answer:Let's denote the weight of the basket alone as B kg and the total weight of the persimmons as P kg. Initially, the basket full of persimmons weighed 62 kg, so we have: B + P = 62 kg After removing half of the persimmons, the weight of the remaining persimmons is P/2. When you weighed the basket with half of the persimmons, it was 34 kg, so we have: B + P/2 = 34 kg Now we have two equations: 1) B + P = 62 2) B + P/2 = 34 We can solve these equations to find the weight of the basket alone (B). Let's subtract equation 2 from equation 1: (B + P) - (B + P/2) = 62 - 34 B + P - B - P/2 = 28 P - P/2 = 28 This simplifies to: P/2 = 28 Now we know that half of the persimmons weigh 28 kg, so the full weight of the persimmons (P) is: P = 28 * 2 P = 56 kg Now we can use the value of P to find the weight of the basket (B) using equation 1: B + P = 62 B + 56 = 62 B = 62 - 56 B = 6 kg Therefore, the basket alone weighs boxed{6} kg.

answer:Since the projection of begin{pmatrix} 2 -4 end{pmatrix} is begin{pmatrix} 3 -3 end{pmatrix}, the vector being projected onto can be assumed to be a scalar multiple of begin{pmatrix} 3 -3 end{pmatrix}. We will assume the vector being projected onto is begin{pmatrix} 1 -1 end{pmatrix} after normalizing begin{pmatrix} 3 -3 end{pmatrix}. The projection of begin{pmatrix} -8 2 end{pmatrix} onto begin{pmatrix} 1 -1 end{pmatrix} is calculated as follows: [ operatorname{proj}_{begin{pmatrix} 1 -1 end{pmatrix}} begin{pmatrix} -8 2 end{pmatrix} = frac{begin{pmatrix} -8 2 end{pmatrix} cdot begin{pmatrix} 1 -1 end{pmatrix}}{begin{pmatrix} 1 -1 end{pmatrix} cdot begin{pmatrix} 1 -1 end{pmatrix}} begin{pmatrix} 1 -1 end{pmatrix} = frac{-10}{2} begin{pmatrix} 1 -1 end{pmatrix} = boxed{begin{pmatrix} -5 5 end{pmatrix}}. ]

answer:To find the value of cot alpha given that sin 2 alpha = -sin alpha and alpha in left(frac{pi}{2}, piright), we can proceed as follows: 1. **Apply the Double Angle Identity for Sine**: The double angle identity for sine states that: [ sin 2 alpha = 2 sin alpha cos alpha ] Given in the problem that: [ sin 2 alpha = -sin alpha ] we substiture the double angle identity: [ 2 sin alpha cos alpha = -sin alpha ] 2. **Solve for cos alpha**: Since sin alpha neq 0, we can divide both sides of the equation by sin alpha: [ 2 cos alpha = -1 ] Solving for cos alpha, we get: [ cos alpha = -frac{1}{2} ] 3. **Find sin alpha**: Recall that alpha in left(frac{pi}{2}, piright), which places alpha in the second quadrant where sin alpha is positive and cos alpha is negative. We use the Pythagorean identity: [ sin^2 alpha + cos^2 alpha = 1 ] Substituting cos alpha = -frac{1}{2}: [ sin^2 alpha + left(-frac{1}{2}right)^2 = 1 ] [ sin^2 alpha + frac{1}{4} = 1 ] [ sin^2 alpha = 1 - frac{1}{4} ] [ sin^2 alpha = frac{3}{4} ] [ sin alpha = sqrt{frac{3}{4}} = frac{sqrt{3}}{2} ] 4. **Calculate cot alpha**: Cotangent is the ratio of cosine to sine: [ cot alpha = frac{cos alpha}{sin alpha} ] Substituting the values we have found: [ cot alpha = frac{-frac{1}{2}}{frac{sqrt{3}}{2}} ] [ cot alpha = -frac{1}{2} cdot frac{2}{sqrt{3}} ] [ cot alpha = -frac{1}{sqrt{3}} ] Rationalizing the denominator: [ cot alpha = -frac{1}{sqrt{3}} cdot frac{sqrt{3}}{sqrt{3}} ] [ cot alpha = -frac{sqrt{3}}{3} ] # Conclusion: [ boxed{-frac{sqrt{3}}{3}} ]

answer:Let's denote the side of the initial square as ( x ). The total number of bottles when arranged in a square is ( x^2 ), and there are 36 bottles left over. So the total number of bottles is ( x^2 + 36 ). When you increase the width and height of the square by one column each, the new arrangement forms a square with a side of ( x + 1 ). The total number of bottles in this new square is ( (x + 1)^2 ), and there are 3 bottles left over. So the total number of bottles is also ( (x + 1)^2 + 3 ). Since the total number of bottles doesn't change, we can set these two expressions equal to each other: [ x^2 + 36 = (x + 1)^2 + 3 ] Expanding the right side of the equation: [ x^2 + 36 = x^2 + 2x + 1 + 3 ] Simplify the equation: [ x^2 + 36 = x^2 + 2x + 4 ] Subtract ( x^2 ) from both sides: [ 36 = 2x + 4 ] Subtract 4 from both sides: [ 32 = 2x ] Divide both sides by 2: [ x = 16 ] Now that we have the value of ( x ), we can find the total number of bottles by plugging it back into either of the original expressions: [ x^2 + 36 = 16^2 + 36 = 256 + 36 = 292 ] So there are boxed{292} bottles in total.