💡Alright, so I've got these three math problems to solve, and I need to use the completing the square method or something similar. Let me take them one by one.**Problem (1):** Given ( a - b = 2 ) and ( ab = 3 ), find the value of ( a^2 + b^2 ).Hmm, okay. I remember from the example that ( a^2 + b^2 ) can be expressed in terms of ( (a + b)^2 ) or ( (a - b)^2 ). Since I have ( a - b ) given, maybe I can use that.The formula they provided is ( a^2 + b^2 = (a - b)^2 + 2ab ). Let me plug in the values I have.So, ( (a - b)^2 = 2^2 = 4 ), and ( 2ab = 2 times 3 = 6 ). Adding those together, ( 4 + 6 = 10 ). So, ( a^2 + b^2 = 10 ). That seems straightforward.**Problem (2):** If ( a + frac{1}{a} = 6 ), find the value of ( a^2 + frac{1}{a^2} ).Alright, this one looks a bit trickier because of the fraction, but I think I can handle it. Maybe I can square both sides to get rid of the fraction.So, squaring both sides: ( left( a + frac{1}{a} right)^2 = 6^2 ). That gives me ( a^2 + 2 times a times frac{1}{a} + frac{1}{a^2} = 36 ).Simplifying the middle term: ( 2 times a times frac{1}{a} = 2 ). So now, the equation is ( a^2 + 2 + frac{1}{a^2} = 36 ).Subtracting 2 from both sides: ( a^2 + frac{1}{a^2} = 34 ). Okay, that worked out nicely.**Problem (3):** If ( n ) satisfies ( left( n - 2024 right)^2 + left( 2023 - n right)^2 = 1 ), find the value of the expression ( left( n - 2024 right) left( 2023 - n right) ).This one looks more complex because of the large numbers, but maybe I can simplify it somehow.Let me denote ( x = n - 2024 ) and ( y = 2023 - n ). Then, the equation becomes ( x^2 + y^2 = 1 ), and I need to find ( xy ).Wait, I remember that ( (x + y)^2 = x^2 + 2xy + y^2 ). Since I know ( x^2 + y^2 = 1 ), maybe I can find ( x + y ) and use that.Calculating ( x + y ): ( x + y = (n - 2024) + (2023 - n) = n - 2024 + 2023 - n = (-2024 + 2023) + (n - n) = -1 + 0 = -1 ).So, ( x + y = -1 ). Now, squaring both sides: ( (x + y)^2 = (-1)^2 = 1 ).Expanding the left side: ( x^2 + 2xy + y^2 = 1 ). But we already know that ( x^2 + y^2 = 1 ), so substituting that in:( 1 + 2xy = 1 ).Subtracting 1 from both sides: ( 2xy = 0 ).Dividing both sides by 2: ( xy = 0 ).So, ( left( n - 2024 right) left( 2023 - n right) = 0 ).That makes sense because if the sum of their squares is 1 and their sum is -1, their product has to be zero to satisfy the equation.Okay, I think I've got all three problems solved. Let me just recap:1. For the first problem, using the given ( a - b ) and ( ab ), I expressed ( a^2 + b^2 ) in terms of those and got 10.2. For the second problem, squaring the given equation helped eliminate the fraction, leading me to ( a^2 + frac{1}{a^2} = 34 ).3. For the third problem, substituting variables and using algebraic identities simplified the problem, resulting in the product being zero.I feel confident with these solutions, but let me double-check each one quickly.**Double-Checking Problem (1):**Given ( a - b = 2 ) and ( ab = 3 ).Using ( a^2 + b^2 = (a - b)^2 + 2ab ):( (2)^2 + 2(3) = 4 + 6 = 10 ). Yep, that's correct.**Double-Checking Problem (2):**Given ( a + frac{1}{a} = 6 ).Squaring both sides:( left( a + frac{1}{a} right)^2 = 36 ).Expanding:( a^2 + 2 + frac{1}{a^2} = 36 ).Subtracting 2:( a^2 + frac{1}{a^2} = 34 ). That checks out.**Double-Checking Problem (3):**Given ( left( n - 2024 right)^2 + left( 2023 - n right)^2 = 1 ).Let ( x = n - 2024 ) and ( y = 2023 - n ).Then, ( x + y = -1 ) and ( x^2 + y^2 = 1 ).Using ( (x + y)^2 = x^2 + 2xy + y^2 ):( (-1)^2 = 1 + 2xy ).So, ( 1 = 1 + 2xy ) implies ( 2xy = 0 ), hence ( xy = 0 ). Correct.Alright, all my answers seem solid. I think I'm ready to present them.

💡First, I need to determine the difference in speed when the bus is moving without stoppages and when it includes stoppages. The speed without stoppages is 32 km/hr, and with stoppages, it is 16 km/hr.Next, I'll assume the bus travels for one hour without any stoppages. In this time, it would cover 32 km.Now, considering the stoppages, the bus travels at an average speed of 16 km/hr. This means that in one hour, the bus only covers 16 km.The difference in distance covered is 16 km, which represents the distance the bus would have covered if it hadn't stopped. To find out how much time this distance would take at the original speed, I'll use the formula:Time = Distance / SpeedPlugging in the numbers:Time = 16 km / 32 km/hr = 0.5 hoursFinally, I'll convert 0.5 hours into minutes:0.5 hours × 60 minutes/hour = 30 minutesTherefore, the bus stops for 30 minutes per hour.

💡First, I need to determine the least common multiple (LCM) of the numbers 25, 40, and 75. This will help me find the smallest number that is divisible by all three.I'll start by finding the prime factorization of each number:- 25 can be factored into (5^2).- 40 can be factored into (2^3 times 5).- 75 can be factored into (3 times 5^2).Next, I'll identify the highest powers of all prime factors present in these factorizations:- The highest power of 2 is (2^3) (from 40).- The highest power of 3 is (3^1) (from 75).- The highest power of 5 is (5^2) (from 25 and 75).Now, I'll calculate the LCM by multiplying these highest powers together:[text{LCM} = 2^3 times 3^1 times 5^2 = 8 times 3 times 25 = 600]With the LCM determined as 600, I need to find the first 4-digit number that is divisible by 600. The smallest 4-digit number is 1000. I'll divide 1000 by 600 to see how many times 600 fits into 1000:[1000 div 600 = 1 text{ with a remainder of } 400]Since there's a remainder, the next multiple of 600 after 1000 is:[1000 + (600 - 400) = 1000 + 200 = 1200]Therefore, the first 4-digit number that is divisible by 25, 40, and 75 is 1200.

💡Okay, so I have this function f(α) that I need to simplify and then evaluate at α = -31π/3. Let me try to work through this step by step.First, looking at the function:f(α) = [sin(π - α) * cos(2π - α) * tan(π + α)] / [tan(-π - α) * sin(-π - α)]Hmm, that's a bit complicated, but maybe I can use some trigonometric identities to simplify each part.Starting with the numerator:1. sin(π - α): I remember that sin(π - α) is equal to sin α. So that simplifies to sin α.2. cos(2π - α): Cosine has a period of 2π, so cos(2π - α) is the same as cos(-α). And since cosine is an even function, cos(-α) = cos α. So that becomes cos α.3. tan(π + α): Tangent has a period of π, so tan(π + α) is the same as tan α. So that simplifies to tan α.Putting it all together, the numerator becomes sin α * cos α * tan α.Now, the denominator:1. tan(-π - α): Let's break this down. tan(-π - α) can be written as tan(-(π + α)). Tangent is an odd function, so tan(-x) = -tan x. Therefore, tan(-(π + α)) = -tan(π + α). But tan(π + α) is equal to tan α because tangent has a period of π. So this becomes -tan α.2. sin(-π - α): Similarly, sin(-π - α) can be written as sin(-(π + α)). Sine is also an odd function, so sin(-x) = -sin x. Therefore, sin(-(π + α)) = -sin(π + α). Now, sin(π + α) is equal to -sin α because sine is negative in the third quadrant. So this becomes -(-sin α) = sin α.Wait, hold on, let me double-check that. sin(π + α) is indeed -sin α because it's in the third quadrant where sine is negative. So sin(-π - α) = -sin(π + α) = -(-sin α) = sin α. Okay, that seems right.So the denominator becomes (-tan α) * sin α.Putting it all together, the function f(α) simplifies to:[sin α * cos α * tan α] / [(-tan α) * sin α]Now, I can cancel out some terms. Both the numerator and the denominator have sin α and tan α. Let's see:- sin α cancels out from numerator and denominator.- tan α cancels out as well, but there's a negative sign in the denominator.So after canceling, we're left with cos α divided by -1, which is -cos α.So f(α) simplifies to -cos α.Alright, that wasn't too bad. Now, moving on to part 2: evaluating f(α) at α = -31π/3.So f(-31π/3) = -cos(-31π/3)But cosine is an even function, so cos(-θ) = cos θ. Therefore, this simplifies to -cos(31π/3).Now, 31π/3 is a bit more than 10π, which is 10 full rotations. But since cosine has a period of 2π, I can subtract multiples of 2π to find an equivalent angle between 0 and 2π.Let me calculate how many times 2π goes into 31π/3.First, 2π is equal to 6π/3. So, 31π/3 divided by 6π/3 is 31/6, which is approximately 5.1667. So that means 5 full periods of 2π, which is 10π, and then a remainder.So, 31π/3 - 10π = 31π/3 - 30π/3 = π/3.Therefore, cos(31π/3) = cos(π/3).And cos(π/3) is a known value; it's 1/2.So, putting it all together:f(-31π/3) = -cos(31π/3) = -cos(π/3) = -1/2.So the value of f(α) at α = -31π/3 is -1/2.I think that makes sense. Let me just recap to make sure I didn't make any mistakes.1. Simplified each trigonometric function using identities.2. Canceled out common terms in numerator and denominator.3. Reduced the angle by subtracting multiples of 2π to find an equivalent angle within one period.4. Evaluated the cosine at the reduced angle and applied the negative sign.Everything seems to check out. I don't see any errors in my steps.**Final Answer**The value of ( fleft(-frac{31pi}{3}right) ) is boxed{-frac{1}{2}}.

💡First, I need to find the value of log 101600 using the given information that log 102 is 0.3010.I can start by expressing 101600 as a product of 102 and 1000, since 101600 equals 102 multiplied by 1000.Next, I'll use the logarithm property that states log(a * b) equals log(a) plus log(b). Applying this property, I can write log(101600) as log(102) plus log(1000).I already know that log(102) is 0.3010. Now, I need to determine log(1000). Since 1000 is 10 raised to the power of 3, log(1000) is equal to 3.Finally, I'll add the two logarithms together: 0.3010 plus 3 equals 3.3010. Therefore, the value of log 101600 is 3.3010.

💡First, I need to understand the problem by visualizing the diagram. Since there are marked angles, I assume they are part of a geometric figure, possibly triangles.Next, I recall that the sum of the interior angles of a single triangle is always (180^circ). This is a fundamental geometric principle.If the marked angles belong to two separate triangles, I can apply this principle to each triangle individually. Adding the sums of both triangles gives:[180^circ + 180^circ = 360^circ]Therefore, the total sum of the marked angles is (360^circ).