terms of ( y ).Alright, so I'll set up the equation:[ y = frac{x - 5}{x - 6} ]Now, I need to solve for ( x ). To do that, I can cross-multiply to get rid of the denominator. Multiplying both sides by ( x - 6 ) gives:[ y(x - 6) = x - 5 ]Expanding the left side:[ yx - 6y = x - 5 ]Now, I want to collect all the terms containing ( x ) on one side and the rest on the other side. So, I'll subtract ( x ) from both sides and add ( 6y ) to both sides:[ yx - x = 6y - 5 ]Factor out ( x ) from the left side:[ x(y - 1) = 6y - 5 ]Now, solve for ( x ) by dividing both sides by ( y - 1 ):[ x = frac{6y - 5}{y - 1} ]So, this expression is the inverse function ( g^{-1}(y) ). To write it in terms of ( x ), I can replace ( y ) with ( x ):[ g^{-1}(x) = frac{6x - 5}{x - 1} ]Now, I need to determine when this inverse function is undefined. A function is undefined when its denominator equals zero because division by zero is not allowed. So, I'll set the denominator equal to zero and solve for ( x ):[ x - 1 = 0 ]Adding 1 to both sides:[ x = 1 ]Therefore, ( g^{-1}(x) ) is undefined when ( x = 1 ).Wait, let me double-check that. If I plug ( x = 1 ) into ( g^{-1}(x) ), the denominator becomes zero, which makes the function undefined. That seems correct.Just to be thorough, I'll verify by checking the original function ( g(x) ). The inverse function ( g^{-1}(x) ) should map outputs of ( g(x) ) back to their inputs. So, if ( g^{-1}(1) ) is undefined, that means there's no ( x ) such that ( g(x) = 1 ). Let me check that:Set ( g(x) = 1 ):[ frac{x - 5}{x - 6} = 1 ]Multiply both sides by ( x - 6 ):[ x - 5 = x - 6 ]Subtract ( x ) from both sides:[ -5 = -6 ]That's not true, which means there's no solution. So, indeed, ( g(x) ) never equals 1, which means ( g^{-1}(1) ) is undefined. That confirms my earlier result.I think I've got it right. The inverse function ( g^{-1}(x) ) is undefined at ( x = 1 ).

💡Okay, so I have this problem here: I'm given a hyperbola with the equation (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where (a > 0) and (b > 0). I'm also told that the line (2x - sqrt{3}y = 0) is an asymptote of this hyperbola. I need to find the eccentricity of the hyperbola.Hmm, let me recall what I know about hyperbolas. The standard form of a hyperbola centered at the origin opening horizontally is indeed (frac{x^2}{a^2} - frac{y^2}{b^2} = 1). The asymptotes of such a hyperbola are the lines that the hyperbola approaches but never touches. For this standard hyperbola, the equations of the asymptotes are (y = pm frac{b}{a}x). So, the given asymptote is (2x - sqrt{3}y = 0). Let me rewrite that in slope-intercept form to make it easier to compare with the asymptote equations I know. Starting with (2x - sqrt{3}y = 0), I can solve for (y):[2x = sqrt{3}y y = frac{2}{sqrt{3}}x]So, the slope of this asymptote is (frac{2}{sqrt{3}}). Now, comparing this to the asymptotes of the hyperbola, which are (y = pm frac{b}{a}x), I can see that the slopes must be equal. Therefore, (frac{b}{a} = frac{2}{sqrt{3}}). Let me write that down:[frac{b}{a} = frac{2}{sqrt{3}}]I can solve for (b) in terms of (a):[b = a cdot frac{2}{sqrt{3}} b = frac{2a}{sqrt{3}}]Now, I need to find the eccentricity of the hyperbola. The formula for the eccentricity (e) of a hyperbola is:[e = sqrt{1 + frac{b^2}{a^2}}]I already have (frac{b}{a}), so I can square that to find (frac{b^2}{a^2}):[left(frac{b}{a}right)^2 = left(frac{2}{sqrt{3}}right)^2 = frac{4}{3}]So, substituting back into the eccentricity formula:[e = sqrt{1 + frac{4}{3}} = sqrt{frac{3}{3} + frac{4}{3}} = sqrt{frac{7}{3}}]Wait, (sqrt{frac{7}{3}}) can be simplified. Let me rationalize the denominator:[sqrt{frac{7}{3}} = frac{sqrt{7}}{sqrt{3}} = frac{sqrt{7} cdot sqrt{3}}{sqrt{3} cdot sqrt{3}} = frac{sqrt{21}}{3}]So, the eccentricity (e) is (frac{sqrt{21}}{3}).Let me just double-check my steps to make sure I didn't make any mistakes. 1. I started by recalling the standard form of the hyperbola and its asymptotes.2. I converted the given asymptote equation into slope-intercept form to find its slope.3. I set the slope equal to (frac{b}{a}) because that's the slope of the asymptotes for the hyperbola.4. I solved for (b) in terms of (a).5. I used the formula for eccentricity, substituting (frac{b^2}{a^2}) with the square of the slope.6. I simplified the square root to get the final answer.Everything seems to check out. I don't think I made any calculation errors. So, I'm confident that the eccentricity is (frac{sqrt{21}}{3}).**Final Answer**The eccentricity of the hyperbola is (boxed{dfrac{sqrt{21}}{3}}).