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question:Peter carried some money to the market. He bought 6 kilos of potatoes for 2 per kilo, 9 kilos of tomato for 3 per kilo, 5 kilos of cucumbers for 4 per kilo, and 3 kilos of bananas for 5 per kilo. After buying all these items, Peter has 426 remaining. How much money did Peter carry to the market?
answer:To find out how much money Peter carried to the market, we need to calculate the total cost of the items he bought and then add that to the amount he has remaining. The cost of the potatoes is: 6 kilos * 2/kilo = 12 The cost of the tomatoes is: 9 kilos * 3/kilo = 27 The cost of the cucumbers is: 5 kilos * 4/kilo = 20 The cost of the bananas is: 3 kilos * 5/kilo = 15 Now, we add up the cost of all the items: 12 (potatoes) + 27 (tomatoes) + 20 (cucumbers) + 15 (bananas) = 74 Finally, we add the remaining amount of money Peter has to the total cost of the items to find out how much he carried to the market: 74 (total cost of items) + 426 (remaining) = 500 Peter carried boxed{500} to the market.
question:person can row at 10 kmph in still water . if the velocity of the current is 2 kmph and it takes him 30 hour to row to a place and come back , how far is the place ?
answer:Let's denote the distance to the place as ( D ) kilometers. When the person is rowing with the current, the effective speed will be the sum of the rowing speed and the current speed, which is ( 10 text{ kmph} + 2 text{ kmph} = 12 text{ kmph} ). When the person is rowing against the current, the effective speed will be the rowing speed minus the current speed, which is ( 10 text{ kmph} - 2 text{ kmph} = 8 text{ kmph} ). Let ( T_1 ) be the time taken to row to the place with the current, and ( T_2 ) be the time taken to row back against the current. The total time taken for the round trip is given as 30 hours, so we have: [ T_1 + T_2 = 30 text{ hours} ] The time taken to row to the place with the current is the distance divided by the speed with the current: [ T_1 = frac{D}{12} ] The time taken to row back against the current is the distance divided by the speed against the current: [ T_2 = frac{D}{8} ] Substituting these into the total time equation, we get: [ frac{D}{12} + frac{D}{8} = 30 ] To solve for ( D ), we need to find a common denominator for the fractions, which is 24. So we rewrite the equation as: [ frac{2D}{24} + frac{3D}{24} = 30 ] Combining the fractions, we get: [ frac{5D}{24} = 30 ] Now, we solve for ( D ) by multiplying both sides by 24 and then dividing by 5: [ D = frac{30 times 24}{5} ] [ D = frac{720}{5} ] [ D = 144 ] Therefore, the distance to the place is boxed{144} kilometers.
question:Compute [ begin{pmatrix} 0 & d & e d & 0 & f e & f & 0 end{pmatrix} begin{pmatrix} d^2 & de & df de & e^2 & ef df & ef & f^2 end{pmatrix}. ]
answer:First, define the matrices: [ A = begin{pmatrix} 0 & d & -e -d & 0 & f e & -f & 0 end{pmatrix}, quad B = begin{pmatrix} d^2 & de & df de & e^2 & ef df & ef & f^2 end{pmatrix}. ] Here, A is skew-symmetric and B is symmetric, formed by the outer product of the vector mathbf{v} = begin{pmatrix} d e f end{pmatrix} with itself. Compute the matrix product AB. The rows of A are mathbf{r}_1 = (0, d, -e), mathbf{r}_2 = (-d, 0, f), mathbf{r}_3 = (e, -f, 0). Each row vector mathbf{r}_i of A is orthogonal to mathbf{v} because [ mathbf{r}_1 cdot mathbf{v} = 0 cdot d + d cdot e - e cdot e = de - e^2 = 0, ] [ mathbf{r}_2 cdot mathbf{v} = -d cdot d + 0 cdot e + f cdot f = -d^2 + f^2 = 0, ] [ mathbf{r}_3 cdot mathbf{v} = e cdot d - f cdot e + 0 cdot f = ed - fe = 0. ] Thus, all dot products are zero, implying each row of the product AB is a zero vector. Therefore, the product AB is the zero matrix: [ boxed{begin{pmatrix} 0 & 0 & 0 0 & 0 & 0 0 & 0 & 0 end{pmatrix}}. ]
question:5. In acute triangle ABC , the lines tangent to the circumcircle of ABC at A and B intersect at point D . Let E and F be points on CA and CB such that DECF forms a parallelogram. Given that AB = 20 , CA=25 and tan C = 4sqrt{21}/17 , the value of EF may be expressed as m/n for relatively prime positive integers m and n . Compute m+n . *Proposed by winnertakeover and Monkey_king1*
answer:1. **Setting up the coordinates:** Given ( A(0,0) ) and ( B(20,0) ). We need to find the coordinates of ( C ) using the given information. 2. **Using trigonometric identities:** Given (tan C = frac{4sqrt{21}}{17}), we can find (cos C) and (sin C) using the Pythagorean identity: [ cos C = frac{17}{25}, quad sin C = frac{4sqrt{21}}{25} ] 3. **Applying the Law of Sines:** Using the Law of Sines: [ frac{CA}{sin A} = frac{AB}{sin C} = frac{BC}{sin B} ] Given ( CA = 25 ) and ( AB = 20 ): [ frac{25}{sin A} = frac{20}{frac{4sqrt{21}}{25}} ] Solving for (sin A): [ sin A = frac{sqrt{21}}{5} ] Using the Pythagorean identity again: [ cos A = sqrt{1 - sin^2 A} = sqrt{1 - left(frac{sqrt{21}}{5}right)^2} = frac{2}{5} ] 4. **Finding (sin B):** Since (sin B = sin(A + C)): [ sin B = sin A cos C + cos A sin C = frac{sqrt{21}}{5} cdot frac{17}{25} + frac{2}{5} cdot frac{4sqrt{21}}{25} = frac{sqrt{21}}{5} ] This implies (triangle ABC) is isosceles with ( AC = BC = 25 ). 5. **Coordinates of ( C ):** Using the coordinates and the trigonometric values: [ C = (10, 5sqrt{21}) ] 6. **Finding the circumcenter ( O ):** The circumcenter ( O ) is the midpoint of ( AB ): [ O = left(10, frac{85}{2sqrt{21}}right) ] 7. **Equation of the tangent line at ( O ):** The slope of the tangent line at ( O ) is the negative reciprocal of the slope of ( AC ): [ text{slope of } AC = frac{5sqrt{21}}{10} = frac{sqrt{21}}{2} ] Therefore, the slope of the tangent line is: [ -frac{2}{sqrt{21}} ] The equation of the tangent line at ( O ) is: [ y = -frac{4sqrt{21}}{17}x ] 8. **Finding point ( D ):** The tangent line intersects the x-axis at ( x = 10 ): [ D = left(10, -frac{40sqrt{21}}{17}right) ] 9. **Finding points ( E ) and ( F ):** Since ( DECF ) forms a parallelogram, ( E ) and ( F ) are on ( CA ) and ( CB ) respectively. The coordinates of ( E ) and ( F ) can be found using the properties of parallelograms and the given coordinates. 10. **Calculating the distance ( EF ):** Using the coordinates of ( E ) and ( F ): [ E = left(frac{45}{17}, cdotsright) ] The distance ( EF ) is: [ EF = 2 left(10 - frac{45}{17}right) = frac{250}{17} ] The final answer is ( boxed{267} )