💡To solve the expression (frac{10^7}{5 times 10^4}), we can break it down into simpler steps.First, separate the powers of ten and the constant:[frac{10^7}{5 times 10^4} = frac{10^7}{10^4} times frac{1}{5}]Next, simplify the powers of ten using the laws of exponents:[frac{10^7}{10^4} = 10^{7-4} = 10^3]Now, substitute back into the expression:[frac{10^3}{5} = frac{1000}{5} = 200]Therefore, the value of the expression is 200, which corresponds to choice (D).

💡Okay, so I have this problem here where I need to find the maximum value of the product xy, given that x and y are both positive numbers and they satisfy the equation x + 2y = 1. Hmm, let me think about how to approach this.First, I know that when dealing with optimization problems like finding maximum or minimum values, calculus is often useful. Maybe I can express one variable in terms of the other using the given equation and then substitute it into the product xy to get a function of a single variable. Then, I can take the derivative and find the critical points to determine the maximum.Let me try that. The equation given is x + 2y = 1. I can solve for x in terms of y: x = 1 - 2y. Since both x and y are positive, 1 - 2y must be greater than 0, which means y < 1/2. So y has to be between 0 and 1/2.Now, substitute x = 1 - 2y into the product xy:xy = (1 - 2y)y = y - 2y²So, the product is a quadratic function in terms of y: f(y) = y - 2y². To find its maximum, I can take the derivative with respect to y and set it equal to zero.f'(y) = d/dy (y - 2y²) = 1 - 4ySetting the derivative equal to zero for critical points:1 - 4y = 04y = 1y = 1/4Okay, so y = 1/4 is a critical point. Since the function f(y) is a quadratic opening downward (because the coefficient of y² is negative), this critical point must be a maximum.Now, let's find the corresponding x value. From x = 1 - 2y:x = 1 - 2*(1/4) = 1 - 1/2 = 1/2So, when y = 1/4, x = 1/2. Now, let's compute the product xy:xy = (1/2)*(1/4) = 1/8Hmm, so the maximum value of xy is 1/8. Let me double-check to make sure I didn't make any mistakes.Alternatively, I remember that for quadratic functions, the vertex occurs at y = -b/(2a). In this case, the quadratic is f(y) = -2y² + y, so a = -2 and b = 1. Plugging into the vertex formula:y = -1/(2*(-2)) = -1/(-4) = 1/4So, that confirms that the maximum occurs at y = 1/4, and the maximum value is indeed 1/8.Wait, another thought: maybe I can use the AM-GM inequality here? The Arithmetic Mean-Geometric Mean inequality states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. Let me see if that applies here.Given x + 2y = 1, I can consider the two terms x and 2y. Applying AM-GM:(x + 2y)/2 ≥ √(x * 2y)Which simplifies to:1/2 ≥ √(2xy)Squaring both sides:1/4 ≥ 2xySo,xy ≤ 1/8That's the same result as before! So, using AM-GM also gives me that the maximum value of xy is 1/8, achieved when x = 2y. Since x + 2y = 1, that means x = 2y, so substituting back, x = 2y and x + 2y = 1 gives 2y + 2y = 1, so 4y = 1, y = 1/4, and x = 1/2. Yep, that matches my earlier calculation.I think I'm confident now that 1/8 is the correct maximum value of xy under the given constraints. So, the answer should be option B: 1/8.**Final Answer**boxed{B}

💡Okay, so I have this quadratic function here: y equals 2 times (x minus 2) squared minus 5. I need to shift its graph 2 units to the right and then 3 units up. Hmm, let me think about how to do this step by step.First, I remember that when you shift a graph horizontally, it affects the x inside the function. Specifically, shifting to the right by a certain number of units means you replace x with (x minus that number). So, if I want to shift it 2 units to the right, I should replace x with (x minus 2) in the function.Let me write that down. The original function is y = 2(x - 2)^2 - 5. If I shift it 2 units to the right, I substitute x with (x - 2). So, it becomes y = 2((x - 2) - 2)^2 - 5. Simplifying inside the parentheses, that's (x - 4). So now the function is y = 2(x - 4)^2 - 5.Okay, that seems right. Shifting to the right by 2 units increased the value being subtracted from x from 2 to 4. That makes sense because shifting right should make the graph appear further along the x-axis.Now, the next transformation is shifting the graph 3 units up. I recall that shifting up affects the constant term at the end of the function. Specifically, shifting up by a certain number means you add that number to the entire function. So, in this case, I need to add 3 to the current function.So, taking the function after the horizontal shift, which is y = 2(x - 4)^2 - 5, and adding 3 to it, it becomes y = 2(x - 4)^2 - 5 + 3. Simplifying that, -5 plus 3 is -2. So the final function is y = 2(x - 4)^2 - 2.Wait, let me double-check that. Starting with y = 2(x - 2)^2 - 5, shifting 2 units right gives y = 2(x - 4)^2 - 5, and then shifting 3 units up adds 3, resulting in y = 2(x - 4)^2 - 2. Yeah, that seems correct.Just to make sure I didn't make any mistakes, let me think about what each transformation does. Shifting right by 2 units moves the vertex of the parabola from (2, -5) to (4, -5). Then, shifting up by 3 units moves the vertex from (4, -5) to (4, -2). So, the new vertex is at (4, -2), which matches the function y = 2(x - 4)^2 - 2.Alright, I think I got it right. The transformations were applied correctly, and the final function reflects both the horizontal and vertical shifts.

💡Okay, so I need to figure out whether the function ( f(x) = 2^{x^2 - 4} + x ) is even, odd, or neither. Hmm, I remember that even functions satisfy the condition ( f(-x) = f(x) ) for all ( x ) in their domain, and odd functions satisfy ( f(-x) = -f(x) ). If neither of these conditions hold, then the function is neither even nor odd.Alright, let me start by finding ( f(-x) ). So, I'll substitute (-x) into the function wherever there's an ( x ). That gives me:[ f(-x) = 2^{(-x)^2 - 4} + (-x) ]Wait, let me simplify that. The exponent part is ((-x)^2 - 4). Since squaring a negative number gives a positive result, ((-x)^2) is the same as (x^2). So, the exponent simplifies to (x^2 - 4). That means the first term becomes (2^{x^2 - 4}), just like in the original function.Now, the second term is (-x). So putting it all together, I have:[ f(-x) = 2^{x^2 - 4} - x ]Okay, so now I need to compare this to the original function ( f(x) = 2^{x^2 - 4} + x ). Let's see:1. Is ( f(-x) = f(x) )? - ( f(-x) = 2^{x^2 - 4} - x ) - ( f(x) = 2^{x^2 - 4} + x ) - These are not equal because of the (-x) versus (+x). So, it's not even.2. Is ( f(-x) = -f(x) )? - Let's compute (-f(x)): [ -f(x) = - (2^{x^2 - 4} + x) = -2^{x^2 - 4} - x ] - Compare this to ( f(-x) = 2^{x^2 - 4} - x ) - These are not equal either because the first term has a negative exponent in (-f(x)) but not in ( f(-x) ). So, it's not odd.Since neither condition is satisfied, the function must be neither even nor odd.Wait, let me double-check my steps to make sure I didn't make a mistake. I substituted (-x) correctly into the function, simplified the exponent correctly, and then compared the results. Everything seems to check out. So, yeah, I think I'm confident that the function is neither even nor odd.

💡First, determine Marie's daily income from selling bread and cakes. Let ( P ) be the price of each loaf of bread.Marie sells 40 loaves of bread daily, so her income from bread is ( 40P ).She sells 6 cakes at 12 each, so her income from cakes is ( 6 times 12 = 72 ).Her total daily income is ( 40P + 72 ).Next, calculate her daily expenses. Marie pays 20 for rent and 2 for electricity, totaling 22.Her net daily profit is her total income minus expenses:[text{Net profit} = (40P + 72) - 22 = 40P + 50]It takes 8 days to cover the cost of the cash register, which is 1040. Therefore, the total profit over 8 days is:[8 times (40P + 50) = 1040]Solve for ( P ):[320P + 400 = 1040][320P = 640][P = frac{640}{320} = 2]Marie sells each loaf of bread for 2.

💡Alright, so I have this problem about an ellipse, and I need to find the value of ( m ). The equation given is ( frac{x^2}{m} + frac{y^2}{4} = 1 ), and the eccentricity is 2. Hmm, okay. Let me try to remember what I know about ellipses.First, I recall that the standard form of an ellipse is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where ( a ) and ( b ) are the lengths of the semi-major and semi-minor axes, respectively. Depending on whether the major axis is along the x-axis or y-axis, ( a ) and ( b ) switch places. If ( a > b ), the major axis is along the x-axis, and if ( b > a ), it's along the y-axis.In this problem, the equation is ( frac{x^2}{m} + frac{y^2}{4} = 1 ). So, comparing this to the standard form, ( a^2 = m ) and ( b^2 = 4 ). That means ( a = sqrt{m} ) and ( b = 2 ). Now, the eccentricity ( e ) of an ellipse is given by ( e = frac{c}{a} ), where ( c ) is the distance from the center to each focus. I also remember that for an ellipse, ( c^2 = a^2 - b^2 ) if ( a > b ), or ( c^2 = b^2 - a^2 ) if ( b > a ). Wait, the problem states that the eccentricity is 2. So, ( e = 2 ). But hold on, isn't the eccentricity of an ellipse always less than 1? Because for an ellipse, ( e ) is between 0 and 1, right? If ( e ) is 2, that would actually be a hyperbola, not an ellipse. Hmm, this is confusing. Maybe I'm missing something.Let me double-check. The equation is ( frac{x^2}{m} + frac{y^2}{4} = 1 ). If ( m ) is positive, this is an ellipse. But if ( m ) is negative, it would be a hyperbola. But the problem says it's an ellipse, so ( m ) must be positive. So, maybe the eccentricity being 2 is a typo, or perhaps there's a different interpretation.Wait, maybe I misread the problem. Let me check again. It says, "The eccentricity of the ellipse ( frac{x^2}{m} + frac{y^2}{4} = 1 ) is 2. What is the value of ( m )?" So, it's definitely stated as an ellipse with eccentricity 2. That seems contradictory because, as I recall, for an ellipse, ( e < 1 ). Is it possible that the problem is referring to a different kind of conic section? Or maybe there's a miscalculation here. Let me think. If ( e = 2 ), which is greater than 1, that would make it a hyperbola. So, perhaps the problem has a mistake, or maybe I'm supposed to consider it as a hyperbola? But the problem explicitly says it's an ellipse.Alternatively, maybe the equation is written in a different form. Let me consider both possibilities: whether the major axis is along the x-axis or the y-axis.Case 1: Major axis along the x-axis. Then, ( a^2 = m ) and ( b^2 = 4 ). So, ( c^2 = a^2 - b^2 ). Then, ( e = frac{c}{a} = 2 ). So, ( c = 2a ). Then, ( c^2 = 4a^2 ). But ( c^2 = a^2 - b^2 ), so ( 4a^2 = a^2 - b^2 ). That simplifies to ( 3a^2 = -b^2 ). But ( a^2 ) and ( b^2 ) are both positive, so this would imply a negative equals a positive, which is impossible. So, this case doesn't work.Case 2: Major axis along the y-axis. Then, ( a^2 = 4 ) and ( b^2 = m ). So, ( c^2 = a^2 - b^2 ). Then, ( e = frac{c}{a} = 2 ). So, ( c = 2a ). Then, ( c^2 = 4a^2 ). But ( c^2 = a^2 - b^2 ), so ( 4a^2 = a^2 - b^2 ). That simplifies to ( 3a^2 = -b^2 ). Again, same problem: positive equals negative, which is impossible.Wait, so both cases lead to impossible equations. That suggests that there's no solution where the eccentricity is 2 for this ellipse. But the problem says there is a solution, so maybe I'm making a mistake.Let me try again. Maybe I got the formula for eccentricity wrong. For an ellipse, ( e = sqrt{1 - frac{b^2}{a^2}} ) if ( a > b ), and ( e = sqrt{1 - frac{a^2}{b^2}} ) if ( b > a ). So, let's use that formula.Case 1: ( a^2 = m ), ( b^2 = 4 ), assuming ( m > 4 ). Then, ( e = sqrt{1 - frac{4}{m}} = 2 ). So, squaring both sides: ( 1 - frac{4}{m} = 4 ). Then, ( -frac{4}{m} = 3 ), which gives ( m = -frac{4}{3} ). But ( m ) has to be positive, so this is invalid.Case 2: ( a^2 = 4 ), ( b^2 = m ), assuming ( 4 > m ). Then, ( e = sqrt{1 - frac{m}{4}} = 2 ). Squaring: ( 1 - frac{m}{4} = 4 ). Then, ( -frac{m}{4} = 3 ), so ( m = -12 ). Again, negative, which is invalid.Hmm, so both cases give negative ( m ), which is impossible for an ellipse. That suggests that there is no such ellipse with eccentricity 2. But the problem is giving options, so maybe I'm misunderstanding something.Wait, perhaps the equation is written differently. Maybe it's ( frac{x^2}{m} + frac{y^2}{4} = 1 ), and ( m ) could be less than 4 or greater than 4. But as I saw earlier, both cases lead to negative ( m ). Maybe the problem is misstated, or perhaps it's a hyperbola.Alternatively, maybe the eccentricity is given as 2, but in reality, it's supposed to be 0.5 or something else. But the problem says 2. Alternatively, maybe the formula for eccentricity is different.Wait, another thought: maybe the eccentricity is given as 2, but in reality, it's supposed to be ( sqrt{2} ). Because ( sqrt{2} ) is approximately 1.414, which is less than 2 but still greater than 1, which would still make it a hyperbola. Hmm, but the problem says it's an ellipse.Alternatively, maybe I'm supposed to consider that ( e = 2 ), but then ( m ) would have to be negative, making it a hyperbola. But the problem says it's an ellipse. So, this is conflicting.Wait, let me check the options given: A: 5 or 3, B: 8, C: 5, D: sqrt(5) or sqrt(3). So, all options are positive numbers, which is fine for an ellipse. So, maybe I made a mistake in my earlier calculations.Let me try again. Let's assume that the major axis is along the y-axis, so ( a^2 = 4 ), ( b^2 = m ). Then, ( c^2 = a^2 - b^2 ). Eccentricity ( e = c/a = 2 ). So, ( c = 2a ). Therefore, ( c^2 = 4a^2 ). But ( c^2 = a^2 - b^2 ), so ( 4a^2 = a^2 - b^2 ). Then, ( 3a^2 = -b^2 ). So, ( b^2 = -3a^2 ). But ( b^2 ) must be positive, so this is impossible.Alternatively, if the major axis is along the x-axis, ( a^2 = m ), ( b^2 = 4 ). Then, ( c^2 = a^2 - b^2 ). Eccentricity ( e = c/a = 2 ). So, ( c = 2a ). Therefore, ( c^2 = 4a^2 ). But ( c^2 = a^2 - b^2 ), so ( 4a^2 = a^2 - b^2 ). Then, ( 3a^2 = -b^2 ). Again, impossible.Wait, so both cases lead to contradictions. That suggests that there is no such ellipse with eccentricity 2. But the problem is giving options, so perhaps I'm misunderstanding the formula for eccentricity.Wait, another thought: maybe the formula for eccentricity is ( e = sqrt{1 + frac{b^2}{a^2}} ) for hyperbolas, but for ellipses, it's ( e = sqrt{1 - frac{b^2}{a^2}} ). So, if ( e = 2 ), then for an ellipse, ( 2 = sqrt{1 - frac{b^2}{a^2}} ). Squaring both sides, ( 4 = 1 - frac{b^2}{a^2} ). Then, ( frac{b^2}{a^2} = -3 ). Which is impossible because ( b^2 ) and ( a^2 ) are positive.So, this confirms that it's impossible for an ellipse to have an eccentricity of 2. Therefore, the problem might have a typo, or perhaps it's referring to a hyperbola. If it's a hyperbola, then the formula for eccentricity is ( e = sqrt{1 + frac{b^2}{a^2}} ). Let's try that.Assuming it's a hyperbola, and the equation is ( frac{x^2}{m} - frac{y^2}{4} = 1 ). Then, ( a^2 = m ), ( b^2 = 4 ). Eccentricity ( e = sqrt{1 + frac{b^2}{a^2}} = 2 ). So, ( sqrt{1 + frac{4}{m}} = 2 ). Squaring both sides: ( 1 + frac{4}{m} = 4 ). Then, ( frac{4}{m} = 3 ), so ( m = frac{4}{3} ). But that's not one of the options.Alternatively, if the hyperbola is ( frac{y^2}{4} - frac{x^2}{m} = 1 ), then ( a^2 = 4 ), ( b^2 = m ). Eccentricity ( e = sqrt{1 + frac{m}{4}} = 2 ). Squaring: ( 1 + frac{m}{4} = 4 ). Then, ( frac{m}{4} = 3 ), so ( m = 12 ). Again, not an option.Hmm, so even if it's a hyperbola, the value of ( m ) doesn't match the given options. So, perhaps the problem is correct as an ellipse, but I'm missing something.Wait, maybe the eccentricity is given as 2, but in reality, it's supposed to be ( frac{2}{3} ) or something else. But the problem says 2. Alternatively, maybe the formula for eccentricity is different.Wait, another thought: maybe the problem is using a different definition of eccentricity. I know that for an ellipse, ( e = frac{c}{a} ), but perhaps in some contexts, it's defined differently. Let me check.No, I think the standard definition is ( e = frac{c}{a} ), which is less than 1 for ellipses. So, if ( e = 2 ), it's definitely a hyperbola. Therefore, the problem might have a mistake, or perhaps it's a trick question.Alternatively, maybe the equation is written differently. Let me consider that ( m ) could be negative, making it a hyperbola, but the problem says it's an ellipse. So, that's conflicting.Wait, perhaps the problem is in a different form. Let me try to rearrange the equation. If ( frac{x^2}{m} + frac{y^2}{4} = 1 ), and it's an ellipse, then ( m ) must be positive. But as we saw, both cases lead to contradictions when trying to get ( e = 2 ).Wait, maybe I made a mistake in the formula. Let me try again. If the major axis is along the y-axis, then ( a^2 = 4 ), ( b^2 = m ). Then, ( c^2 = a^2 - b^2 ). Eccentricity ( e = c/a = 2 ). So, ( c = 2a ). Therefore, ( c^2 = 4a^2 ). But ( c^2 = a^2 - b^2 ), so ( 4a^2 = a^2 - b^2 ). Then, ( 3a^2 = -b^2 ). So, ( b^2 = -3a^2 ). Which is impossible because ( b^2 ) is positive.Similarly, if the major axis is along the x-axis, ( a^2 = m ), ( b^2 = 4 ). Then, ( c^2 = a^2 - b^2 ). Eccentricity ( e = c/a = 2 ). So, ( c = 2a ). Therefore, ( c^2 = 4a^2 ). But ( c^2 = a^2 - b^2 ), so ( 4a^2 = a^2 - b^2 ). Then, ( 3a^2 = -b^2 ). Again, impossible.So, both cases lead to contradictions. Therefore, there is no such ellipse with eccentricity 2. But the problem is giving options, so perhaps I'm misunderstanding the problem.Wait, maybe the equation is written as ( frac{x^2}{m} + frac{y^2}{4} = 1 ), and ( m ) could be negative, making it a hyperbola, but the problem says it's an ellipse. So, that's conflicting.Alternatively, maybe the problem is referring to the absolute value of the eccentricity, but that doesn't make sense because eccentricity is always positive.Wait, another thought: maybe the problem is referring to the reciprocal of the eccentricity. If ( e = 2 ), then the reciprocal is ( 1/2 ), which is less than 1, making it an ellipse. But that's just speculation.Alternatively, maybe the problem is referring to the distance between the foci, which is ( 2c ), and that's given as 2. So, ( 2c = 2 ), so ( c = 1 ). Then, using ( c^2 = a^2 - b^2 ), we can find ( m ).Let me try that. If ( 2c = 2 ), then ( c = 1 ). Then, ( c^2 = 1 ). Now, if the major axis is along the y-axis, ( a^2 = 4 ), so ( c^2 = a^2 - b^2 ). Therefore, ( 1 = 4 - b^2 ), so ( b^2 = 3 ). Therefore, ( m = b^2 = 3 ).Alternatively, if the major axis is along the x-axis, ( a^2 = m ), so ( c^2 = a^2 - b^2 ). Therefore, ( 1 = m - 4 ), so ( m = 5 ).So, in this case, ( m ) could be either 3 or 5, depending on whether the major axis is along the y-axis or x-axis. Therefore, the answer is A: 5 or 3.Wait, but earlier I thought that the eccentricity being 2 was impossible for an ellipse. But if the problem is referring to the distance between the foci as 2, then ( 2c = 2 ), so ( c = 1 ), and then we can find ( m ) as either 3 or 5. So, maybe the problem was referring to the distance between the foci as 2, not the eccentricity. That would make sense because the eccentricity is a ratio, not a distance.So, perhaps the problem has a misstatement, and it's referring to the distance between the foci being 2, not the eccentricity. In that case, the value of ( m ) is 3 or 5, which is option A.Alternatively, if the problem is correctly stating the eccentricity as 2, then there is no solution because it's impossible for an ellipse. But since the problem gives options, and option A is 5 or 3, which matches the calculation if we consider the distance between the foci as 2, I think that's the intended interpretation.Therefore, the value of ( m ) is either 3 or 5, so the correct answer is A.