💡First, I need to evaluate the expression (left(frac{1}{4}right)^{-1}). A negative exponent means taking the reciprocal of the base, so (left(frac{1}{4}right)^{-1} = 4).Next, I'll calculate the absolute value (|sqrt{3} - 2|). Since (sqrt{3}) is approximately 1.732, which is less than 2, the expression inside the absolute value is negative. Therefore, (|sqrt{3} - 2| = 2 - sqrt{3}).Then, I'll multiply 2 by (-sqrt{3}), which gives (2 times (-sqrt{3}) = -2sqrt{3}).Now, I'll substitute these values back into the original expression:[4 - (2 - sqrt{3}) - 2sqrt{3}]Simplifying further:[4 - 2 + sqrt{3} - 2sqrt{3} = 2 - sqrt{3}]So, the final result of the expression is (2 - sqrt{3}).

💡Okay, so I need to find the sum of this geometric series: 1 + (1/3) + (1/3)^2 + (1/3)^3 + ... and so on. Hmm, I remember that geometric series have something to do with a common ratio between the terms. Let me think about what that means.First, let's identify the first term of the series. The first term is clearly 1. Now, what's the common ratio? To find the common ratio, I can divide the second term by the first term. So, (1/3) divided by 1 is just 1/3. Let me check the next term to make sure. The third term is (1/3)^2, which is 1/9. If I divide that by the second term, which is 1/3, I get (1/9) / (1/3) = 1/3. Okay, so the common ratio r is 1/3.Now, I recall that the sum of an infinite geometric series can be found using the formula S = a / (1 - r), where a is the first term and r is the common ratio. But wait, this formula only works if the absolute value of r is less than 1, right? In this case, r is 1/3, which is definitely less than 1, so we're good to use this formula.Let me plug in the values. The first term a is 1, and the common ratio r is 1/3. So, S = 1 / (1 - 1/3). Let's compute the denominator first: 1 - 1/3 is equal to 2/3. So now, S = 1 / (2/3). Dividing by a fraction is the same as multiplying by its reciprocal, so 1 divided by (2/3) is the same as 1 multiplied by 3/2, which equals 3/2.Wait, let me double-check to make sure I didn't make a mistake. If I add up the first few terms: 1 + 1/3 is 4/3, plus 1/9 is 13/9, plus 1/27 is 40/27, and so on. It seems like the sum is approaching 3/2, which is 1.5. Yeah, that makes sense because each term is getting smaller and smaller, so the total sum converges to a finite number.I think I did everything correctly. I identified the first term and the common ratio, checked that the ratio is less than 1, applied the formula, and even did a quick check by adding the first few terms. So, I'm pretty confident that the sum of this infinite geometric series is 3/2.Just to recap, the steps were:1. Identify the first term (a) and the common ratio (r).2. Confirm that |r| < 1 to use the infinite series sum formula.3. Apply the formula S = a / (1 - r).4. Simplify the expression to get the sum as a common fraction.Everything seems to add up, so I think I'm done!

💡First, I recognize that the total cost of purchasing both commodities is 827.Given that the first commodity costs 477, I can find the cost of the second commodity by subtracting the cost of the first commodity from the total cost.Once I have the cost of the second commodity, I can determine the difference between the purchase prices of the two commodities by subtracting the cost of the second commodity from the cost of the first commodity.This straightforward approach allows me to calculate the desired difference efficiently.

💡First, I need to determine which set of line segments cannot form a triangle. To do this, I'll use the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side.I'll examine each set one by one:1. **Set A (4cm, 5cm, 6cm):** - (4 + 5 = 9) which is greater than 6. - This set satisfies the triangle inequality theorem, so it can form a triangle.2. **Set B (3cm, 4cm, 5cm):** - (3 + 4 = 7) which is greater than 5. - This set also satisfies the triangle inequality theorem, so it can form a triangle.3. **Set C (2cm, 3cm, 4cm):** - (2 + 3 = 5) which is greater than 4. - This set satisfies the triangle inequality theorem, so it can form a triangle.4. **Set D (1cm, 2cm, 3cm):** - (1 + 2 = 3) which is equal to 3, not greater than 3. - This set does not satisfy the triangle inequality theorem, so it cannot form a triangle.After evaluating all the sets, I conclude that only Set D cannot form a triangle.

💡First, I need to determine the width of the rectangle. Since the length is four times the width, I'll represent the width as ( w ) and the length as ( 4w ).Next, I'll use the formula for the perimeter of a rectangle, which is ( 2 times (text{length} + text{width}) ). Plugging in the expressions for length and width, the equation becomes:[2(4w + w) = 200]Simplifying inside the parentheses:[2(5w) = 200]Multiplying:[10w = 200]Solving for ( w ):[w = 20 text{ cm}]Now that I have the width, I can find the length:[text{Length} = 4w = 4 times 20 = 80 text{ cm}]Finally, to find the area of the rectangle, I'll multiply the length by the width:[text{Area} = 80 times 20 = 1600 text{ square centimeters}]

💡Alright, so I have this problem here: Given that ( e ) is the base of the natural logarithm, if for any ( x in [0,1] ), there exists a unique ( y in [-1,1] ) such that ( x + y^2 e^y - a = 0 ) holds, then I need to find the range of the real number ( a ).Hmm, okay. Let me try to unpack this step by step. The equation given is ( x + y^2 e^y - a = 0 ). So, rearranging that, I get ( y^2 e^y = a - x ). That means for each ( x ) in the interval [0,1], there must be exactly one ( y ) in [-1,1] such that this equation holds.So, essentially, for every ( x ) between 0 and 1, the value ( a - x ) must be equal to ( y^2 e^y ) for some unique ( y ) in [-1,1]. Therefore, ( a - x ) must lie within the range of the function ( f(y) = y^2 e^y ) over the interval ( y in [-1,1] ).Wait, so if I can find the range of ( f(y) = y^2 e^y ) over ( y in [-1,1] ), then ( a - x ) must lie within that range for all ( x in [0,1] ). Moreover, since for each ( x ), there must be a unique ( y ), the function ( f(y) ) must be injective (one-to-one) over the interval [-1,1]. Hmm, is that the case?Let me think. The function ( f(y) = y^2 e^y ). Let's analyze its behavior on [-1,1]. First, let's compute its derivative to see if it's increasing or decreasing.The derivative ( f'(y) ) is ( 2y e^y + y^2 e^y ) by the product rule. So, ( f'(y) = e^y (2y + y^2) ). Let's factor that: ( f'(y) = e^y y (2 + y) ).Now, since ( e^y ) is always positive, the sign of ( f'(y) ) depends on ( y (2 + y) ). Let's analyze the critical points where ( f'(y) = 0 ). That occurs when ( y = 0 ) or ( 2 + y = 0 ), so ( y = -2 ). But ( y = -2 ) is outside our interval [-1,1], so within [-1,1], the only critical point is at ( y = 0 ).So, let's see the behavior around ( y = 0 ). For ( y < 0 ), say ( y = -1 ), ( y (2 + y) = (-1)(2 -1) = (-1)(1) = -1 ), which is negative. For ( y > 0 ), say ( y = 1 ), ( y (2 + y) = (1)(3) = 3 ), which is positive. So, the function ( f(y) ) is decreasing on [-1,0) and increasing on (0,1]. Therefore, it has a minimum at ( y = 0 ).So, the function ( f(y) ) has a minimum at ( y = 0 ), which is ( f(0) = 0^2 e^0 = 0 ). Wait, but ( f(y) = y^2 e^y ), so at ( y = 0 ), it's 0. But at ( y = -1 ), ( f(-1) = (-1)^2 e^{-1} = 1 times frac{1}{e} = frac{1}{e} ). At ( y = 1 ), ( f(1) = 1^2 e^1 = e ).So, the function ( f(y) ) starts at ( y = -1 ) with ( f(-1) = frac{1}{e} ), decreases to 0 at ( y = 0 ), and then increases to ( e ) at ( y = 1 ). Therefore, the range of ( f(y) ) over [-1,1] is [0, e]. But wait, at ( y = -1 ), it's ( frac{1}{e} ), which is approximately 0.3679, and at ( y = 0 ), it's 0. So, actually, the range is [0, e], but the function doesn't take all values between 0 and ( frac{1}{e} ) only once?Wait, no. Because from ( y = -1 ) to ( y = 0 ), the function decreases from ( frac{1}{e} ) to 0, and from ( y = 0 ) to ( y = 1 ), it increases from 0 to ( e ). So, the function is not injective over the entire interval [-1,1] because it's decreasing on [-1,0] and increasing on [0,1]. So, for values of ( f(y) ) between 0 and ( frac{1}{e} ), there are two corresponding ( y ) values: one in [-1,0) and another in (0,1]. But for values above ( frac{1}{e} ), there is only one ( y ) in (0,1].But in our problem, for each ( x in [0,1] ), there must exist a unique ( y in [-1,1] ) such that ( x + y^2 e^y = a ). So, ( a - x = y^2 e^y ). Therefore, ( a - x ) must be in the range of ( f(y) ), which is [0, e]. But more importantly, for each ( x ), ( a - x ) must lie in the range of ( f(y) ) such that there's only one ( y ) corresponding to it.But since ( f(y) ) is not injective over [-1,1], except when ( a - x ) is greater than ( frac{1}{e} ), because for ( a - x > frac{1}{e} ), there's only one ( y ) in (0,1] that satisfies ( f(y) = a - x ). For ( a - x leq frac{1}{e} ), there are two possible ( y ) values: one negative and one positive.But the problem states that for any ( x in [0,1] ), there exists a unique ( y in [-1,1] ). So, we must ensure that for each ( x ), ( a - x ) is such that there's only one ( y ) in [-1,1] satisfying ( f(y) = a - x ). Therefore, ( a - x ) must lie in the range where ( f(y) ) is injective, which is when ( a - x geq frac{1}{e} ). Because for ( a - x geq frac{1}{e} ), the equation ( f(y) = a - x ) has only one solution in (0,1].But wait, let's think about this. If ( a - x geq frac{1}{e} ), then ( y ) must be in (0,1], because for ( y ) in [-1,0), ( f(y) ) would be less than or equal to ( frac{1}{e} ). So, to have a unique ( y ), ( a - x ) must be greater than ( frac{1}{e} ), so that only the positive ( y ) is a solution.But ( x ) can vary from 0 to 1. So, ( a - x ) must be greater than or equal to ( frac{1}{e} ) for all ( x in [0,1] ). The smallest ( a - x ) occurs when ( x ) is largest, which is ( x = 1 ). So, ( a - 1 geq frac{1}{e} ), which implies ( a geq 1 + frac{1}{e} ).On the other hand, ( a - x ) must also be less than or equal to the maximum value of ( f(y) ), which is ( e ) at ( y = 1 ). So, ( a - x leq e ) for all ( x in [0,1] ). The largest ( a - x ) occurs when ( x ) is smallest, which is ( x = 0 ). So, ( a - 0 leq e ), which implies ( a leq e ).Therefore, combining these two inequalities, ( a ) must satisfy ( 1 + frac{1}{e} leq a leq e ).But wait, let me double-check. If ( a = 1 + frac{1}{e} ), then when ( x = 1 ), ( a - x = frac{1}{e} ). But at ( y = -1 ), ( f(-1) = frac{1}{e} ), so there would be two solutions: ( y = -1 ) and ( y ) such that ( f(y) = frac{1}{e} ). But wait, is there another ( y ) in (0,1] such that ( f(y) = frac{1}{e} )?Let me see. ( f(y) = y^2 e^y ). Let's set ( y^2 e^y = frac{1}{e} ). So, ( y^2 e^{y + 1} = 1 ). Let's see if there's a solution in (0,1]. Let's try ( y = 0 ): ( 0 times e^1 = 0 neq 1 ). ( y = 1 ): ( 1 times e^2 approx 7.389 neq 1 ). So, maybe somewhere between 0 and 1. Let's try ( y = -1 ): ( (-1)^2 e^{-1 + 1} = 1 times e^0 = 1 ). So, ( y = -1 ) is a solution. But in (0,1], is there another solution?Wait, if ( y ) is in (0,1], then ( y^2 e^y ) is increasing from 0 to ( e ). So, ( y^2 e^y = frac{1}{e} ) would have a solution in (0,1] only if ( frac{1}{e} ) is in the range of ( f(y) ) over (0,1]. But ( f(y) ) starts at 0 and goes up to ( e ), so ( frac{1}{e} ) is indeed in that range. Therefore, when ( a = 1 + frac{1}{e} ), for ( x = 1 ), ( a - x = frac{1}{e} ), which has two solutions: ( y = -1 ) and some ( y ) in (0,1]. Therefore, ( a = 1 + frac{1}{e} ) would not satisfy the uniqueness condition because there are two ( y ) values. So, ( a ) must be strictly greater than ( 1 + frac{1}{e} ).Similarly, when ( a = e ), then ( a - x = e - x ). The maximum value of ( e - x ) is ( e ) when ( x = 0 ), and the minimum is ( e - 1 ) when ( x = 1 ). Since ( e - 1 approx 1.718 ), which is greater than ( frac{1}{e} approx 0.3679 ). So, for ( a = e ), ( a - x ) ranges from ( e - 1 ) to ( e ), both of which are greater than ( frac{1}{e} ). Therefore, for each ( x ), there is a unique ( y ) in (0,1] such that ( f(y) = a - x ). So, ( a = e ) is acceptable.Therefore, the range of ( a ) is ( (1 + frac{1}{e}, e] ).Let me just recap to make sure I didn't miss anything. The function ( f(y) = y^2 e^y ) has a minimum at ( y = 0 ) with ( f(0) = 0 ), and it's symmetric in a way around ( y = 0 ) but not exactly because of the exponential term. The key point is that for ( a - x ) to have a unique solution ( y ), ( a - x ) must lie in the range where ( f(y) ) is injective, which is when ( a - x > frac{1}{e} ). Since ( x ) can be as large as 1, ( a ) must be at least ( 1 + frac{1}{e} ). But since at ( a = 1 + frac{1}{e} ), there's a case where two ( y ) values satisfy the equation, we exclude this point. The upper bound is ( a leq e ) because ( f(y) ) can't exceed ( e ) in the interval [-1,1].So, putting it all together, the range of ( a ) is ( (1 + frac{1}{e}, e] ).**Final Answer**The range of the real number ( a ) is boxed{left(1 + frac{1}{e}, eright]}.