💡Okay, so I have this problem where I need to find the projection of the vector (4, -2) onto another vector b, given that vectors a and b are orthogonal. I also know the projection of (4, -2) onto a is (-2/5, -4/5). Hmm, let me think about how to approach this.First, I remember that when you project a vector onto two orthogonal vectors, the projections should add up to the original vector. That is, if a and b are orthogonal, then the projection of any vector onto a and the projection onto b should sum to the original vector. So, maybe I can use that property here.Let me write that down. If I have a vector v, then:proj_a(v) + proj_b(v) = vIs that correct? Wait, actually, I think this is true when a and b form an orthogonal basis for the space, which in this case, since they are orthogonal vectors in 2D, they should form such a basis. So, yes, the sum of the projections onto a and b should give me back the original vector.So, if I know proj_a(v), I can subtract that from v to get proj_b(v). Let me try that.Given that v is (4, -2) and proj_a(v) is (-2/5, -4/5), then:proj_b(v) = v - proj_a(v)So, let's compute that.First, subtract the x-components: 4 - (-2/5). That would be 4 + 2/5. 4 is 20/5, so 20/5 + 2/5 is 22/5.Next, subtract the y-components: -2 - (-4/5). That's -2 + 4/5. -2 is -10/5, so -10/5 + 4/5 is -6/5.So, putting that together, proj_b(v) should be (22/5, -6/5). Hmm, that seems straightforward.Wait, but let me make sure I didn't make a mistake. Maybe I should double-check my reasoning.I know that if a and b are orthogonal, then the projections onto each should be independent. So, subtracting proj_a(v) from v should indeed give me proj_b(v). Let me verify this with another method.Another way to think about it is that since a and b are orthogonal, any vector in the plane can be expressed as a linear combination of a and b. So, v = c*a + d*b, where c and d are scalars. Then, the projection of v onto a would be c*a, and the projection onto b would be d*b. Therefore, subtracting proj_a(v) from v should give me d*b, which is proj_b(v).So, yes, that seems consistent. Therefore, my initial calculation should be correct.But just to be thorough, let me also consider the properties of projections. The projection of v onto a is given by (v · a / ||a||²) * a. Similarly, the projection onto b would be (v · b / ||b||²) * b. Since a and b are orthogonal, their dot product is zero. So, if I add proj_a(v) and proj_b(v), I should get v.Let me write that out:proj_a(v) + proj_b(v) = (v · a / ||a||²) * a + (v · b / ||b||²) * b = vSo, if I subtract proj_a(v) from both sides, I get:proj_b(v) = v - proj_a(v)Which is exactly what I did earlier. So, this confirms that my approach is correct.Therefore, I can confidently say that the projection of (4, -2) onto b is (22/5, -6/5).Wait, just to make sure, let me compute the vector (22/5, -6/5) and see if it makes sense. 22/5 is 4.4, and -6/5 is -1.2. So, the vector is (4.4, -1.2). If I add this to (-0.4, -0.8), which is (-2/5, -4/5), I should get back (4, -2).Let's check:4.4 + (-0.4) = 4.0-1.2 + (-0.8) = -2.0Yes, that adds up correctly. So, this further confirms that my answer is correct.I think I've covered all the bases here. I used the property of orthogonal projections, verified it with another method, and even did a numerical check. So, I'm pretty sure that the projection onto b is (22/5, -6/5).**Final Answer**The projection of the vector onto (mathbf{b}) is boxed{begin{pmatrix} dfrac{22}{5} -dfrac{6}{5} end{pmatrix}}.

💡First, I need to simplify the left side of the equation (23 - 5 = 3 + x). Subtracting 5 from 23 gives me 18, so the equation becomes (18 = 3 + x).Next, to solve for (x), I should isolate it on one side of the equation. I can do this by subtracting 3 from both sides of the equation. Subtracting 3 from 18 gives me 15, so (x = 15).Therefore, the solution to the equation is (x = 15).

💡Okay, so I have this function f(x) = -x³ + 3x² + 9x + 1, and I need to do two things: first, find the intervals where the function is decreasing, and second, find the equation of the tangent line at the point (-2, f(-2)). Hmm, let me start with the first part.Alright, to find where the function is decreasing, I remember that I need to look at the derivative of the function. The derivative will tell me the slope of the function at any point x. If the derivative is negative, the function is decreasing there. So, first step: find f'(x).Let me compute the derivative term by term. The derivative of -x³ is -3x², right? Then, the derivative of 3x² is 6x. The derivative of 9x is 9, and the derivative of the constant 1 is 0. So putting it all together, f'(x) = -3x² + 6x + 9. Got that down.Now, I need to find where this derivative is less than zero because that's where the function is decreasing. So, I need to solve the inequality -3x² + 6x + 9 < 0. Hmm, this is a quadratic inequality. Maybe I should first find the roots of the quadratic equation -3x² + 6x + 9 = 0 to determine the critical points.Let me solve for x. I can factor out a -3 first to make it simpler: -3(x² - 2x - 3) = 0. So, x² - 2x - 3 = 0. Now, let's factor this quadratic. Looking for two numbers that multiply to -3 and add to -2. Hmm, that would be -3 and +1. So, (x - 3)(x + 1) = 0. Therefore, the roots are x = 3 and x = -1.Okay, so the critical points are at x = -1 and x = 3. These divide the real number line into three intervals: (-∞, -1), (-1, 3), and (3, ∞). I need to test each interval to see where the derivative is negative.Let me pick a test point in each interval. For (-∞, -1), let's choose x = -2. Plugging into f'(x): -3(-2)² + 6(-2) + 9 = -3(4) -12 + 9 = -12 -12 + 9 = -15. That's negative, so the function is decreasing here.For (-1, 3), let's pick x = 0. Plugging into f'(x): -3(0)² + 6(0) + 9 = 0 + 0 + 9 = 9. That's positive, so the function is increasing here.For (3, ∞), let's choose x = 4. Plugging into f'(x): -3(4)² + 6(4) + 9 = -3(16) + 24 + 9 = -48 + 24 + 9 = -15. That's negative, so the function is decreasing here as well.So, putting it all together, the function is decreasing on the intervals (-∞, -1) and (3, ∞). That should answer the first part.Now, moving on to the second part: finding the equation of the tangent line at the point (-2, f(-2)). I remember that the equation of a tangent line at a point (a, f(a)) is given by y = f(a) + f'(a)(x - a). So, I need to compute f(-2) and f'(-2).First, let's compute f(-2). Plugging x = -2 into f(x):f(-2) = -(-2)³ + 3(-2)² + 9(-2) + 1.Calculating each term:-(-2)³ = -(-8) = 8.3(-2)² = 3(4) = 12.9(-2) = -18.Adding them up: 8 + 12 - 18 + 1 = (8 + 12) + (-18 + 1) = 20 - 17 = 3.So, f(-2) = 3. Therefore, the point is (-2, 3).Next, compute f'(-2). From earlier, f'(x) = -3x² + 6x + 9. Plugging in x = -2:f'(-2) = -3(-2)² + 6(-2) + 9.Calculating each term:-3(-2)² = -3(4) = -12.6(-2) = -12.Adding them up: -12 -12 + 9 = (-12 -12) + 9 = -24 + 9 = -15.So, the slope of the tangent line at x = -2 is -15.Now, using the point-slope form of the tangent line: y - f(a) = f'(a)(x - a). Plugging in a = -2, f(a) = 3, and f'(a) = -15:y - 3 = -15(x - (-2)) => y - 3 = -15(x + 2).Let me simplify this equation. Distribute the -15 on the right side:y - 3 = -15x - 30.Now, add 3 to both sides to solve for y:y = -15x - 30 + 3 => y = -15x - 27.Alternatively, if I want to write this in standard form (Ax + By + C = 0), I can rearrange the equation:15x + y + 27 = 0.So, that's the equation of the tangent line.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, f(-2): -(-2)^3 is 8, 3*(-2)^2 is 12, 9*(-2) is -18, and +1. So, 8 + 12 is 20, minus 18 is 2, plus 1 is 3. That seems correct.f'(-2): -3*(-2)^2 is -12, 6*(-2) is -12, plus 9. So, -12 -12 is -24, plus 9 is -15. That also seems correct.So, the tangent line equation is y = -15x -27, or 15x + y +27 = 0. Yeah, that looks right.I think I've got both parts done. The intervals where the function is decreasing are (-∞, -1) and (3, ∞), and the tangent line at (-2, 3) is 15x + y +27 = 0.**Final Answer**(1) The intervals of decrease for ( f(x) ) are (boxed{(-infty, -1) text{ and } (3, +infty)}).(2) The equation of the tangent line to ( f(x) ) at the point ((-2, f(-2))) is (boxed{15x + y + 27 = 0}).

💡Okay, so I have this proposition: "For all x > 0, x² - 3x + 2 < 0." I need to find its negation. Hmm, let's break this down step by step.First, I remember that in logic, the negation of a universal statement ("for all") is an existential statement ("there exists"). So, if the original statement is saying that something is true for every x > 0, the negation would say that there exists at least one x > 0 where that something isn't true.The original statement is: For all x > 0, x² - 3x + 2 < 0.So, the negation should be: There exists an x > 0 such that x² - 3x + 2 is not less than 0. In other words, x² - 3x + 2 ≥ 0.Let me write that out to make sure I got it right: ∃x > 0, x² - 3x + 2 ≥ 0.Wait, is that all? It seems straightforward, but maybe I should double-check. Let's think about what the original statement is claiming. It's saying that for every positive x, the quadratic expression x² - 3x + 2 is negative. So, if I negate that, I'm saying that there is at least one positive x where the expression isn't negative, meaning it's either zero or positive.To be thorough, maybe I should solve the inequality x² - 3x + 2 < 0 to see where it holds true and where it doesn't. That might help me understand the negation better.Let's factor the quadratic: x² - 3x + 2. Factoring, we get (x - 1)(x - 2). So, the roots are x = 1 and x = 2.Now, to find where the quadratic is less than zero, I can analyze the intervals determined by the roots: (-∞, 1), (1, 2), and (2, ∞).Since the coefficient of x² is positive, the parabola opens upwards. That means the quadratic will be below the x-axis (negative) between the roots. So, for x between 1 and 2, the expression is negative.Therefore, the original statement "For all x > 0, x² - 3x + 2 < 0" is actually not true because when x is greater than 2, the expression becomes positive again. So, the negation should reflect that there exists an x > 0 (specifically, x > 2) where the expression is not less than zero.Wait, but in the original problem, the domain is x > 0, not all real numbers. So, even though the quadratic is negative between 1 and 2, it's positive for x > 2. Therefore, there are values of x > 0 where the expression is not less than zero.So, the negation is indeed "There exists an x > 0 such that x² - 3x + 2 ≥ 0." That makes sense because the original statement claims the expression is always negative for positive x, but we've found that it's positive for x > 2.Just to make sure, let's test a value greater than 2, say x = 3. Plugging into the expression: 3² - 3*3 + 2 = 9 - 9 + 2 = 2, which is greater than zero. Yep, that confirms it.So, putting it all together, the negation of "For all x > 0, x² - 3x + 2 < 0" is "There exists an x > 0 such that x² - 3x + 2 ≥ 0."I think that's solid. I've considered the logical structure, factored the quadratic, analyzed the intervals, and even tested a value to confirm. Everything checks out.

💡To determine the value of ( k ) that makes the quadratic expression ( x^2 - 18x + k ) the square of a binomial, I start by recognizing that a perfect square trinomial has the form ( (x + b)^2 ), which expands to ( x^2 + 2bx + b^2 ).First, I compare the linear term of the given expression with that of the perfect square. The coefficient of ( x ) in the given expression is (-18), and in the perfect square, it is ( 2b ). Setting these equal gives ( 2b = -18 ), which simplifies to ( b = -9 ).Next, I substitute ( b = -9 ) back into the perfect square to find the constant term. Calculating ( (-9)^2 ) gives ( 81 ). Therefore, the constant term ( k ) must be ( 81 ) to make the quadratic expression a perfect square.Finally, I verify by expanding ( (x - 9)^2 ) to ensure it matches the original expression, confirming that ( x^2 - 18x + 81 ) is indeed the square of the binomial ( (x - 9) ).

💡Okay, so I have this problem where I need to find the smallest constant ( c ) such that the inequality[sum_{1 leq i < j leq n} x_{i} x_{j} left(x_{i}^{2} + x_{j}^{2} right) leqslant c left( sum_{1 leq i leq n} x_{i} right)^{4}]holds for all non-negative real numbers ( x_{1}, x_{2}, cdots, x_{n} ). Then, I also need to determine the necessary and sufficient condition for equality to hold.First, I should understand what the left-hand side (LHS) and the right-hand side (RHS) represent. The LHS is a sum over all pairs ( i < j ) of the product ( x_i x_j (x_i^2 + x_j^2) ). The RHS is a constant ( c ) times the fourth power of the sum of all ( x_i ).Since the inequality is homogeneous, I can assume without loss of generality that the sum ( sum_{i=1}^{n} x_i = 1 ). This simplifies the problem because now I just need to maximize the LHS under the constraint that the sum of ( x_i ) is 1.Let me denote the function to maximize as:[F(x_1, x_2, cdots, x_n) = sum_{1 leq i < j leq n} x_i x_j (x_i^2 + x_j^2)]I need to find the maximum value of ( F ) when ( sum_{i=1}^{n} x_i = 1 ). The smallest constant ( c ) will then be this maximum value because the inequality must hold for all non-negative ( x_i ).To maximize ( F ), I might need to use techniques from optimization, perhaps Lagrange multipliers, but since all ( x_i ) are non-negative, maybe there's a symmetry or a specific configuration where the maximum occurs.Let me consider the case when all ( x_i ) are equal. If ( x_i = frac{1}{n} ) for all ( i ), then:[F = sum_{1 leq i < j leq n} left( frac{1}{n} cdot frac{1}{n} cdot left( left( frac{1}{n} right)^2 + left( frac{1}{n} right)^2 right) right) = sum_{1 leq i < j leq n} frac{1}{n^4} cdot 2 cdot frac{1}{n^2}]Wait, that seems off. Let me recalculate:Each term ( x_i x_j (x_i^2 + x_j^2) ) when ( x_i = x_j = frac{1}{n} ) becomes:[frac{1}{n} cdot frac{1}{n} cdot left( left( frac{1}{n} right)^2 + left( frac{1}{n} right)^2 right) = frac{1}{n^2} cdot frac{2}{n^2} = frac{2}{n^4}]The number of terms in the sum is ( binom{n}{2} = frac{n(n-1)}{2} ). So,[F = frac{n(n-1)}{2} cdot frac{2}{n^4} = frac{n(n-1)}{n^4} = frac{n-1}{n^3}]So, when all ( x_i ) are equal, ( F = frac{n-1}{n^3} ). But is this the maximum? Maybe not. Let me test with smaller ( n ).Take ( n = 2 ). Then, the sum reduces to a single term:[F = x_1 x_2 (x_1^2 + x_2^2)]With ( x_1 + x_2 = 1 ), so ( x_2 = 1 - x_1 ). Then,[F = x_1 (1 - x_1) (x_1^2 + (1 - x_1)^2)]Let me compute this:First, ( x_1^2 + (1 - x_1)^2 = 2x_1^2 - 2x_1 + 1 )So,[F = x_1 (1 - x_1) (2x_1^2 - 2x_1 + 1)]Let me set ( t = x_1 ), then ( F(t) = t(1 - t)(2t^2 - 2t + 1) )To find the maximum, take derivative:First, expand ( F(t) ):( F(t) = t(1 - t)(2t^2 - 2t + 1) )Multiply ( t(1 - t) = t - t^2 )Then,( F(t) = (t - t^2)(2t^2 - 2t + 1) )Multiply term by term:( t cdot 2t^2 = 2t^3 )( t cdot (-2t) = -2t^2 )( t cdot 1 = t )( -t^2 cdot 2t^2 = -2t^4 )( -t^2 cdot (-2t) = 2t^3 )( -t^2 cdot 1 = -t^2 )Combine all terms:( 2t^3 - 2t^2 + t - 2t^4 + 2t^3 - t^2 )Combine like terms:- ( t^4 ): -2t^4- ( t^3 ): 2t^3 + 2t^3 = 4t^3- ( t^2 ): -2t^2 - t^2 = -3t^2- ( t ): tSo,( F(t) = -2t^4 + 4t^3 - 3t^2 + t )Now, take derivative:( F'(t) = -8t^3 + 12t^2 - 6t + 1 )Set derivative to zero:( -8t^3 + 12t^2 - 6t + 1 = 0 )This is a cubic equation. Let me try to find rational roots using Rational Root Theorem. Possible roots are ( pm1, pmfrac{1}{2}, pmfrac{1}{4}, pmfrac{1}{8} ).Test t = 1:( -8 + 12 - 6 + 1 = -1 neq 0 )t = 1/2:( -8*(1/8) + 12*(1/4) - 6*(1/2) + 1 = -1 + 3 - 3 + 1 = 0 )So, t = 1/2 is a root.Factor out (t - 1/2):Use polynomial division or synthetic division.Divide ( -8t^3 + 12t^2 - 6t + 1 ) by (t - 1/2).Using synthetic division:1/2 | -8 12 -6 1 -4 4 -1 ---------------- -8 8 -2 0So, the cubic factors as (t - 1/2)(-8t^2 + 8t - 2)Set the quadratic to zero:( -8t^2 + 8t - 2 = 0 )Multiply both sides by -1:( 8t^2 - 8t + 2 = 0 )Divide by 2:( 4t^2 - 4t + 1 = 0 )Discriminant: ( 16 - 16 = 0 ), so double root at t = (4)/(2*4) = 1/2Thus, all roots are t = 1/2 (double root) and t = 1/2.Wait, that can't be. Wait, the quadratic equation ( 4t^2 -4t +1 =0 ) has discriminant 16 - 16 = 0, so t = (4)/(8) = 1/2.So, all roots are t = 1/2. So, the derivative is zero only at t = 1/2.Thus, the function F(t) has a critical point at t = 1/2.Compute F(1/2):( F(1/2) = (1/2)(1 - 1/2)(2*(1/2)^2 - 2*(1/2) + 1) )Simplify:( (1/2)(1/2)(2*(1/4) - 1 + 1) = (1/4)( (1/2) -1 +1 ) = (1/4)(1/2) = 1/8 )So, when n=2, the maximum of F is 1/8, achieved when x1 = x2 = 1/2.So, for n=2, c=1/8.But what about n>2?Let me test n=3.For n=3, we have variables x1, x2, x3, with x1 + x2 + x3 =1.Compute F = x1x2(x1² +x2²) + x1x3(x1² +x3²) + x2x3(x2² +x3²)I need to find the maximum of this expression.Perhaps the maximum occurs when two variables are equal and the third is zero.Let me set x3=0, then F reduces to x1x2(x1² +x2²), which is the same as the n=2 case, so maximum 1/8.Alternatively, if all three variables are equal, x1=x2=x3=1/3.Compute F:Each term is (1/3)(1/3)( (1/3)^2 + (1/3)^2 ) = (1/9)(2/9) = 2/81.There are 3 terms, so total F= 3*(2/81)=6/81=2/27≈0.074, which is less than 1/8≈0.125.So, in n=3, the maximum is still 1/8, achieved when two variables are 1/2 and the third is 0.Similarly, for n=4, if I set two variables to 1/2 and the rest to 0, F=1/8, and if I spread the variables more, F decreases.Therefore, it seems that regardless of n≥2, the maximum of F is 1/8, achieved when two variables are 1/2 and the rest are 0.Thus, the smallest constant c is 1/8.For the equality condition, it must hold when exactly two variables are equal to 1/2 and the rest are zero.So, summarizing:1. The smallest constant c is 1/8.2. Equality holds if and only if exactly two of the variables are equal to 1/2 and the rest are zero.**Final Answer**The smallest constant ( c ) is (boxed{dfrac{1}{8}}), and equality holds when exactly two of the ( x_i ) are equal to (dfrac{1}{2}) and the rest are zero.