💡Okay, so I have this system of equations with four variables: a, b, c, and d. The equations are:1. ( a^3 + c^3 = 2 )2. ( a^2 b + c^2 d = 0 )3. ( b^3 + d^3 = 1 )4. ( a b^2 + c d^2 = -6 )Hmm, I need to find all quadruples (a, b, c, d) that satisfy these equations simultaneously. Let me see how to approach this.First, I notice that the equations involve cubes and products of variables. Maybe I can find a way to relate these equations or express some variables in terms of others. Let me look for patterns or symmetries.Looking at equations 1 and 3, both are sums of cubes. Maybe I can use the identity for the sum of cubes: ( x^3 + y^3 = (x + y)(x^2 - xy + y^2) ). But I'm not sure if that will help directly here. Let me see.Equation 1: ( a^3 + c^3 = 2 ). If I factor this, it becomes ( (a + c)(a^2 - a c + c^2) = 2 ).Equation 3: ( b^3 + d^3 = 1 ). Similarly, this factors to ( (b + d)(b^2 - b d + d^2) = 1 ).I don't know if that helps yet. Maybe I can consider equations 2 and 4 together. Equation 2 is ( a^2 b + c^2 d = 0 ) and equation 4 is ( a b^2 + c d^2 = -6 ).These look similar in structure. Maybe I can think of them as dot products or something. Alternatively, perhaps I can express one variable in terms of others.Let me consider equations 2 and 4. If I write them as:( a^2 b = -c^2 d ) (from equation 2)( a b^2 = -c d^2 - 6 ) (from equation 4)Hmm, maybe I can solve for b and d in terms of a and c. Let me try.From equation 2: ( a^2 b = -c^2 d ) => ( b = -frac{c^2}{a^2} d ), assuming ( a neq 0 ).Similarly, from equation 4: ( a b^2 = -c d^2 - 6 ). If I substitute b from equation 2 into this, I get:( a left( -frac{c^2}{a^2} d right)^2 = -c d^2 - 6 )Let me compute the left side:( a left( frac{c^4}{a^4} d^2 right) = frac{c^4}{a^3} d^2 )So, equation 4 becomes:( frac{c^4}{a^3} d^2 = -c d^2 - 6 )Let me multiply both sides by ( a^3 ) to eliminate the denominator:( c^4 d^2 = -c a^3 d^2 - 6 a^3 )Hmm, this seems complicated. Maybe I can factor out ( c d^2 ):( c d^2 (c^3 + a^3) = -6 a^3 )Wait, from equation 1, ( a^3 + c^3 = 2 ). So, ( c^3 + a^3 = 2 ). Therefore, the equation becomes:( c d^2 (2) = -6 a^3 )Simplify:( 2 c d^2 = -6 a^3 ) => ( c d^2 = -3 a^3 )So, ( c d^2 = -3 a^3 ). Let me note this as equation 5.Now, going back to equation 3: ( b^3 + d^3 = 1 ). From equation 2, we have ( b = -frac{c^2}{a^2} d ). Let me substitute this into equation 3.So, ( left( -frac{c^2}{a^2} d right)^3 + d^3 = 1 )Compute the first term:( -frac{c^6}{a^6} d^3 + d^3 = 1 )Factor out ( d^3 ):( d^3 left( 1 - frac{c^6}{a^6} right) = 1 )Hmm, this is getting more complicated. Maybe I need a different approach.Let me think about the original equations again. Maybe I can consider the variables in pairs, like (a, c) and (b, d). Equations 1 and 3 involve only a, c and b, d respectively, while equations 2 and 4 mix them.Another idea: Maybe I can use substitution or elimination. Let me see if I can express b and d in terms of a and c, then substitute into equations 1 and 3.From equation 2: ( a^2 b + c^2 d = 0 ) => ( a^2 b = -c^2 d ) => ( b = -frac{c^2}{a^2} d ) (assuming ( a neq 0 ))From equation 4: ( a b^2 + c d^2 = -6 ). Substitute b from above:( a left( -frac{c^2}{a^2} d right)^2 + c d^2 = -6 )Compute the first term:( a cdot frac{c^4}{a^4} d^2 = frac{c^4}{a^3} d^2 )So, equation 4 becomes:( frac{c^4}{a^3} d^2 + c d^2 = -6 )Factor out ( c d^2 ):( c d^2 left( frac{c^3}{a^3} + 1 right) = -6 )But from equation 1, ( a^3 + c^3 = 2 ), so ( frac{c^3}{a^3} = frac{2 - a^3}{a^3} = frac{2}{a^3} - 1 )Substitute this into the equation:( c d^2 left( frac{2}{a^3} - 1 + 1 right) = -6 )Simplify inside the parentheses:( frac{2}{a^3} )So, equation becomes:( c d^2 cdot frac{2}{a^3} = -6 )Multiply both sides by ( a^3 ):( 2 c d^2 = -6 a^3 )Which simplifies to:( c d^2 = -3 a^3 )Wait, this is the same as equation 5 I derived earlier. So, that's consistent.Now, from equation 5: ( c d^2 = -3 a^3 )Let me see if I can express d in terms of a and c. From equation 5:( d^2 = -frac{3 a^3}{c} )But ( d^2 ) must be non-negative since it's a square. So, ( -frac{3 a^3}{c} geq 0 )This implies that ( frac{a^3}{c} leq 0 ). So, either ( a^3 ) and c have opposite signs, or one of them is zero.But from equation 1, ( a^3 + c^3 = 2 ). If a or c were zero, the other would have to be cube root of 2 or -2, but let's check.If a = 0, then ( c^3 = 2 ) => c = ( sqrt[3]{2} ). Then from equation 2: ( 0 + c^2 d = 0 ) => ( c^2 d = 0 ). Since c ≠ 0, d must be 0. Then from equation 3: ( b^3 + 0 = 1 ) => b = 1. From equation 4: ( 0 + c d^2 = -6 ). But d = 0, so 0 = -6, which is impossible. So, a cannot be zero.Similarly, if c = 0, then ( a^3 = 2 ) => a = ( sqrt[3]{2} ). From equation 2: ( a^2 b + 0 = 0 ) => b = 0. From equation 3: ( 0 + d^3 = 1 ) => d = 1. From equation 4: ( a b^2 + 0 = -6 ). But b = 0, so 0 = -6, which is impossible. So, c cannot be zero.Therefore, both a and c are non-zero, and from ( frac{a^3}{c} leq 0 ), we have that a and c have opposite signs.So, either a > 0 and c < 0, or a < 0 and c > 0.Let me proceed.From equation 5: ( c d^2 = -3 a^3 )From equation 2: ( b = -frac{c^2}{a^2} d )From equation 3: ( b^3 + d^3 = 1 ). Substitute b from equation 2:( left( -frac{c^2}{a^2} d right)^3 + d^3 = 1 )Compute the first term:( -frac{c^6}{a^6} d^3 + d^3 = 1 )Factor out ( d^3 ):( d^3 left( 1 - frac{c^6}{a^6} right) = 1 )Let me write this as:( d^3 = frac{1}{1 - frac{c^6}{a^6}} = frac{a^6}{a^6 - c^6} )But from equation 1, ( a^3 + c^3 = 2 ). Let me denote ( a^3 = x ), so ( c^3 = 2 - x ). Then ( a^6 = x^2 ), ( c^6 = (2 - x)^2 ).So, ( a^6 - c^6 = x^2 - (2 - x)^2 = x^2 - (4 - 4x + x^2) = 4x - 4 )Therefore, ( d^3 = frac{x^2}{4x - 4} = frac{x^2}{4(x - 1)} )So, ( d = sqrt[3]{frac{x^2}{4(x - 1)}} )But x = a^3, so:( d = sqrt[3]{frac{a^6}{4(a^3 - 1)}} )Hmm, this is getting quite involved. Maybe I can express everything in terms of a single variable.Let me try to express c in terms of a from equation 1: ( c^3 = 2 - a^3 ) => ( c = sqrt[3]{2 - a^3} )Similarly, from equation 5: ( c d^2 = -3 a^3 ) => ( d^2 = -frac{3 a^3}{c} )But c is expressed in terms of a, so:( d^2 = -frac{3 a^3}{sqrt[3]{2 - a^3}} )Since d^2 must be non-negative, the right-hand side must be non-negative. So:( -frac{3 a^3}{sqrt[3]{2 - a^3}} geq 0 )This implies that ( frac{a^3}{sqrt[3]{2 - a^3}} leq 0 )So, either both numerator and denominator are negative, or both are positive.But let's analyze:Case 1: ( a^3 > 0 ) and ( sqrt[3]{2 - a^3} < 0 )This would require ( a > 0 ) and ( 2 - a^3 < 0 ) => ( a^3 > 2 ) => ( a > sqrt[3]{2} )Case 2: ( a^3 < 0 ) and ( sqrt[3]{2 - a^3} > 0 )This would require ( a < 0 ) and ( 2 - a^3 > 0 ). Since a is negative, ( a^3 ) is negative, so ( 2 - a^3 > 2 ), which is always positive.So, possible cases are:1. ( a > sqrt[3]{2} )2. ( a < 0 )Let me consider these cases separately.Case 1: ( a > sqrt[3]{2} )In this case, ( c = sqrt[3]{2 - a^3} ). Since ( a^3 > 2 ), ( 2 - a^3 < 0 ), so c is negative.From equation 5: ( c d^2 = -3 a^3 ). Since c is negative and a^3 is positive, the right-hand side is negative. But d^2 is non-negative, so c must be negative to make the left-hand side negative, which is consistent.From equation 5: ( d^2 = -frac{3 a^3}{c} = -frac{3 a^3}{sqrt[3]{2 - a^3}} )But since ( 2 - a^3 < 0 ), ( sqrt[3]{2 - a^3} ) is negative, so ( -frac{3 a^3}{sqrt[3]{2 - a^3}} ) is positive, as it should be.So, d is real.From equation 3: ( b^3 + d^3 = 1 ). From equation 2: ( b = -frac{c^2}{a^2} d )So, ( b = -frac{(sqrt[3]{2 - a^3})^2}{a^2} d )Let me compute ( b^3 ):( b^3 = -frac{(2 - a^3)^{2/3}}{a^6} d^3 )From equation 3: ( b^3 + d^3 = 1 ), so:( -frac{(2 - a^3)^{2/3}}{a^6} d^3 + d^3 = 1 )Factor out ( d^3 ):( d^3 left( 1 - frac{(2 - a^3)^{2/3}}{a^6} right) = 1 )Let me express ( (2 - a^3)^{2/3} ) as ( left( (2 - a^3)^{1/3} right)^2 = c^2 )So, equation becomes:( d^3 left( 1 - frac{c^2}{a^6} right) = 1 )But from equation 5: ( c d^2 = -3 a^3 ) => ( d^2 = -frac{3 a^3}{c} ) => ( d = sqrt{ -frac{3 a^3}{c} } ). Wait, but d is real, so the expression inside the square root must be non-negative, which we already considered.But this seems circular. Maybe I need to find a relationship between a and d.Alternatively, let me consider that from equation 5: ( c d^2 = -3 a^3 ). Since c = ( sqrt[3]{2 - a^3} ), we have:( sqrt[3]{2 - a^3} cdot d^2 = -3 a^3 )Let me cube both sides to eliminate the cube root:( (2 - a^3) cdot d^6 = (-3 a^3)^3 = -27 a^9 )So,( (2 - a^3) d^6 = -27 a^9 )But from equation 3: ( b^3 + d^3 = 1 ). From equation 2: ( b = -frac{c^2}{a^2} d ), so ( b^3 = -frac{c^6}{a^6} d^3 )Thus, equation 3 becomes:( -frac{c^6}{a^6} d^3 + d^3 = 1 )Factor out ( d^3 ):( d^3 left( 1 - frac{c^6}{a^6} right) = 1 )Again, ( c^6 = (2 - a^3)^2 ), so:( d^3 left( 1 - frac{(2 - a^3)^2}{a^6} right) = 1 )Let me denote ( t = a^3 ), so ( c^3 = 2 - t ), and ( c^6 = (2 - t)^2 ). Then, equation becomes:( d^3 left( 1 - frac{(2 - t)^2}{t^2} right) = 1 )Simplify the expression inside the parentheses:( 1 - frac{(2 - t)^2}{t^2} = frac{t^2 - (4 - 4t + t^2)}{t^2} = frac{4t - 4}{t^2} = frac{4(t - 1)}{t^2} )So, equation becomes:( d^3 cdot frac{4(t - 1)}{t^2} = 1 ) => ( d^3 = frac{t^2}{4(t - 1)} )But from earlier, we have:( (2 - t) d^6 = -27 t^3 )From ( d^3 = frac{t^2}{4(t - 1)} ), we can write ( d^6 = left( frac{t^2}{4(t - 1)} right)^2 = frac{t^4}{16(t - 1)^2} )Substitute into the equation:( (2 - t) cdot frac{t^4}{16(t - 1)^2} = -27 t^3 )Multiply both sides by 16(t - 1)^2:( (2 - t) t^4 = -27 t^3 cdot 16(t - 1)^2 )Simplify:( (2 - t) t^4 = -432 t^3 (t - 1)^2 )Divide both sides by t^3 (assuming t ≠ 0, which is true since a ≠ 0):( (2 - t) t = -432 (t - 1)^2 )Expand both sides:Left side: ( 2t - t^2 )Right side: ( -432(t^2 - 2t + 1) = -432 t^2 + 864 t - 432 )Bring all terms to one side:( 2t - t^2 + 432 t^2 - 864 t + 432 = 0 )Combine like terms:( ( -t^2 + 432 t^2 ) + (2t - 864 t) + 432 = 0 )( 431 t^2 - 862 t + 432 = 0 )Simplify by dividing all terms by 431:( t^2 - 2 t + frac{432}{431} = 0 )Wait, 432 / 431 is approximately 1.0023, but let me check the calculation again because this seems messy.Wait, let me double-check the previous steps.After expanding:Left side: ( 2t - t^2 )Right side: ( -432 t^2 + 864 t - 432 )Bring all terms to left:( 2t - t^2 + 432 t^2 - 864 t + 432 = 0 )Combine like terms:- t^2 + 432 t^2 = 431 t^22t - 864 t = -862 tSo, equation is:( 431 t^2 - 862 t + 432 = 0 )Divide by 431:( t^2 - 2 t + frac{432}{431} = 0 )Compute discriminant:( D = ( -2 )^2 - 4 cdot 1 cdot frac{432}{431} = 4 - frac{1728}{431} )Convert 4 to ( frac{1724}{431} ):( D = frac{1724}{431} - frac{1728}{431} = -frac{4}{431} )Negative discriminant, so no real solutions in this case.So, Case 1: ( a > sqrt[3]{2} ) leads to no real solutions.Case 2: ( a < 0 )In this case, from equation 1: ( c^3 = 2 - a^3 ). Since a < 0, ( a^3 < 0 ), so ( c^3 = 2 - a^3 > 2 ), so c > ( sqrt[3]{2} ).From equation 5: ( c d^2 = -3 a^3 ). Since c > 0 and a^3 < 0, the right-hand side is positive, so d^2 is positive, which is fine.From equation 2: ( b = -frac{c^2}{a^2} d ). Since c > 0, a^2 > 0, so b = -k d, where k > 0.From equation 3: ( b^3 + d^3 = 1 ). Substitute b:( (-k d)^3 + d^3 = 1 ) => ( -k^3 d^3 + d^3 = 1 ) => ( d^3 (1 - k^3) = 1 )From equation 5: ( c d^2 = -3 a^3 ). Since a < 0, let me write a = -m, where m > 0.So, a = -m, m > 0.Then, ( a^3 = -m^3 ), so ( c^3 = 2 - (-m^3) = 2 + m^3 ) => c = ( sqrt[3]{2 + m^3} )From equation 5: ( c d^2 = -3 a^3 = -3 (-m^3) = 3 m^3 )So, ( d^2 = frac{3 m^3}{c} = frac{3 m^3}{sqrt[3]{2 + m^3}} )From equation 2: ( b = -frac{c^2}{a^2} d = -frac{(sqrt[3]{2 + m^3})^2}{m^2} d )Let me compute ( k = frac{c^2}{m^2} = frac{(2 + m^3)^{2/3}}{m^2} )So, b = -k dFrom equation 3: ( b^3 + d^3 = 1 ) => ( (-k d)^3 + d^3 = 1 ) => ( -k^3 d^3 + d^3 = 1 ) => ( d^3 (1 - k^3) = 1 )So, ( d^3 = frac{1}{1 - k^3} )But k = ( frac{(2 + m^3)^{2/3}}{m^2} ), so ( k^3 = frac{(2 + m^3)^2}{m^6} )Thus, ( d^3 = frac{1}{1 - frac{(2 + m^3)^2}{m^6}} = frac{m^6}{m^6 - (2 + m^3)^2} )Simplify denominator:( m^6 - (4 + 4 m^3 + m^6) = -4 - 4 m^3 )So, ( d^3 = frac{m^6}{-4 - 4 m^3} = -frac{m^6}{4(m^3 + 1)} )Thus, ( d = sqrt[3]{ -frac{m^6}{4(m^3 + 1)} } = -sqrt[3]{ frac{m^6}{4(m^3 + 1)} } )Since d is real, this is fine.From equation 5: ( d^2 = frac{3 m^3}{c} = frac{3 m^3}{sqrt[3]{2 + m^3}} )But from above, ( d^3 = -frac{m^6}{4(m^3 + 1)} ), so ( d^2 = left( d^3 right)^{2/3} = left( -frac{m^6}{4(m^3 + 1)} right)^{2/3} )But this seems complicated. Alternatively, let me express d in terms of m.From ( d^3 = -frac{m^6}{4(m^3 + 1)} ), we have:( d = -sqrt[3]{frac{m^6}{4(m^3 + 1)}} = -frac{m^2}{sqrt[3]{4(m^3 + 1)}} )So, ( d = -frac{m^2}{sqrt[3]{4(m^3 + 1)}} )From equation 5: ( d^2 = frac{3 m^3}{sqrt[3]{2 + m^3}} )Compute ( d^2 ):( d^2 = left( -frac{m^2}{sqrt[3]{4(m^3 + 1)}} right)^2 = frac{m^4}{sqrt[3]{16(m^3 + 1)^2}} )Set equal to equation 5:( frac{m^4}{sqrt[3]{16(m^3 + 1)^2}} = frac{3 m^3}{sqrt[3]{2 + m^3}} )Multiply both sides by ( sqrt[3]{16(m^3 + 1)^2} ):( m^4 = 3 m^3 cdot frac{sqrt[3]{16(m^3 + 1)^2}}{sqrt[3]{2 + m^3}} )Simplify the fraction:( frac{sqrt[3]{16(m^3 + 1)^2}}{sqrt[3]{2 + m^3}} = sqrt[3]{frac{16(m^3 + 1)^2}{2 + m^3}} )Note that ( m^3 + 1 = (m + 1)(m^2 - m + 1) ), but not sure if that helps.Let me denote ( n = m^3 ), so ( n > 0 ). Then, equation becomes:( m^4 = 3 m^3 cdot sqrt[3]{frac{16(n + 1)^2}{n + 2}} )Divide both sides by ( m^3 ) (since m > 0):( m = 3 cdot sqrt[3]{frac{16(n + 1)^2}{n + 2}} )But ( n = m^3 ), so:( m = 3 cdot sqrt[3]{frac{16(m^3 + 1)^2}{m^3 + 2}} )Let me cube both sides:( m^3 = 27 cdot frac{16(m^3 + 1)^2}{m^3 + 2} )Simplify:( m^3 (m^3 + 2) = 432 (m^3 + 1)^2 )Expand both sides:Left side: ( m^6 + 2 m^3 )Right side: ( 432 (m^6 + 2 m^3 + 1) )Bring all terms to left:( m^6 + 2 m^3 - 432 m^6 - 864 m^3 - 432 = 0 )Combine like terms:( (1 - 432) m^6 + (2 - 864) m^3 - 432 = 0 )( -431 m^6 - 862 m^3 - 432 = 0 )Multiply both sides by -1:( 431 m^6 + 862 m^3 + 432 = 0 )Let me denote ( p = m^3 ), so equation becomes:( 431 p^2 + 862 p + 432 = 0 )Divide all terms by 431:( p^2 + 2 p + frac{432}{431} = 0 )Compute discriminant:( D = 4 - 4 cdot 1 cdot frac{432}{431} = 4 - frac{1728}{431} )Convert 4 to ( frac{1724}{431} ):( D = frac{1724}{431} - frac{1728}{431} = -frac{4}{431} )Negative discriminant again, so no real solutions in this case.Therefore, both cases lead to no real solutions. Hence, the system has no real solutions.

💡Okay, so I have this problem here: I need to find the phase shift and amplitude of the graph of y equals 2 sine of (4x minus pi over 2). Hmm, alright. Let me think about how to approach this.First, I remember that the general form of a sine function is y equals A sine(Bx minus C). In this form, A represents the amplitude, B affects the period, and C is related to the phase shift. So, comparing this to my equation, y equals 2 sine(4x minus pi/2), I can identify A, B, and C.Let me write that down:y = 2 sin(4x - π/2)So, A is 2, B is 4, and C is π/2. Got that.Now, amplitude is straightforward, right? The amplitude is just the absolute value of A. So, in this case, A is 2, so the amplitude is 2. That seems simple enough.But what about the phase shift? I think the phase shift is calculated by dividing C by B, but I need to make sure I get the direction right. I recall that if the function is y = A sin(Bx - C), the phase shift is C divided by B. But wait, is it to the right or to the left?Let me think. If it's y = A sin(Bx - C), it's equivalent to y = A sin(B(x - C/B)). So, this means the graph is shifted to the right by C/B. So, yes, the phase shift is C divided by B, and it's a shift to the right.So, in my case, C is π/2 and B is 4. Therefore, the phase shift should be (π/2) divided by 4, which is π/8. So, the phase shift is π/8 to the right.Wait, let me double-check that. If I rewrite the equation as y = 2 sin(4(x - π/8)), then yes, it's shifted π/8 units to the right. That makes sense.So, putting it all together, the amplitude is 2, and the phase shift is π/8 to the right.I think that's it. But just to be thorough, let me review the steps:1. Identify the general form: y = A sin(Bx - C).2. Compare with the given equation: y = 2 sin(4x - π/2).3. Extract A, B, and C: A = 2, B = 4, C = π/2.4. Calculate amplitude: |A| = 2.5. Calculate phase shift: C/B = (π/2)/4 = π/8.6. Determine direction: Since it's in the form (Bx - C), it's a shift to the right.Yep, that all checks out. I don't think I made any mistakes here. It seems pretty straightforward once I break it down step by step.Just to visualize, the amplitude of 2 means the graph will go up to 2 and down to -2. The phase shift of π/8 means that instead of starting at the origin, it starts π/8 units to the right. The period, which is 2π divided by B, would be 2π/4 = π/2, but since the question only asks for amplitude and phase shift, I don't need to worry about that right now.So, in summary, the amplitude is 2, and the phase shift is π/8 to the right.

💡Okay, so I need to find the focus of the parabola given by the equation ( y = -2x^2 + 5 ). Hmm, I remember that parabolas have a vertex and a focus, and depending on the direction they open, the focus is located at a certain point relative to the vertex. First, let me recall the standard forms of a parabola. There's the standard form for a vertical parabola, which is ( y = ax^2 + bx + c ), and the vertex form, which is ( y = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. Since the given equation is ( y = -2x^2 + 5 ), it looks like it's already in the standard form, but without the ( bx ) term. So, that means ( b = 0 ) here. I think the vertex form will be helpful because it directly gives me the vertex coordinates. Let me try to rewrite the given equation into vertex form. Since there's no ( bx ) term, the equation simplifies to ( y = -2x^2 + 5 ), which is the same as ( y = -2(x - 0)^2 + 5 ). So, the vertex ( (h, k) ) is at ( (0, 5) ). That makes sense because when ( x = 0 ), ( y = 5 ), so the vertex is at the point ( (0, 5) ).Now, to find the focus, I remember that for a parabola in the form ( y = ax^2 + bx + c ), the focus is located at a distance of ( frac{1}{4a} ) from the vertex. But wait, I need to be careful here because the coefficient ( a ) affects the direction the parabola opens. If ( a ) is positive, the parabola opens upwards, and if ( a ) is negative, it opens downwards. In this case, ( a = -2 ), so the parabola opens downward.So, the formula for the focus when the parabola is in the form ( y = ax^2 ) is ( (0, frac{1}{4a}) ) relative to the vertex. But since our vertex is at ( (0, 5) ), I need to adjust this formula accordingly. Let me write down the formula for the focus. If the vertex is ( (h, k) ), then the focus is at ( (h, k + frac{1}{4a}) ). Since ( a = -2 ), plugging that in, the focus should be at ( (0, 5 + frac{1}{4(-2)}) ). Calculating ( frac{1}{4a} ), that's ( frac{1}{4(-2)} = frac{1}{-8} = -frac{1}{8} ). So, the focus is at ( (0, 5 - frac{1}{8}) ). Now, I need to compute ( 5 - frac{1}{8} ). Converting 5 into eighths, that's ( frac{40}{8} ). So, ( frac{40}{8} - frac{1}{8} = frac{39}{8} ). Therefore, the focus is at ( (0, frac{39}{8}) ).Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the equation ( y = -2x^2 + 5 ), I identified the vertex correctly at ( (0, 5) ). Then, using the formula for the focus, which is ( (h, k + frac{1}{4a}) ), I substituted ( a = -2 ), so ( frac{1}{4a} = -frac{1}{8} ). Adding that to the vertex y-coordinate, ( 5 - frac{1}{8} = frac{39}{8} ). Yes, that seems correct. The focus is indeed at ( (0, frac{39}{8}) ). I think I got it right.

💡First, I need to determine the speed of the boat in still water. I know that the man rows 30 km upstream in 3 hours, and the speed of the stream is 7 km/h. Using the formula for speed, which is distance divided by time, I can set up the equation for the upstream journey.Next, I'll solve this equation to find the speed of the boat in still water. Once I have the boat's speed, I can use it to calculate the distance rowed downstream. Since the time taken for the downstream journey is also 3 hours, I'll apply the same formula for speed but this time for the downstream scenario.Finally, by plugging in the known values, I can find the distance the man rowed downstream.

💡Okay, so I need to figure out the range of ( g(x) - 2x ) given the graph of ( y = g(x) ) which is piecewise linear through the points ( (-4, -2) ), ( (-2, -1) ), ( (0, 0) ), ( (2, 3) ), and ( (4, 4) ). Hmm, let me think about how to approach this.First, I know that the range of a function is the set of all possible output values (y-values) it can produce. So, in this case, I need to find all possible values of ( g(x) - 2x ) as ( x ) varies from -4 to 4.Since ( g(x) ) is piecewise linear, it means it's made up of straight line segments connecting the given points. That should make things easier because I can analyze each segment separately.Let me list out the points again to visualize the graph:- ( (-4, -2) )- ( (-2, -1) )- ( (0, 0) )- ( (2, 3) )- ( (4, 4) )So, the graph of ( g(x) ) is a series of connected line segments between these points. Now, I need to consider the function ( h(x) = g(x) - 2x ). Essentially, this is subtracting ( 2x ) from each y-value of ( g(x) ).To find the range of ( h(x) ), I can evaluate ( h(x) ) at each of the given points and also check if there are any maximum or minimum values within the intervals between these points.Let me start by calculating ( h(x) ) at each of the given points:1. At ( x = -4 ): ( h(-4) = g(-4) - 2(-4) = -2 + 8 = 6 )2. At ( x = -2 ): ( h(-2) = g(-2) - 2(-2) = -1 + 4 = 3 )3. At ( x = 0 ): ( h(0) = g(0) - 2(0) = 0 - 0 = 0 )4. At ( x = 2 ): ( h(2) = g(2) - 2(2) = 3 - 4 = -1 )5. At ( x = 4 ): ( h(4) = g(4) - 2(4) = 4 - 8 = -4 )So, at the given points, ( h(x) ) takes the values 6, 3, 0, -1, and -4. That gives me some idea about the range, but I need to make sure there aren't any higher or lower values between these points.Since ( g(x) ) is piecewise linear, each segment between two points is a straight line. Therefore, ( h(x) = g(x) - 2x ) will also be linear on each segment because subtracting ( 2x ) is a linear operation.For a linear function, the maximum and minimum values occur at the endpoints of the interval. That means on each segment between two points, the maximum and minimum of ( h(x) ) will be at the endpoints, which are the points we've already evaluated.Therefore, the maximum value of ( h(x) ) is 6, and the minimum value is -4. So, the range of ( h(x) ) should be from -4 to 6.Wait, let me double-check to make sure I haven't missed anything. Maybe there's a segment where ( h(x) ) could be higher or lower than at the endpoints?Looking at the segments:1. Between ( x = -4 ) and ( x = -2 ): The function goes from ( -4, 6 ) to ( -2, 3 ). Since it's linear, it decreases from 6 to 3. So, no higher or lower values here.2. Between ( x = -2 ) and ( x = 0 ): From ( -2, 3 ) to ( 0, 0 ). It decreases from 3 to 0. Again, no extremes beyond the endpoints.3. Between ( x = 0 ) and ( x = 2 ): From ( 0, 0 ) to ( 2, -1 ). It decreases from 0 to -1. Still, no new extremes.4. Between ( x = 2 ) and ( x = 4 ): From ( 2, -1 ) to ( 4, -4 ). It decreases further to -4. So, the minimum is at x=4.Therefore, my initial conclusion seems correct. The range of ( g(x) - 2x ) is from -4 to 6.I think that's it. I don't see any mistakes in my reasoning. The key was recognizing that since ( h(x) ) is linear on each segment, the extrema occur at the endpoints, which we've already calculated.

💡First, identify the total number of ways to select three different numbers from the set {2, 0, 1, 5}. There are 4 possible combinations: (2, 0, 1), (2, 0, 5), (0, 1, 5), and (2, 1, 5).Next, determine which of these combinations have the number 2 as the median. To be the median, 2 must be the middle number when the three selected numbers are arranged in order. The combinations where 2 is the median are (2, 0, 5) and (2, 1, 5).There are 2 favorable combinations out of 4 total combinations. Therefore, the probability is 2/4, which simplifies to 1/2.