💡First, I recognize that the largest rhombus that can be inscribed in a circle is actually a square. This is because, in a circle, the maximum distance between any two points is the diameter, and a square maximizes the area among all rhombuses inscribed in a circle.Next, I recall that the diagonal of a square inscribed in a circle is equal to the diameter of the circle. Given that the radius of the circle is 10 centimeters, the diameter is twice the radius, which is 20 centimeters.Therefore, the diagonal of the largest rhombus (which is a square) inscribed in the circle is 20 centimeters.

💡First, I'll assume Kelly's original salary is ( S ).After the first pay cut of 8%, her salary becomes:[S times (1 - 0.08) = S times 0.92]Let the second pay cut be ( x% ). After the second pay cut, her salary becomes:[S times 0.92 times (1 - frac{x}{100})]After the third pay cut of 18%, her salary becomes:[S times 0.92 times (1 - frac{x}{100}) times (1 - 0.18) = S times 0.92 times (1 - frac{x}{100}) times 0.82]We are given that the overall decrease is 35.1216%, which means her final salary is:[S times (1 - 0.351216) = S times 0.648784]Setting up the equation:[S times 0.92 times (1 - frac{x}{100}) times 0.82 = S times 0.648784]Canceling out ( S ) from both sides:[0.92 times (1 - frac{x}{100}) times 0.82 = 0.648784]Solving for ( (1 - frac{x}{100}) ):[1 - frac{x}{100} = frac{0.648784}{0.92 times 0.82} = frac{0.648784}{0.7544} approx 0.8602]Solving for ( x ):[frac{x}{100} = 1 - 0.8602 = 0.1398][x = 0.1398 times 100 = 13.98]Therefore, the percentage of the second pay cut was approximately ( boxed{13.98%} ).

💡Okay, so I have this problem here: I need to find the value of m such that the tangent line to the curve y = e^x at x = 1 is perpendicular to the line given by 2x + my + 1 = 0. The options are A: -2e, B: 2e, C: -2/e, and D: 2/e. Hmm, let me try to figure this out step by step.First, I remember that the derivative of a function gives the slope of the tangent line at a particular point. So, since the curve is y = e^x, its derivative should be y' = e^x, right? Yeah, that makes sense because the derivative of e^x is itself. So, at x = 1, the slope of the tangent line would be e^1, which is just e. Got that down.Now, the next part is about the line 2x + my + 1 = 0. I need to find its slope. I think to find the slope of a line in the form Ax + By + C = 0, the formula is -A/B. So, in this case, A is 2 and B is m, so the slope should be -2/m. Let me write that down: slope = -2/m.The problem says that these two lines are perpendicular. I remember that if two lines are perpendicular, the product of their slopes is -1. So, if one slope is k, the other should be -1/k. Let me verify that. If two lines are perpendicular, then yes, their slopes multiply to -1.So, the slope of the tangent line is e, and the slope of the given line is -2/m. Therefore, according to the perpendicular condition, e * (-2/m) should equal -1. Let me write that equation:e * (-2/m) = -1Hmm, let me solve for m. First, I can multiply both sides by m to get rid of the denominator:e * (-2) = -1 * mWhich simplifies to:-2e = -mNow, if I multiply both sides by -1, I get:2e = mSo, m is equal to 2e. Let me check the options: A is -2e, B is 2e, C is -2/e, and D is 2/e. So, B is 2e, which matches my result. That seems straightforward.Wait, let me double-check my steps to make sure I didn't make a mistake. First, derivative of e^x is e^x, correct. At x=1, it's e, that's right. The line equation: 2x + my + 1 = 0. Converting to slope-intercept form, y = (-2/m)x - 1/m, so the slope is indeed -2/m. Then, for perpendicularity, the product of the slopes should be -1. So, e * (-2/m) = -1. Solving that gives m = 2e. Yeah, that seems correct.I think I might have confused myself if I thought the slope of the tangent was different, but no, I think I did it right. The key was remembering how to find the slope of a line in standard form and the condition for perpendicularity. Yeah, I think B is the correct answer.**Final Answer**boxed{B}

💡First, I need to determine how many questions each student answered correctly based on their scores.For Xiao Ming, who scored 87 points:1. Calculate the total possible points if all answers were correct: 10 questions × 10 points = 100 points.2. Determine the points lost: 100 points - 87 points = 13 points.3. Since each incorrect answer results in a loss of 13 points (10 points for not answering correctly and 3 points deducted), the number of incorrect answers is 13 points ÷ 13 points per incorrect answer = 1 incorrect answer.4. Therefore, Xiao Ming answered 10 total questions - 1 incorrect answer = 9 correct answers.For Xiao Hong, who scored 74 points:1. Calculate the total possible points if all answers were correct: 10 questions × 10 points = 100 points.2. Determine the points lost: 100 points - 74 points = 26 points.3. Since each incorrect answer results in a loss of 13 points, the number of incorrect answers is 26 points ÷ 13 points per incorrect answer = 2 incorrect answers.4. Therefore, Xiao Hong answered 10 total questions - 2 incorrect answers = 8 correct answers.For Xiao Hua, who scored 9 points:1. Calculate the total possible points if all answers were correct: 10 questions × 10 points = 100 points.2. Determine the points lost: 100 points - 9 points = 91 points.3. Since each incorrect answer results in a loss of 13 points, the number of incorrect answers is 91 points ÷ 13 points per incorrect answer = 7 incorrect answers.4. Therefore, Xiao Hua answered 10 total questions - 7 incorrect answers = 3 correct answers.Finally, sum the correct answers from all three students:9 (Xiao Ming) + 8 (Xiao Hong) + 3 (Xiao Hua) = 20 correct answers in total.

💡Okay, so I have this function ( f(x) = lg left[ (a^2 - 1)x^2 + (a + 1)x + 1 right] ). I need to figure out two things: first, the range of values for ( a ) such that the domain of ( f(x) ) is all real numbers, and second, the range of values for ( a ) such that the range of ( f(x) ) is all real numbers. Starting with the first part: the domain of ( f(x) ) is all real numbers. Since ( f(x) ) is a logarithmic function, the argument inside the log must be positive for all ( x ). So, the quadratic expression ( (a^2 - 1)x^2 + (a + 1)x + 1 ) must be greater than zero for all ( x in mathbb{R} ).Hmm, for a quadratic ( Ax^2 + Bx + C ) to be positive for all ( x ), two conditions must be satisfied: the leading coefficient ( A ) must be positive, and the discriminant ( B^2 - 4AC ) must be negative. So, applying this to our quadratic:1. The leading coefficient ( A = a^2 - 1 ) must be positive. So, ( a^2 - 1 > 0 ). This implies ( a^2 > 1 ), so ( a > 1 ) or ( a < -1 ).2. The discriminant must be negative. Let's compute the discriminant: ( B^2 - 4AC = (a + 1)^2 - 4(a^2 - 1)(1) ) Expanding this: ( (a + 1)^2 = a^2 + 2a + 1 ) ( 4(a^2 - 1)(1) = 4a^2 - 4 ) So, discriminant ( D = a^2 + 2a + 1 - 4a^2 + 4 = -3a^2 + 2a + 5 ) We need ( D < 0 ), so: ( -3a^2 + 2a + 5 < 0 ) Let's solve this inequality. Multiply both sides by -1 (remembering to reverse the inequality sign): ( 3a^2 - 2a - 5 > 0 ) Now, solve ( 3a^2 - 2a - 5 = 0 ): Using quadratic formula: ( a = frac{2 pm sqrt{4 + 60}}{6} = frac{2 pm 8}{6} ) So, ( a = frac{10}{6} = frac{5}{3} ) or ( a = frac{-6}{6} = -1 ) So, the quadratic ( 3a^2 - 2a - 5 ) is positive outside the roots, i.e., ( a < -1 ) or ( a > frac{5}{3} ).But wait, from the first condition, we had ( a > 1 ) or ( a < -1 ). So, combining both conditions:- For ( a < -1 ), both conditions are satisfied.- For ( a > 1 ), the discriminant condition requires ( a > frac{5}{3} ).So, the range of ( a ) for the domain to be all real numbers is ( a in (-infty, -1] cup (frac{5}{3}, infty) ).Now, moving on to the second part: the range of ( f(x) ) is all real numbers. For the range of a logarithmic function ( lg(g(x)) ) to be all real numbers, ( g(x) ) must take all positive real values. That is, ( g(x) ) must be able to take any positive value as ( x ) varies over its domain.So, ( (a^2 - 1)x^2 + (a + 1)x + 1 ) must be able to take any positive value. This requires that the quadratic can attain all positive values, which happens if the quadratic is a parabola opening upwards and has a minimum value that is zero or negative. Wait, but since the argument of the log must be positive, the quadratic must actually be able to take all positive values, which would require that the quadratic is surjective onto ( mathbb{R}^+ ).For a quadratic to be surjective onto ( mathbb{R}^+ ), it must open upwards (so ( a^2 - 1 > 0 )) and its minimum value must be less than or equal to zero. The minimum value of a quadratic ( Ax^2 + Bx + C ) is at ( x = -frac{B}{2A} ), and the minimum value is ( C - frac{B^2}{4A} ).So, let's compute the minimum value:Minimum value ( = 1 - frac{(a + 1)^2}{4(a^2 - 1)} )We need this minimum value to be less than or equal to zero:( 1 - frac{(a + 1)^2}{4(a^2 - 1)} leq 0 )Multiply both sides by ( 4(a^2 - 1) ) (which is positive since ( a^2 - 1 > 0 )):( 4(a^2 - 1) - (a + 1)^2 leq 0 )Expand:( 4a^2 - 4 - (a^2 + 2a + 1) leq 0 )Simplify:( 4a^2 - 4 - a^2 - 2a - 1 leq 0 )( 3a^2 - 2a - 5 leq 0 )We've seen this quadratic before. Solving ( 3a^2 - 2a - 5 = 0 ):Roots are ( a = frac{5}{3} ) and ( a = -1 ). So, the quadratic ( 3a^2 - 2a - 5 ) is less than or equal to zero between its roots:( -1 leq a leq frac{5}{3} )But remember, we also need ( a^2 - 1 > 0 ), which implies ( a > 1 ) or ( a < -1 ). So, intersecting this with ( -1 leq a leq frac{5}{3} ), we get ( 1 < a leq frac{5}{3} ).Wait, but what about ( a = 1 )? If ( a = 1 ), the quadratic becomes ( 0x^2 + 2x + 1 = 2x + 1 ), which is a linear function. For the range of ( f(x) ) to be all real numbers, the linear function ( 2x + 1 ) must take all positive values, which it does as ( x ) varies. So, ( a = 1 ) is also acceptable.Similarly, at ( a = frac{5}{3} ), the quadratic will have its minimum value equal to zero, so ( f(x) ) can take all real numbers as ( x ) approaches the point where the quadratic is zero.So, combining these, the range of ( a ) is ( 1 leq a leq frac{5}{3} ).Wait, but earlier when solving for the domain, we had ( a in (-infty, -1] cup (frac{5}{3}, infty) ). So, for the range to be all real numbers, ( a ) must be between 1 and ( frac{5}{3} ), inclusive.Let me just double-check:- If ( a = 1 ), the quadratic becomes linear, which is fine.- If ( a = frac{5}{3} ), the quadratic touches the x-axis, so the log can take all real values as ( x ) approaches the root.- For ( a ) between 1 and ( frac{5}{3} ), the quadratic opens upwards and has a minimum below zero, so the log can take all real values.Yes, that seems correct.So, summarizing:1. For the domain of ( f(x) ) to be all real numbers, ( a ) must be in ( (-infty, -1] cup (frac{5}{3}, infty) ).2. For the range of ( f(x) ) to be all real numbers, ( a ) must be in ( [1, frac{5}{3}] ).**Final Answer**1. The range of values for ( a ) is ( boxed{(-infty, -1] cup left( frac{5}{3}, +infty right)} ).2. The range of values for ( a ) is ( boxed{left[1, frac{5}{3}right]} ).

💡First, I need to understand what a reciprocal is. The reciprocal of a number is a value that, when multiplied by the original number, equals 1.Given the fraction (frac{5}{8}), to find its reciprocal, I will flip the numerator and the denominator.So, the reciprocal of (frac{5}{8}) is (frac{8}{5}).Next, I will compare this result with the provided options:- **A: (frac{5}{8})** – This is the original fraction, not its reciprocal.- **B: (-frac{5}{8})** – This is the negative of the original fraction, not the reciprocal.- **C: (frac{8}{5})** – This matches the reciprocal I calculated.- **D: (-frac{8}{5})** – This is the negative of the reciprocal, which is not what we need.Therefore, the correct answer is **C: (frac{8}{5})**.