of a golf ball whose diameter is 4.267cm","","1. The formula for the volume of a sphere is given by \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V = \\frac{4}{3} \\pi r^3\u003C/math-field>\u003C/math-field> .\u003Cbr>\u003Cbr>2. The diameter of the golf ball is given as 4.267 cm, so the radius is half of that: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>r = \\frac{4.267}{2} = 2.1335 \\, \\text{cm}\u003C/math-field>\u003C/math-field> .\u003Cbr>\u003Cbr>3. Substitute the radius into the volume formula: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V = \\frac{4}{3} \\pi $2.1335$^3\u003C/math-field>\u003C/math-field> .\u003Cbr>\u003Cbr>4. Calculate the cube of the radius: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>$2.1335$^3 = 9.707432537375\u003C/math-field>\u003C/math-field> .\u003Cbr>\u003Cbr>5. Substitute this back into the formula: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V=\\frac{4}{3}\\pi\\times9.707432537375\\approx40.7\\,\\text{cm}^3\u003C/math-field>\u003C/math-field> .\u003Cbr>\u003Cbr>6. The volume of the golf ball is approximately \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>40.7\\,\\text{cm}^3\u003C/math-field>\u003C/math-field> .",1440,288,"calculate-the-volume-to-the-nearest-tenth-of-a-cubic-centimeter-of-a-golf-ball-whose-diameter-is-4-267cm",{"id":55,"category":47,"text_question":56,"photo_question":49,"text_answer":57,"step_text_answer":8,"step_photo_answer":8,"views":58,"likes":59,"slug":60},537970,"Express 69% as a decimal","Solution:\u003Cbr />\n1. Start with the percentage value: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>69\\%\u003C/math-field>\u003C/math-field>.\u003Cbr />\n2. Convert the percentage to a decimal by dividing by 100:\u003Cbr />\n* \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>69\\% = \\frac{69}{100}\u003C/math-field>\u003C/math-field>.\u003Cbr />\n3. Perform the division:\u003Cbr />\n* \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{69}{100} = 0.69\u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\nTherefore, the decimal representation of 69% is 0.69.",813,163,"express-69-as-a-decimal",{"id":62,"category":47,"text_question":63,"photo_question":49,"text_answer":64,"step_text_answer":8,"step_photo_answer":8,"views":65,"likes":66,"slug":67},537956,"The volume of a cylindrical tank that is 23 feet high with a radius of 6.25 feet is _____. $Round your ANSWER: to the nearest cubic foot$","Solution:\u003Cbr />\n1. Given:\u003Cbr />\n * Height, \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>h = 23\u003C/math-field>\u003C/math-field> feet\u003Cbr />\n * Radius, \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>r = 6.25\u003C/math-field>\u003C/math-field> feet\u003Cbr />\n\u003Cbr />\n2. Formula for the volume of a cylinder:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V = \\pi r^2 h\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Substitute the given values into the formula:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V = \\pi \\times $6.25$^2 \\times 23\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Calculate \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>$6.25$^2\u003C/math-field>\u003C/math-field>:\u003Cbr />\n * \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>$6.25$^2 = 39.0625\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Substitute back into the volume formula:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V = \\pi \\times 39.0625 \\times 23\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n6. Calculate \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>39.0625 \\times 23\u003C/math-field>\u003C/math-field>:\u003Cbr />\n * \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>39.0625 \\times 23 = 898.4375\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n7. Multiply by \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\pi\u003C/math-field>\u003C/math-field> $approximately \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3.14159\u003C/math-field>\u003C/math-field>:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V = 3.14159 \\times 898.4375\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n8. Calculate the volume \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V\u003C/math-field>\u003C/math-field>:\u003Cbr />\n * \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V \\approx 2823.6363\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n9. Round the answer to the nearest cubic foot:\u003Cbr />\n * The volume of the cylindrical tank is approximately \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2824\u003C/math-field>\u003C/math-field> cubic feet.",732,146,"the-volume-of-a-cylindrical-tank-that-is-23-feet-high-with-a-radius-of-6-25-feet-is-round-your-answer-to-the-nearest-cubic-foot",{"id":69,"category":47,"text_question":70,"photo_question":49,"text_answer":71,"step_text_answer":8,"step_photo_answer":8,"views":72,"likes":73,"slug":74},537951,"150 is what percent of 625\nRound your answer","Solution:\u003Cbr />\n1. Use the formula for percentage:\u003Cbr />\n- Percentage = \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\left( \\frac{\\text{part}}{\\text{whole}} \\right$ \\times 100\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Substitute the given values:\u003Cbr />\n- Part = 150\u003Cbr />\n- Whole = 625\u003Cbr />\n- Percentage = \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\left$\\frac{150}{625} \\right$ \\times 100\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Calculate the fraction:\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{150}{625} = 0.24\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Convert to a percentage:\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>0.24 \\times 100 = 24 \\%\u003C/math-field>\u003C/math-field>",1295,259,"150-is-what-percent-of-625-round-your-answer",{"id":76,"category":47,"text_question":77,"photo_question":49,"text_answer":78,"step_text_answer":8,"step_photo_answer":8,"views":79,"likes":80,"slug":81},537906,"561-$t+154$=137","1. Start with the equation:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>561-$t+154$=137\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Distribute the negative sign:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>561-t-154=137\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Simplify the left side by combining like terms:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>407-t=137\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Subtract 407 from both sides:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>-t=137-407\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Simplify the right side:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>-t=-270\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n6. Multiply both sides by -1:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>t=270\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nAnswer: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>t = 270\u003C/math-field>\u003C/math-field>",1257,251,"561-t-154-137",{"id":83,"category":47,"text_question":84,"photo_question":49,"text_answer":85,"step_text_answer":8,"step_photo_answer":8,"views":86,"likes":87,"slug":88},537849,"Factorize \n2x2+11x+12","1. Identify the quadratic expression: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2x^2 + 11x + 12\u003C/math-field>\u003C/math-field>.\u003Cbr />\n2. To factor, seek two numbers whose product is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2 \\cdot 12 = 24\u003C/math-field>\u003C/math-field> and sum is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>11\u003C/math-field>\u003C/math-field>.\u003Cbr />\n3. The numbers are \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3\u003C/math-field>\u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>8\u003C/math-field>\u003C/math-field>.\u003Cbr />\n4. Rewrite the middle term using these numbers: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2x^2 + 3x + 8x + 12\u003C/math-field>\u003C/math-field>.\u003Cbr />\n5. Group terms: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>$2x^2 + 3x$ + $8x + 12$\u003C/math-field>\u003C/math-field>.\u003Cbr />\n6. Factor out the greatest common factor in each group: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x$2x + 3$ + 4$2x + 3$\u003C/math-field>\u003C/math-field>.\u003Cbr />\n7. Factor by grouping: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>$x + 4$$2x + 3$\u003C/math-field>\u003C/math-field>.\u003Cbr />\n8. The factorized form is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>$x + 4$$2x + 3$\u003C/math-field>\u003C/math-field>.",1386,277,"factorize-2x2-11x-12",{"id":90,"category":47,"text_question":91,"photo_question":49,"text_answer":92,"step_text_answer":8,"step_photo_answer":8,"views":93,"likes":94,"slug":95},537831,"Someone cuts a rope 13 into thirteen pieces How many times did they cut it","1. Start with 1 uncut rope.\u003Cbr />\n2. To create 13 pieces, you need to make cuts: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x + 1 = 13\u003C/math-field>\u003C/math-field>, where \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x\u003C/math-field>\u003C/math-field> is the number of cuts.\u003Cbr />\n3. Solve for \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x: x = 13 - 1\u003C/math-field>\u003C/math-field>.\u003Cbr />\n4. Therefore, \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x = 12\u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\nThe answer is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>12\u003C/math-field>\u003C/math-field> cuts.",406,81,"someone-cuts-a-rope-13-into-thirteen-pieces-how-many-times-did-they-cut-it",{"id":97,"category":47,"text_question":98,"photo_question":49,"text_answer":99,"step_text_answer":8,"step_photo_answer":8,"views":100,"likes":101,"slug":102},537788,"How to slove 2.4 : 1.5 step by step","1. Write the ratio as a fraction: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{2.4}{1.5}\u003C/math-field>\u003C/math-field> .\u003Cbr>\u003Cbr>2. To simplify, divide both the numerator and the denominator of the fraction by the same number. Here, we can divide both by 0.3:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\[\\frac{2.4 \\div 0.3}{1.5 \\div 0.3} = \\frac{8}{5}\\]\u003C/math-field>\u003C/math-field>\u003Cbr>\u003Cbr>3. The simplified form of the ratio 2.4 : 1.5 is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{8}{5}\u003C/math-field>\u003C/math-field> .\u003Cbr>\u003Cbr>Therefore, the solution is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{8}{5}\u003C/math-field>\u003C/math-field> .",498,100,"how-to-slove-2-4-1-5-step-by-step",{"id":104,"category":47,"text_question":105,"photo_question":49,"text_answer":106,"step_text_answer":8,"step_photo_answer":8,"views":107,"likes":108,"slug":109},537690,"2x+3y=12\n4x-3y=5\n\nThis is for y=mx+b","1. Add the two equations to eliminate \\( y \\):\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 2x + 3y = 12 \u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 4x - 3y = 5 \u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> (2x + 3y) + (4x - 3y) = 12 + 5 \u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 6x = 17 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Solve for \\( x \\):\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> x = \\frac{17}{6} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Substitute \\( x = \\frac{17}{6} \\) into the first equation to find \\( y \\):\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 2\\left(\\frac{17}{6}\\right) + 3y = 12 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{34}{6} + 3y = 12 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 3y = 12 - \\frac{34}{6} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 3y = \\frac{72}{6} - \\frac{34}{6} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 3y = \\frac{38}{6} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> y = \\frac{38}{18} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> y = \\frac{19}{9} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. The equation for \\( y = mx + b \\) therefore becomes:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> y = \\frac{17}{6}x - \\frac{19}{6} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Answer: \u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> y = \\frac{17}{6}x - \\frac{19}{6} \u003C/math-field>\u003C/math-field>",544,109,"2x-3y-12-4x-3y-5-this-is-for-y-mx-b",{"id":111,"category":47,"text_question":112,"photo_question":49,"text_answer":113,"step_text_answer":8,"step_photo_answer":8,"views":114,"likes":115,"slug":116},537669,"A 5 kW water heater uses a copper wire resistance that is 50 cm long and has a radius of 0.15 cm. To heat the water, the resistance is completely submerged. Determine the convection heat transfer coefficient when heating pure water to its boiling point using a resistance whose surface temperature is 150°C.","1. Identificar as variáveis relevantes para o problema:\u003Cbr>\u003Cbr>- Potência elétrica: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> P = 5 \\, \\text{kW} = 5000 \\, \\text{W} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>- Comprimento do fio: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> L = 0.5 \\, \\text{m} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>- Raio do fio: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> r = 0.15 \\, \\text{cm} = 0.0015 \\, \\text{m} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>- Temperatura da superfície: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> T_s = 150 \\, \\text{°C} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>- Temperatura da água em ebulição: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> T_\\infty = 100 \\, \\text{°C} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>2. Determinar a área de superfície do fio de cobre:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A_s = 2 \\pi r L = 2 \\pi (0.0015) (0.5) \\approx 0.00471 \\, \\text{m}^2 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>3. A utilização da fórmula da transferência de calor por convecção:\u003Cbr>\u003Cbr>A relação básica para transferência de calor por convecção é dada por:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> Q = h A_s (T_s - T_\\infty) \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>4. Determinar a taxa de transferência de calor, que é também a potência fornecida:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> Q = P = 5000 \\, \\text{W} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>5. Substituir na equação de convecção para calcular o coeficiente de convecção:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 5000 = h \\times 0.00471 \\times (150 - 100) \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>6. Resolver para \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> h \u003C/math-field>\u003C/math-field> :\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>h=\\frac{5000}{0.00471 \\times50}\\approx21222\\,\\text{W/m}^2\\text{K}\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>Portanto, o coeficiente de transferência de calor por convecção é aproximadamente \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>21222\\,\\text{W/m}^2\\text{K}\u003C/math-field>\u003C/math-field> 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