💡Okay, so I need to evaluate the definite integral from 0 to 2 of (sqrt(4 - x²) - x) dx. Hmm, let me think about how to approach this. I remember that integrals can sometimes be interpreted geometrically, especially when dealing with functions like sqrt(4 - x²), which looks like part of a circle. First, maybe I can split the integral into two separate integrals to make it easier. So, I can write it as the integral from 0 to 2 of sqrt(4 - x²) dx minus the integral from 0 to 2 of x dx. That seems manageable.Starting with the first integral, the integral of sqrt(4 - x²) dx from 0 to 2. I recall that the integral of sqrt(a² - x²) dx from 0 to a is (πa²)/4, which represents the area of a quarter-circle. In this case, a is 2 because sqrt(4 - x²) is the equation of a semicircle with radius 2. So, if I plug in a = 2, the integral should be (π*(2)²)/4 = (4π)/4 = π. That makes sense because it's a quarter of the area of the full circle, which would be πr² = 4π, so a quarter of that is π.Now, moving on to the second integral, the integral of x dx from 0 to 2. I think the antiderivative of x is (1/2)x². So, evaluating that from 0 to 2, it would be (1/2)(2)² - (1/2)(0)² = (1/2)(4) - 0 = 2. That seems straightforward.Putting it all together, the original integral is the first integral minus the second integral, so π - 2. That should be the value of the definite integral from 0 to 2 of (sqrt(4 - x²) - x) dx.Wait, let me double-check to make sure I didn't make any mistakes. For the first integral, I used the geometric interpretation as a quarter-circle, which seems correct. The radius is 2, so the area is π. For the second integral, integrating x gives (1/2)x², and evaluating from 0 to 2 gives 2. Subtracting those, π - 2, seems right.I think that's the correct answer. It combines both the geometric interpretation and basic integration techniques, which fits with what I've learned so far. I don't see any errors in my reasoning, so I feel confident about this solution.

💡Okay, so I have this problem where I need to find how many integers n, between 1 and 150 inclusive, make the fraction n/(n+1) a repeating decimal. Hmm, I remember that whether a fraction is terminating or repeating depends on the denominator after simplifying the fraction. First, I should recall that a fraction will be a terminating decimal if and only if the denominator has no prime factors other than 2 or 5. That means if the denominator can be factored into 2s and 5s only, the decimal will terminate; otherwise, it will repeat. In this case, the fraction is n/(n+1). Since n and n+1 are consecutive integers, they are coprime, right? So, the fraction is already in its simplest form. That means the denominator is n+1, and I need to check if n+1 has any prime factors other than 2 or 5. If it does, then the decimal will repeat. So, my task reduces to finding how many numbers between 2 and 151 (since n+1 ranges from 2 to 151 when n is from 1 to 150) are not of the form 2^a * 5^b, where a and b are non-negative integers. Let me list out all the numbers between 2 and 151 that are only divisible by 2 and 5. These would be numbers like 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128. Let me count them: 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128. That's 16 numbers. Wait, let me make sure I didn't miss any. Starting from the smallest:- Powers of 2: 2, 4, 8, 16, 32, 64, 128. That's 7 numbers.- Powers of 5: 5, 25, 125. That's 3 numbers.- Now, combinations of 2 and 5: 10 (2*5), 20 (4*5), 40 (8*5), 80 (16*5), 160 (32*5) but 160 is beyond 151, so stop at 80. Similarly, 50 (2*25), 100 (4*25), 200 (8*25) which is beyond 151. So, 50, 100. Wait, I think I missed 25*2=50, 25*4=100, 25*8=200 which is too big. So, 50 and 100. So, combining all these, I have 7 (powers of 2) + 3 (powers of 5) + 4 (combinations) = 14. But earlier I thought it was 16. Hmm, maybe I double-counted some. Let me list them again:From powers of 2: 2, 4, 8, 16, 32, 64, 128.From powers of 5: 5, 25, 125.From combinations: 10, 20, 40, 50, 80, 100.Wait, that's 7 + 3 + 6 = 16. So, 16 numbers in total. So, there are 16 numbers between 2 and 151 that are only divisible by 2 and 5. Therefore, the number of n+1 that are not of this form is 151 - 16 = 135. Hence, there are 135 integers n between 1 and 150 for which n/(n+1) is a repeating decimal.

💡First, I need to determine how much David earned from mowing lawns during the week. He mowed for 2 hours each day for 7 days, so the total hours he worked are 2 multiplied by 7, which equals 14 hours.Next, I'll calculate his total earnings by multiplying the number of hours he worked by the hourly rate of 14. This gives me 14 hours multiplied by 14 per hour, resulting in 196.David then spent half of his earnings on a pair of shoes. Half of 196 is 98, so he spent 98 on shoes. This leaves him with 196 minus 98, which is 98 remaining.Finally, David gave half of the remaining 98 to his mom. Half of 98 is 49, so after giving 49 to his mom, he is left with 49.

💡Alright, so I have this equation to solve: ( x = left|3x - |80 - 3x|right| ). Hmm, it looks a bit complicated with absolute values inside absolute values. I remember that absolute value equations can have multiple solutions, so I need to consider different cases based on the expressions inside the absolute values.First, let me focus on the inner absolute value: ( |80 - 3x| ). The expression inside this absolute value, ( 80 - 3x ), can be either positive or negative. So, I'll split this into two cases.**Case 1:** ( 80 - 3x geq 0 )If ( 80 - 3x geq 0 ), then ( |80 - 3x| = 80 - 3x ). Substituting this back into the original equation, we get:( x = |3x - (80 - 3x)| )Simplify inside the absolute value:( x = |3x - 80 + 3x| = |6x - 80| )Now, this is another absolute value equation, so I'll have to consider two sub-cases here as well.**Sub-case 1.1:** ( 6x - 80 geq 0 )If ( 6x - 80 geq 0 ), then ( |6x - 80| = 6x - 80 ). So, the equation becomes:( x = 6x - 80 )Subtract ( x ) from both sides:( 0 = 5x - 80 )Add 80 to both sides:( 5x = 80 )Divide by 5:( x = 16 )Now, I need to check if this solution satisfies the conditions of both Case 1 and Sub-case 1.1.First, for Case 1: ( 80 - 3x geq 0 )Plugging ( x = 16 ):( 80 - 3(16) = 80 - 48 = 32 geq 0 ) ✔️Next, for Sub-case 1.1: ( 6x - 80 geq 0 )Plugging ( x = 16 ):( 6(16) - 80 = 96 - 80 = 16 geq 0 ) ✔️So, ( x = 16 ) is a valid solution.**Sub-case 1.2:** ( 6x - 80 < 0 )If ( 6x - 80 < 0 ), then ( |6x - 80| = -(6x - 80) = -6x + 80 ). So, the equation becomes:( x = -6x + 80 )Add ( 6x ) to both sides:( 7x = 80 )Divide by 7:( x = frac{80}{7} ) ≈ 11.4286Again, I need to check if this solution satisfies the conditions of both Case 1 and Sub-case 1.2.First, for Case 1: ( 80 - 3x geq 0 )Plugging ( x = frac{80}{7} ):( 80 - 3left(frac{80}{7}right) = 80 - frac{240}{7} = frac{560}{7} - frac{240}{7} = frac{320}{7} approx 45.714 geq 0 ) ✔️Next, for Sub-case 1.2: ( 6x - 80 < 0 )Plugging ( x = frac{80}{7} ):( 6left(frac{80}{7}right) - 80 = frac{480}{7} - 80 = frac{480}{7} - frac{560}{7} = -frac{80}{7} approx -11.4286 < 0 ) ✔️So, ( x = frac{80}{7} ) is also a valid solution.**Case 2:** ( 80 - 3x < 0 )If ( 80 - 3x < 0 ), then ( |80 - 3x| = -(80 - 3x) = 3x - 80 ). Substituting this back into the original equation, we get:( x = |3x - (3x - 80)| )Simplify inside the absolute value:( x = |3x - 3x + 80| = |80| = 80 )So, the equation simplifies to ( x = 80 ).Now, I need to check if this solution satisfies the condition of Case 2.For Case 2: ( 80 - 3x < 0 )Plugging ( x = 80 ):( 80 - 3(80) = 80 - 240 = -160 < 0 ) ✔️So, ( x = 80 ) is a valid solution.Now, I have three solutions: ( x = 16 ), ( x = frac{80}{7} ), and ( x = 80 ). The problem asks for the sum of all the solutions.Let me calculate the sum:( 16 + frac{80}{7} + 80 )First, add 16 and 80:( 16 + 80 = 96 )Now, add ( frac{80}{7} ):( 96 + frac{80}{7} )To add these, convert 96 to a fraction with denominator 7:( 96 = frac{672}{7} )So,( frac{672}{7} + frac{80}{7} = frac{752}{7} )Therefore, the sum of all solutions is ( frac{752}{7} ).I should double-check my calculations to make sure I didn't make any mistakes.First, in Case 1, Sub-case 1.1: ( x = 16 ) is valid.Sub-case 1.2: ( x = frac{80}{7} ) is valid.Case 2: ( x = 80 ) is valid.Sum: ( 16 + frac{80}{7} + 80 = frac{112}{7} + frac{80}{7} + frac{560}{7} = frac{112 + 80 + 560}{7} = frac{752}{7} ). Yep, that looks correct.So, I think I've covered all possible cases and correctly found the solutions and their sum.

💡To determine how many integers ( n ) between 1 and 150 have a greatest common divisor (gcd) of 3 with 21, we need to consider the prime factors of 21, which are 3 and 7. 1. **Multiples of 3**: The numbers that have a gcd of at least 3 with 21 are the multiples of 3. There are ( frac{150}{3} = 50 ) multiples of 3 between 1 and 150.2. **Excluding multiples of 21**: To ensure the gcd is exactly 3 and not higher (i.e., not 21), we must exclude the multiples of 21. There are ( frac{150}{21} approx 7 ) multiples of 21 between 1 and 150.3. **Final Count**: Subtract the number of multiples of 21 from the number of multiples of 3 to get the count of numbers where gcd(21, ( n )) = 3.Thus, the number of integers ( n ) between 1 and 150 with gcd(21, ( n )) = 3 is ( 50 - 7 = 43 ).

💡First, I need to substitute ( x = -2 ) into the polynomial ( f(x) = x^2 + 3x - 5 ).Calculating each term step by step:- ( (-2)^2 = 4 )- ( 3 times (-2) = -6 )- The constant term is ( -5 )Next, I add these results together:( 4 - 6 - 5 = -7 )Therefore, the value of ( f(-2) ) is (-7).