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question:Due to the restrictions of Coronavirus, Mr. Trent, the local store owner, increased the prices of goods at his store by 15% each to make the same income he was making before. If buying an orange was 40 and the price of purchasing a mango was 50, calculate how much it costs in total now to buy ten oranges and ten mangoes at the new prices.
answer:First, we need to calculate the new price for each orange and each mango after the 15% increase. For an orange: The original price of an orange is 40. The price increase is 15% of 40, which is (15/100) * 40 = 6. So, the new price for an orange is 40 + 6 = 46. For a mango: The original price of a mango is 50. The price increase is 15% of 50, which is (15/100) * 50 = 7.50. So, the new price for a mango is 50 + 7.50 = 57.50. Now, let's calculate the total cost for ten oranges and ten mangoes at the new prices. For ten oranges: 10 oranges * 46 per orange = 460. For ten mangoes: 10 mangoes * 57.50 per mango = 575. Finally, we add the total cost for oranges and mangoes together to get the total cost. Total cost = 460 (for oranges) + 575 (for mangoes) = 1035. So, it now costs boxed{1035} in total to buy ten oranges and ten mangoes at the new prices.
question:Define a new operation Y such that a Y b = b + 12a a^3. Calculate the value of 3Y5.
answer:Using the defined operation Y, we compute 3 Y 5. 1. Substitute a = 3 and b = 5 into the operation a Y b = b + 12a - a^3. 2. Calculate the expression: 3 Y 5 = 5 + 12(3) - 3^3 = 5 + 36 - 27 = 41 - 27 = 14 Thus, the value of 3Y5 is boxed{14}.
question:Three mutually perpendicular rays O_x,O_y,O_z and three points A,B,C on O_x,O_y,O_z , respectively. A variable sphere є through A, B,C meets O_x,O_y,O_z again at A', B',C' , respectively. Let M and M' be the centroids of triangles ABC and A'B'C' . Find the locus of the midpoint of MM' .
answer:1. Let the coordinates of points ( A, B, C ) be ( (a, 0, 0), (0, b, 0), (0, 0, c) ) respectively. These points lie on the coordinate axes ( O_x, O_y, O_z ). 2. The sphere passing through ( A, B, C ) intersects the coordinate axes again at points ( A', B', C' ) with coordinates ( (a', 0, 0), (0, b', 0), (0, 0, c') ) respectively. 3. The centroid ( M ) of triangle ( ABC ) is given by: [ M = left( frac{a}{3}, frac{b}{3}, frac{c}{3} right) ] 4. The centroid ( M' ) of triangle ( A'B'C' ) is given by: [ M' = left( frac{a'}{3}, frac{b'}{3}, frac{c'}{3} right) ] 5. The midpoint ( S ) of segment ( MM' ) is given by: [ S = left( frac{frac{a}{3} + frac{a'}{3}}{2}, frac{frac{b}{3} + frac{b'}{3}}{2}, frac{frac{c}{3} + frac{c'}{3}}{2} right) ] Simplifying, we get: [ S = left( frac{a + a'}{6}, frac{b + b'}{6}, frac{c + c'}{6} right) ] 6. Let ( A'', B'', C'' ) be the midpoints of ( AA', BB', CC' ) respectively. Then: [ A'' = left( frac{a + a'}{2}, 0, 0 right), quad B'' = left( 0, frac{b + b'}{2}, 0 right), quad C'' = left( 0, 0, frac{c + c'}{2} right) ] 7. The coordinates of ( S ) can be rewritten as: [ S = left( frac{a''}{3}, frac{b''}{3}, frac{c''}{3} right) ] 8. Since ( A'', B'', C'' ) are the midpoints of ( AA', BB', CC' ), they lie on the coordinate axes. Therefore, the coordinates of ( S ) are: [ S = left( frac{x}{3}, frac{y}{3}, frac{z}{3} right) ] where ( x, y, z ) are the coordinates of the midpoints of ( AA', BB', CC' ) respectively. Conclusion: The locus of the midpoint of ( MM' ) is a line passing through the origin and is given by: [ S = left( frac{x}{3}, frac{y}{3}, frac{z}{3} right) ] The final answer is ( boxed{ left( frac{x}{3}, frac{y}{3}, frac{z}{3} right) } ).
question:Write the decomposition of the vector x in terms of the vectors p, q, r: x = {5 ; 9 ; 13} p = {0 ; 1 ; 2} q = {3 ; 1 ; 1} r = {4 ; 1 ; 0}
answer:To express the vector (x) in terms of the vectors (p, q, , text{and} , r), we need to find scalars (alpha), (beta), and (gamma) such that: [ x = alpha p + beta q + gamma r ] Given: [ x = begin{pmatrix} -5 9 -13 end{pmatrix}, quad p = begin{pmatrix} 0 1 -2 end{pmatrix}, quad q = begin{pmatrix} 3 -1 1 end{pmatrix}, quad r = begin{pmatrix} 4 1 0 end{pmatrix} ] We can set up the following system of equations based on the components of the vectors: [ left{ begin{array}{l} alpha cdot 0 + beta cdot 3 + gamma cdot 4 = -5 alpha cdot 1 + beta cdot (-1) + gamma cdot 1 = 9 alpha cdot (-2) + beta cdot 1 + gamma cdot 0 = -13 end{array} right. ] Simplifying, we obtain: [ left{ begin{array}{l} 3beta + 4gamma = -5 alpha - beta + gamma = 9 -2alpha + beta = -13 end{array} right. ] Now we will solve this system step-by-step: 1. From the third equation, solve for (beta): [ -2alpha + beta = -13 implies beta = 2alpha - 13 ] 2. Substitute (beta) into the first equation: [ 3(2alpha - 13) + 4gamma = -5 6alpha - 39 + 4gamma = -5 6alpha + 4gamma = 34 3alpha + 2gamma = 17 ] 3. Substitute (beta) into the second equation: [ alpha - (2alpha - 13) + gamma = 9 alpha - 2alpha + 13 + gamma = 9 -alpha + 13 + gamma = 9 -alpha + gamma = -4 gamma = alpha - 4 ] 4. Plug in (gamma = alpha - 4) into (3alpha + 2gamma = 17): [ 3alpha + 2(alpha - 4) = 17 3alpha + 2alpha - 8 = 17 5alpha = 25 alpha = 5 ] 5. Substitute (alpha = 5) back to find (gamma): [ gamma = 5 - 4 = 1 ] 6. Substitute (alpha = 5) back to find (beta): [ beta = 2(5) - 13 beta = 10 - 13 = -3 ] The solution to the system is: [ alpha = 5, quad beta = -3, quad gamma = 1 ] Thus, the vector (x) can be written as: [ x = 5p - 3q + r ] Finally: [ boxed{x = 5p - 3q + r} ]