💡To solve the equation (frac{x+4}{x-3} = frac{x-2}{x+2}), I'll start by cross-multiplying to eliminate the denominators. This gives me ((x+4)(x+2) = (x-2)(x-3)).Next, I'll expand both sides of the equation. On the left side, multiplying (x+4) by (x+2) results in (x^2 + 6x + 8). On the right side, multiplying (x-2) by (x-3) gives (x^2 - 5x + 6).Now, I'll simplify the equation by subtracting (x^2) from both sides, which leaves me with (6x + 8 = -5x + 6). To isolate (x), I'll add (5x) to both sides, resulting in (11x + 8 = 6). Finally, subtracting 8 from both sides gives (11x = -2), and dividing by 11 yields (x = -frac{2}{11}).Thus, the solution to the equation is (x = -frac{2}{11}).

💡First, I need to determine the amount of chromium in each of the original alloys. The first alloy has 12% chromium, and I have 10 kg of it. To find the chromium content, I'll calculate 12% of 10 kg.Next, the second alloy contains 8% chromium, and I have 30 kg of this alloy. I'll calculate 8% of 30 kg to find the chromium content in the second alloy.After finding the chromium amounts in both alloys, I'll add them together to get the total chromium in the new alloy. Additionally, I'll add the weights of the two original alloys to determine the total weight of the new alloy.Finally, to find the percentage of chromium in the new alloy, I'll divide the total chromium by the total weight of the new alloy and multiply by 100%.

💡Okay, so I have this problem here with four options, A through D, and I need to figure out which one is correct. Let me take it step by step because I want to make sure I understand each operation properly.Starting with **Option A**: It says ( x^{2} cdot x^{3} = x^{6} ). Hmm, I remember that when you multiply two exponents with the same base, you add the exponents. So, ( x^{2} cdot x^{3} ) should be ( x^{2+3} = x^{5} ). But here it's saying ( x^{6} ), which is not right. So, A is incorrect.Moving on to **Option B**: ( x^{12} div x^{2} = x^{6} ). I think when you divide exponents with the same base, you subtract the exponents. So, ( x^{12} div x^{2} ) should be ( x^{12-2} = x^{10} ). But the option says ( x^{6} ), so that's wrong too. B is incorrect.Now, **Option C**: ( (x + y)^{2} = x^{2} + y^{2} ). Wait a second, I recall that squaring a binomial like ( (x + y)^{2} ) involves more than just squaring each term. It should be ( x^{2} + 2xy + y^{2} ). So, the middle term ( 2xy ) is missing here. That means C is also incorrect.Finally, **Option D**: ( (x^{2}y)^{3} = x^{6}y^{3} ). Let me break this down. When you raise a product to a power, you raise each factor to that power. So, ( (x^{2})^{3} = x^{2 times 3} = x^{6} ) and ( y^{3} ) stays as it is. Putting it together, it's ( x^{6}y^{3} ), which matches what's given in D. So, D looks correct.To double-check, I'll go through each option again quickly:- A: ( x^{2} cdot x^{3} = x^{5} ), not ( x^{6} ).- B: ( x^{12} div x^{2} = x^{10} ), not ( x^{6} ).- C: ( (x + y)^{2} = x^{2} + 2xy + y^{2} ), missing the ( 2xy ) term.- D: ( (x^{2}y)^{3} = x^{6}y^{3} ), which is correct.Yeah, I'm confident now that D is the correct operation.

💡To find the largest four-digit number that is divisible by the sum of its digits, we start by considering the largest four-digit number, which is 9999. We then check each number sequentially downward until we find one that satisfies the condition.1. **Check 9999**: - Sum of digits: (9 + 9 + 9 + 9 = 36) - (9999 div 36 = 277.75) (not an integer)2. **Check 9998**: - Sum of digits: (9 + 9 + 9 + 8 = 35) - (9998 div 35 = 285.6571) (not an integer)3. **Continue this process...**4. **Check 9990**: - Sum of digits: (9 + 9 + 9 + 0 = 27) - (9990 div 27 = 370) (an integer)Thus, the largest four-digit number that is divisible by the sum of its digits is 9990.

💡Okay, so I have this inequality to solve: ( ax^2 - (a + 2)x + 2 < 0 ), where ( a ) is a real number. Hmm, quadratic inequalities can be tricky, especially when the coefficient of ( x^2 ) isn't fixed—it depends on ( a ). I need to figure out how the value of ( a ) affects the solution set for ( x ).First, I remember that to solve a quadratic inequality, it's helpful to factor the quadratic expression if possible. Let me try factoring ( ax^2 - (a + 2)x + 2 ). Looking at the quadratic, I notice that the coefficient of ( x^2 ) is ( a ), and the constant term is 2. Maybe I can factor this by grouping. Let me see:( ax^2 - (a + 2)x + 2 )I can rewrite the middle term ( -(a + 2)x ) as ( -ax - 2x ). So, the expression becomes:( ax^2 - ax - 2x + 2 )Now, let's group the first two terms and the last two terms:( (ax^2 - ax) + (-2x + 2) )Factor out ( a ) from the first group and ( -2 ) from the second group:( a(x^2 - x) - 2(x - 1) )Hmm, that doesn't seem to help much because ( x^2 - x ) and ( x - 1 ) aren't the same. Maybe I should try a different approach. Perhaps factor the quadratic as ( (mx + n)(px + q) ). Let me attempt that.We need ( m times p = a ) and ( n times q = 2 ). Also, the cross terms should add up to ( -(a + 2) ). Let me try ( m = a ) and ( p = 1 ), so that ( m times p = a times 1 = a ). Then, for the constants, I need ( n times q = 2 ). Let's try ( n = 2 ) and ( q = 1 ). So, the factors would be ( (ax + 2)(x + 1) ). Let's check:( (ax + 2)(x + 1) = ax^2 + ax + 2x + 2 = ax^2 + (a + 2)x + 2 )Wait, that's almost the original quadratic, but the middle term is ( +(a + 2)x ) instead of ( -(a + 2)x ). So, I need to adjust the signs. Maybe ( (ax - 2)(x - 1) ). Let's multiply that out:( (ax - 2)(x - 1) = ax^2 - ax - 2x + 2 = ax^2 - (a + 2)x + 2 )Yes! That works. So, the quadratic factors as ( (ax - 2)(x - 1) ). Therefore, the inequality becomes:( (ax - 2)(x - 1) < 0 )Now, to solve this inequality, I need to find the critical points where each factor is zero. So, setting each factor equal to zero:1. ( ax - 2 = 0 ) ⇒ ( x = frac{2}{a} )2. ( x - 1 = 0 ) ⇒ ( x = 1 )These critical points divide the real number line into intervals. The sign of the product ( (ax - 2)(x - 1) ) will alternate between these intervals. However, the order of the critical points depends on the value of ( a ).Let me consider different cases based on the value of ( a ):**Case 1: ( a = 0 )**If ( a = 0 ), the quadratic becomes ( 0x^2 - (0 + 2)x + 2 = -2x + 2 ). So, the inequality simplifies to:( -2x + 2 < 0 )Solving for ( x ):( -2x < -2 )Divide both sides by -2 (remembering to reverse the inequality sign when dividing by a negative):( x > 1 )So, when ( a = 0 ), the solution is ( x > 1 ).**Case 2: ( a > 0 )**When ( a ) is positive, the quadratic opens upwards. The critical points are ( x = frac{2}{a} ) and ( x = 1 ). The order of these points depends on the value of ( a ):- If ( frac{2}{a} > 1 ), which implies ( a < 2 ), then the critical points are ordered as ( 1 < frac{2}{a} ).- If ( frac{2}{a} = 1 ), which implies ( a = 2 ), then both critical points coincide at ( x = 1 ).- If ( frac{2}{a} < 1 ), which implies ( a > 2 ), then the critical points are ordered as ( frac{2}{a} < 1 ).Let's analyze each subcase:**Subcase 2.1: ( 0 < a < 2 )**Here, ( frac{2}{a} > 1 ). The quadratic ( (ax - 2)(x - 1) ) opens upwards, so the product is negative between the roots. Therefore, the solution is:( 1 < x < frac{2}{a} )**Subcase 2.2: ( a = 2 )**Here, ( frac{2}{a} = 1 ). The quadratic becomes ( (2x - 2)(x - 1) = 2(x - 1)(x - 1) = 2(x - 1)^2 ). The inequality is:( 2(x - 1)^2 < 0 )Since ( (x - 1)^2 ) is always non-negative and multiplied by 2, the left side is always non-negative. Therefore, there is no solution— the inequality is never true.**Subcase 2.3: ( a > 2 )**Here, ( frac{2}{a} < 1 ). The quadratic opens upwards, so the product is negative between the roots. Therefore, the solution is:( frac{2}{a} < x < 1 )**Case 3: ( a < 0 )**When ( a ) is negative, the quadratic opens downwards. The critical points are still ( x = frac{2}{a} ) and ( x = 1 ). However, since ( a ) is negative, ( frac{2}{a} ) is negative. So, the order of the critical points is ( frac{2}{a} < 1 ).Since the quadratic opens downwards, the product ( (ax - 2)(x - 1) ) is negative outside the interval between the roots. Therefore, the solution is:( x < frac{2}{a} ) or ( x > 1 )But wait, ( frac{2}{a} ) is negative, so ( x < frac{2}{a} ) is a region to the left of a negative number, and ( x > 1 ) is to the right of 1. So, the solution set is two separate intervals.Let me double-check this. If ( a < 0 ), then ( ax - 2 ) is a line with a negative slope, crossing the x-axis at ( x = frac{2}{a} ) (negative). The other root is at ( x = 1 ). Since the parabola opens downward, the expression is negative when ( x ) is less than the smaller root or greater than the larger root. But since ( frac{2}{a} ) is negative and 1 is positive, the expression is negative for ( x < frac{2}{a} ) or ( x > 1 ).Yes, that makes sense.**Summary of Cases:**- If ( a = 0 ): ( x > 1 )- If ( 0 < a < 2 ): ( 1 < x < frac{2}{a} )- If ( a = 2 ): No solution- If ( a > 2 ): ( frac{2}{a} < x < 1 )- If ( a < 0 ): ( x < frac{2}{a} ) or ( x > 1 )I think I've covered all possible cases for ( a ). Let me just verify with a specific example for each case to ensure my solutions make sense.**Verification:**1. **Case ( a = 0 ):** Let's choose ( a = 0 ). The inequality becomes ( -2x + 2 < 0 ), which simplifies to ( x > 1 ). If I pick ( x = 2 ), the left side is ( -4 + 2 = -2 < 0 ), which satisfies the inequality. If I pick ( x = 0 ), the left side is ( 0 + 2 = 2 ), which does not satisfy the inequality. So, correct.2. **Case ( 0 < a < 2 ):** Let's choose ( a = 1 ). The inequality becomes ( x^2 - 3x + 2 < 0 ), which factors to ( (x - 1)(x - 2) < 0 ). The solution is ( 1 < x < 2 ), which matches my earlier result ( 1 < x < frac{2}{1} = 2 ). Testing ( x = 1.5 ): ( (1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0 ). Correct.3. **Case ( a = 2 ):** The inequality becomes ( 2x^2 - 4x + 2 < 0 ), which factors to ( 2(x - 1)^2 < 0 ). Since ( (x - 1)^2 ) is always non-negative, multiplying by 2 keeps it non-negative. Thus, no solution. Correct.4. **Case ( a > 2 ):** Let's choose ( a = 3 ). The inequality becomes ( 3x^2 - 5x + 2 < 0 ), which factors to ( (3x - 2)(x - 1) < 0 ). The critical points are ( x = frac{2}{3} ) and ( x = 1 ). Since ( a > 2 ), the solution should be ( frac{2}{3} < x < 1 ). Testing ( x = 0.5 ): ( (1.5 - 2)(0.5 - 1) = (-0.5)(-0.5) = 0.25 > 0 ). Testing ( x = 0.75 ): ( (2.25 - 2)(0.75 - 1) = (0.25)(-0.25) = -0.0625 < 0 ). Testing ( x = 1.5 ): ( (4.5 - 2)(1.5 - 1) = (2.5)(0.5) = 1.25 > 0 ). So, the solution is indeed between ( frac{2}{3} ) and 1. Correct.5. **Case ( a < 0 ):** Let's choose ( a = -1 ). The inequality becomes ( -x^2 - ( -1 + 2)x + 2 < 0 ) ⇒ ( -x^2 - x + 2 < 0 ). Let's factor this: ( -(x^2 + x - 2) = -(x + 2)(x - 1) ). So, the inequality is ( -(x + 2)(x - 1) < 0 ) ⇒ ( (x + 2)(x - 1) > 0 ). The critical points are ( x = -2 ) and ( x = 1 ). The product is positive when ( x < -2 ) or ( x > 1 ). According to my earlier solution, it should be ( x < frac{2}{a} = frac{2}{-1} = -2 ) or ( x > 1 ). Testing ( x = -3 ): ( (-3 + 2)(-3 - 1) = (-1)(-4) = 4 > 0 ). Testing ( x = 0 ): ( (0 + 2)(0 - 1) = 2(-1) = -2 < 0 ). Testing ( x = 2 ): ( (2 + 2)(2 - 1) = 4(1) = 4 > 0 ). So, the solution is ( x < -2 ) or ( x > 1 ), which matches my result. Correct.Everything seems to check out. I think I've covered all the cases and verified them with specific examples. So, the final solution depends on the value of ( a ) as outlined above.

💡Okay, so I've got this equation here: 0.75 raised to the power of x, plus 2, equals 8. I need to find the value of x that makes this equation true. Alright, let's break this down step by step.First, I want to isolate the term with x in it. Right now, the equation is 0.75^x + 2 = 8. To get 0.75^x by itself, I should subtract 2 from both sides of the equation. That way, I'm not changing the equality, just simplifying it.So, subtracting 2 from both sides gives me:0.75^x = 8 - 20.75^x = 6Now, I have 0.75 raised to the power of x equals 6. I need to solve for x, which is in the exponent. This seems like a good time to use logarithms because they're useful for solving equations where the variable is in the exponent.I remember that taking the logarithm of both sides can help me bring the exponent down. So, I'll take the natural logarithm (ln) of both sides. The natural logarithm is just one type of logarithm, but it works well here.Taking the natural logarithm of both sides:ln(0.75^x) = ln(6)Now, there's a logarithm property that says ln(a^b) = b * ln(a). This means I can bring the exponent x down in front of the logarithm. Applying this property to the left side:x * ln(0.75) = ln(6)Now, I need to solve for x. To do that, I'll divide both sides of the equation by ln(0.75):x = ln(6) / ln(0.75)At this point, I can plug these values into a calculator to find the numerical value of x. Let me recall that ln(6) is approximately 1.791759 and ln(0.75) is approximately -0.287682.So, plugging these in:x ≈ 1.791759 / -0.287682x ≈ -6.226251635Wait a minute, x is negative? That seems a bit odd because 0.75 is less than 1, and raising a number less than 1 to a negative exponent actually makes it larger. Let me check if that makes sense.If x is negative, say x = -6.226, then 0.75^(-6.226) is the same as 1 / (0.75^6.226). Since 0.75 raised to a positive power would be a number less than 1, taking the reciprocal would make it greater than 1. In this case, 0.75^6.226 is approximately 0.75^6, which is about 0.177, so 1 / 0.177 is roughly 5.65, and adding 2 gives us around 7.65, which is close to 8. So, it does make sense that x is negative.I think I've done everything correctly. I isolated the term with x, used logarithms to bring down the exponent, solved for x, and even checked if the negative value makes sense in the context of the problem. I don't see any mistakes in my calculations, so I'm confident that x is approximately -6.226.