💡Okay, so I have this problem here where I need to find the estimated value of y when x is 25, using the linear regression equation y = 0.50x - 0.81. Alright, let's break this down step by step because I want to make sure I understand it correctly.First, I know that a linear regression equation is a straight line that best fits the data points. It has the form y = mx + b, where m is the slope and b is the y-intercept. In this case, the equation is already given as y = 0.50x - 0.81. So, m is 0.50, and b is -0.81. That means for every unit increase in x, y increases by 0.50 units, and when x is 0, y is -0.81.Now, I need to find the value of y when x is 25. That sounds straightforward enough. I just need to plug in x = 25 into the equation and solve for y. Let me write that out:y = 0.50 * 25 - 0.81Okay, so first, I'll calculate 0.50 multiplied by 25. Hmm, 0.50 is the same as half, so half of 25 is 12.5. So, 0.50 * 25 = 12.5.Now, I need to subtract 0.81 from that result. So, 12.5 minus 0.81. Let me do that subtraction carefully to avoid any mistakes. 12.5 is the same as 12.50, so subtracting 0.81 from 12.50:12.50- 0.81--------11.69So, 12.50 minus 0.81 equals 11.69. Therefore, y should be 11.69 when x is 25.Wait, let me double-check my calculations to make sure I didn't make any errors. Multiplying 0.50 by 25 gives 12.5, which seems correct. Then subtracting 0.81 from 12.5, I get 11.69. That seems right.Just to be thorough, I'll think about what this means in the context of the problem. If I have a linear regression model, it's predicting y based on x. So, when x increases by 1 unit, y increases by 0.50 units. Starting from x = 0, y is -0.81, which is the y-intercept. As x increases, y increases proportionally.So, when x is 25, y is predicted to be 11.69. That seems reasonable given the slope and intercept provided. I don't see any immediate red flags in my calculations, so I think I'm confident with this answer.I guess another way to look at it is to think about the units involved. If x represents, say, years, and y represents some measurement, then each year adds 0.50 units to y. Starting from a baseline of -0.81 units when x is 0, after 25 years, y would have increased by 12.5 units, but since we started below zero, subtracting 0.81 brings it down to 11.69.Yeah, that makes sense. I think I've got it right. The estimated value of y when x is 25 is 11.69.

💡First, I'll define the variables for the amounts each person paid: let J represent Joan's payment, K represent Karl's payment, and L represent Lea's payment.From the problem, I have three pieces of information:1. The total amount paid by Joan, Karl, and Lea is 600.2. Twice what Joan paid is 74 more than what Karl paid.3. The difference between what Lea paid and what Karl paid is 52.I'll translate these statements into mathematical equations:1. J + K + L = 6002. 2J = K + 743. L - K = 52Next, I'll solve these equations step by step. Starting with the third equation, I can express L in terms of K:L = K + 52Substituting this expression for L into the first equation:J + K + (K + 52) = 600Simplifying this, I get:J + 2K + 52 = 600Now, I'll solve the second equation for J:2J = K + 74J = (K + 74) / 2Substituting this expression for J into the equation J + 2K + 52 = 600:(K + 74)/2 + 2K + 52 = 600To eliminate the fraction, I'll multiply the entire equation by 2:K + 74 + 4K + 104 = 1200Combining like terms:5K + 178 = 1200Subtracting 178 from both sides:5K = 1022Dividing both sides by 5:K = 204.4Now that I have the value of K, I'll substitute it back into the expression for J:J = (204.4 + 74) / 2J = 278.4 / 2J = 139.2Therefore, Joan paid 139.20 for her telescope.

💡First, I recognize that the given expression involves squaring a large number and multiplying two numbers that are close to it. To simplify the calculation, I can use the difference of squares formula.I notice that (123456788) and (123456790) are consecutive even integers surrounding (123456789). This allows me to express them as (123456789 - 1) and (123456789 + 1), respectively.By applying the difference of squares formula (a^2 - b^2 = (a - b)(a + b)), I can rewrite the original expression in a simpler form. This transformation will help eliminate the large numbers and make the calculation more manageable.After simplifying the expression using the formula, I will perform the arithmetic operations to find the final result.

💡Okay, so I have this equation here: ((x^2 - k)(x + k) = x^3 + k(x^2 - x - 2)), and I need to find the value of (k) given that (k neq 0). Hmm, let me try to figure this out step by step.First, I think I should expand the left side of the equation to see what it looks like. So, I'll multiply ((x^2 - k)) by ((x + k)). Let me recall how to multiply two binomials. It's like the FOIL method, but since these are a bit more complex, I'll just distribute each term.Starting with (x^2) multiplied by each term in ((x + k)):- (x^2 cdot x = x^3)- (x^2 cdot k = kx^2)Now, moving on to (-k) multiplied by each term in ((x + k)):- (-k cdot x = -kx)- (-k cdot k = -k^2)So, putting it all together, the expanded form of the left side is:[x^3 + kx^2 - kx - k^2]Alright, now let me look at the right side of the equation: (x^3 + k(x^2 - x - 2)). I should also expand this to make it easier to compare with the left side. Let's distribute the (k) inside the parentheses:- (k cdot x^2 = kx^2)- (k cdot (-x) = -kx)- (k cdot (-2) = -2k)So, the expanded right side becomes:[x^3 + kx^2 - kx - 2k]Now, I have both sides expanded:- Left side: (x^3 + kx^2 - kx - k^2)- Right side: (x^3 + kx^2 - kx - 2k)Hmm, they look quite similar except for the last terms. On the left side, the last term is (-k^2), and on the right side, it's (-2k). Since these two expressions are equal for all values of (x), their corresponding coefficients must be equal. That means the coefficients of (x^3), (x^2), (x), and the constant terms must all be the same on both sides.Looking at the coefficients:- For (x^3): Both sides have 1, so that's good.- For (x^2): Both sides have (k), so that's consistent.- For (x): Both sides have (-k), so that's also consistent.- For the constant term: Left side is (-k^2) and right side is (-2k).So, the key equation I need to solve is:[-k^2 = -2k]Let me simplify this equation. I can multiply both sides by (-1) to make it easier:[k^2 = 2k]Now, I'll bring all terms to one side to set the equation to zero:[k^2 - 2k = 0]I can factor out a (k) from both terms:[k(k - 2) = 0]This gives me two possible solutions:1. (k = 0)2. (k - 2 = 0 Rightarrow k = 2)But wait, the problem states that (k neq 0), so the only valid solution is (k = 2).Let me double-check my work to make sure I didn't make any mistakes. I expanded both sides correctly, compared the coefficients, and solved the resulting equation. Everything seems to add up. So, I'm confident that (k = 2) is the correct answer.**Final Answer**The value of (k) is boxed{2}.

💡Okay, so I'm trying to solve this problem about the function ( f(x) = sqrt{3}cos(omega x - frac{pi}{2}) - cos(omega x) ) where ( 0 < omega < 3 ). The graph of this function passes through the point ( Pleft( frac{pi}{3}, 0 right) ). The question is asking how we should shift the graph of ( f(x) ) to make it an odd function. The options are shifting left or right by ( frac{2pi}{3} ) or ( frac{pi}{3} ) units.First, I need to understand what an odd function is. An odd function satisfies the condition ( f(-x) = -f(x) ) for all ( x ) in its domain. So, if I can shift the graph of ( f(x) ) such that this condition holds, then it will be an odd function.Looking at the given function ( f(x) ), it seems like it's a combination of cosine functions with different phases. Maybe I can simplify it using trigonometric identities. I remember that ( cos(theta - frac{pi}{2}) ) is equal to ( sin(theta) ). Let me verify that:Yes, ( cos(theta - frac{pi}{2}) = sin(theta) ) because cosine is shifted by ( frac{pi}{2} ) to become sine. So, substituting that into the function:( f(x) = sqrt{3}sin(omega x) - cos(omega x) )Hmm, this looks like a combination of sine and cosine with the same argument ( omega x ). Maybe I can express this as a single sine or cosine function using the amplitude-phase form. The general form is ( Asin(omega x + phi) ) or ( Acos(omega x + phi) ). Let me try that.The expression ( sqrt{3}sin(omega x) - cos(omega x) ) can be written as ( Rsin(omega x + phi) ), where ( R ) is the amplitude and ( phi ) is the phase shift. To find ( R ) and ( phi ), I can use the identities:( R = sqrt{(sqrt{3})^2 + (-1)^2} = sqrt{3 + 1} = sqrt{4} = 2 )And the phase shift ( phi ) can be found using:( tan(phi) = frac{-1}{sqrt{3}} )So, ( phi = arctanleft( frac{-1}{sqrt{3}} right) ). I know that ( arctanleft( frac{1}{sqrt{3}} right) = frac{pi}{6} ), so since the tangent is negative, the angle is in the fourth quadrant. Therefore, ( phi = -frac{pi}{6} ).Putting it all together, the function becomes:( f(x) = 2sinleft( omega x - frac{pi}{6} right) )Now, the graph passes through the point ( Pleft( frac{pi}{3}, 0 right) ). That means when ( x = frac{pi}{3} ), ( f(x) = 0 ). Let's substitute these values into the equation:( 0 = 2sinleft( omega cdot frac{pi}{3} - frac{pi}{6} right) )Divide both sides by 2:( 0 = sinleft( frac{omega pi}{3} - frac{pi}{6} right) )The sine function is zero when its argument is an integer multiple of ( pi ):( frac{omega pi}{3} - frac{pi}{6} = kpi ) where ( k ) is an integer.Let's solve for ( omega ):( frac{omega pi}{3} = kpi + frac{pi}{6} )Divide both sides by ( pi ):( frac{omega}{3} = k + frac{1}{6} )Multiply both sides by 3:( omega = 3k + frac{1}{2} )Given that ( 0 < omega < 3 ), let's find the possible values of ( k ):If ( k = 0 ), then ( omega = frac{1}{2} ).If ( k = 1 ), then ( omega = 3 + frac{1}{2} = frac{7}{2} ), which is greater than 3, so it's out of the given range.Therefore, the only valid value is ( omega = frac{1}{2} ).So, the function simplifies to:( f(x) = 2sinleft( frac{1}{2}x - frac{pi}{6} right) )Now, we need to make this function odd by shifting it. Let me recall that shifting a function horizontally can affect its symmetry. To make it odd, we need to adjust the phase shift so that the resulting function satisfies ( f(-x) = -f(x) ).Let's denote the shifted function as ( f(x - h) ) where ( h ) is the shift amount. If ( h ) is positive, it's a shift to the right; if negative, to the left.So, ( f(x - h) = 2sinleft( frac{1}{2}(x - h) - frac{pi}{6} right) )Simplify:( f(x - h) = 2sinleft( frac{1}{2}x - frac{h}{2} - frac{pi}{6} right) )We want this function to be odd. For a sine function, ( sin(theta) ) is odd, but if there's a phase shift, it might not be. So, to make the entire function odd, the phase shift should be such that the function becomes symmetric about the origin.Alternatively, another approach is to express the function as ( 2sinleft( frac{1}{2}x - phi right) ) and find ( phi ) such that the function is odd.But perhaps a better way is to consider the general form of an odd function. Since sine is odd, if we can write the function as a sine function without any phase shift, it would be odd. So, perhaps shifting the function so that the phase shift cancels out.Looking back at the function:( f(x) = 2sinleft( frac{1}{2}x - frac{pi}{6} right) )If we can shift it such that the phase shift ( -frac{pi}{6} ) is eliminated, then the function would be ( 2sinleft( frac{1}{2}x right) ), which is odd.To eliminate the phase shift, we need to shift the function to the left by an amount that corresponds to the phase shift. The phase shift in the argument is ( -frac{pi}{6} ), which corresponds to a shift to the right by ( frac{pi}{6} times frac{2}{1} = frac{pi}{3} ) units because the coefficient of ( x ) is ( frac{1}{2} ). Wait, let me think about that.The general form is ( sin(Bx - C) ), which can be written as ( sin(B(x - frac{C}{B})) ). So, the phase shift is ( frac{C}{B} ). In our case, ( B = frac{1}{2} ) and ( C = frac{pi}{6} ). Therefore, the phase shift is ( frac{frac{pi}{6}}{frac{1}{2}} = frac{pi}{3} ). So, the function is shifted to the right by ( frac{pi}{3} ).To make the function odd, we need to reverse this shift, meaning shifting it to the left by ( frac{pi}{3} ) units. Let me verify this.If we shift ( f(x) ) to the left by ( frac{pi}{3} ), the new function becomes:( fleft(x + frac{pi}{3}right) = 2sinleft( frac{1}{2}left(x + frac{pi}{3}right) - frac{pi}{6} right) )Simplify inside the sine:( frac{1}{2}x + frac{pi}{6} - frac{pi}{6} = frac{1}{2}x )So, the function becomes:( 2sinleft( frac{1}{2}x right) )Which is indeed an odd function because ( sin ) is odd and the coefficient 2 doesn't affect the oddness.Therefore, shifting the graph of ( f(x) ) to the left by ( frac{pi}{3} ) units will make it an odd function.Looking back at the options, option C is shifting to the left by ( frac{pi}{3} ) units, which matches our conclusion.I think that's the correct answer. Let me just recap to make sure I didn't make any mistakes.1. Simplified the original function using trigonometric identities.2. Found ( omega ) by substituting the point ( P ).3. Expressed the function in a simpler sine form.4. Determined the phase shift and how to eliminate it by shifting the graph.5. Verified that the resulting function is odd.Everything seems to check out. So, I'm confident that option C is correct.**Final Answer**boxed{C}

💡First, I need to solve the inequality (frac{x - 2}{3} leq x).To eliminate the denominator, I'll multiply both sides of the inequality by 3:[frac{x - 2}{3} times 3 leq x times 3]This simplifies to:[x - 2 leq 3x]Next, I'll move all terms involving (x) to one side by subtracting (x) from both sides:[x - 2 - x leq 3x - x]This simplifies to:[-2 leq 2x]Now, I'll solve for (x) by dividing both sides by 2:[frac{-2}{2} leq frac{2x}{2}]This gives:[-1 leq x]Finally, to express the solution in terms of (x), I'll rewrite it as:[x geq -1]So, the solution set of the inequality is (x geq -1).