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Calculate the air of the surface limited by the parabola which has the equation y=x(x) and the line passing through the points of abscissa -2 and 1 of the parabola.

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Answer to a math question Calculate the air of the surface limited by the parabola which has the equation y=x(x) and the line passing through the points of abscissa -2 and 1 of the parabola.

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Brice
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108 Answers
Pour trouver l'aire de la surface limitée par la parabole y=x^2 et la droite passant par les points d'abscisse -2 et 1 de la parabole, nous devons d'abord trouver les points d'intersection de la parabole et de la droite.

La droite passant par les points d'abscisse -2 et 1 de la parabole est définie par deux points (-2, (-2)^2) et (1, 1^2) . Donc, la droite a pour équation y = \frac{3}{3}x+4 .

Nous devons maintenant trouver les points d'intersection entre la parabole y=x^2 et la droite y = \frac{1}{3}x+4 :
x^2 = \frac{1}{3}x+4
x^2 - \frac{1}{3}x - 4 = 0

En résolvant cette équation quadratique, nous trouvons deux solutions pour x, à savoir x=-3 et x=4.

Pour calculer l'aire de la surface limitée par la parabole et la droite, nous devons trouver les limites d'intégration. Pour ce faire, nous trouvons les ordonnées des points d'intersection : (-3, (-3)^2) et (4, 4^2) .

L'aire recherchée est donc donnée par :
A = \int_{-3}^{4} (x^2 - \frac{1}{3}x - 4) \, dx
A = \left[\frac{x^3}{3} - \frac{x^2}{6} - 4x\right]_{-3}^{4}
A = (\frac{64}{3} - \frac{16}{6} - 16) - (-\frac{27}{3} + \frac{9}{6} + 12)
A = \frac{64}{3} - \frac{8}{3} - 16 + \frac{27}{3} - \frac{3}{2} - 12
A = \frac{64-8-48+27-6-72}{6} = \frac{27}{6} = \frac{9}{2}

\boxed{A = \frac{9}{2}}

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^x+1 +1 describe transformation","","Solution:\u003Cbr />\n1. Given function:\u003Cbr />\n * \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>y = -24^{x+1} + 1\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Base function:\u003Cbr />\n * \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>y = 4^x\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Identify transformations step-by-step:\u003Cbr />\n - **Translation horizontally**: The function has \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x+1\u003C/math-field>\u003C/math-field> as the exponent instead of \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x\u003C/math-field>\u003C/math-field>. This indicates a horizontal shift to the left by 1 unit.\u003Cbr />\n - **Vertical stretch and reflection**: The coefficient before \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>4\u003C/math-field>\u003C/math-field> is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>-2\u003C/math-field>\u003C/math-field>.\u003Cbr />\n - **Vertical stretch**: The factor \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2\u003C/math-field>\u003C/math-field> indicates that the function is stretched vertically by a factor of \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2\u003C/math-field>\u003C/math-field>.\u003Cbr />\n - **Reflection**: The negative sign indicates a reflection across the x-axis.\u003Cbr />\n - **Vertical translation**: The \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>+1\u003C/math-field>\u003C/math-field> outside the function indicates a vertical shift upwards by 1 unit.\u003Cbr />\n\u003Cbr />\n4. Describe the complete transformation:\u003Cbr />\n - The function \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>y = 4^x\u003C/math-field>\u003C/math-field> undergoes the following transformations: a horizontal shift to the left by 1 unit, a vertical stretch by a factor of 2, reflection across the x-axis, and finally a vertical shift upwards by 1 unit.",1255,251,"y-2-4-x-1-1-describe-transformation",{"id":44,"category":36,"text_question":45,"photo_question":38,"text_answer":46,"step_text_answer":8,"step_photo_answer":8,"views":47,"likes":48,"slug":49},538086,"Add the polynomials gx=x3-2x2+3x-1+4x2-x+2","Solution: \u003Cbr />\n1. Write down the given polynomials:\u003Cbr />\n- First polynomial: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>gx = x^3 - 2x^2 + 3x - 1\u003C/math-field>\u003C/math-field>\u003Cbr />\n- Second polynomial: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>4x^2 - x + 2\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Align and add the polynomials term by term:\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>gx = x^3 - 2x^2 + 3x - 1\u003C/math-field>\u003C/math-field>\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>4x^2 - x + 2\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Add the corresponding like terms:\u003Cbr />\n- For \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x^3\u003C/math-field>\u003C/math-field> terms: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x^3\u003C/math-field>\u003C/math-field>\u003Cbr />\n- For \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x^2\u003C/math-field>\u003C/math-field> terms: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>-2x^2 + 4x^2 = 2x^2\u003C/math-field>\u003C/math-field>\u003Cbr />\n- For \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x\u003C/math-field>\u003C/math-field> terms: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3x - x = 2x\u003C/math-field>\u003C/math-field>\u003Cbr />\n- For constant terms: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>-1 + 2 = 1\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. The resulting polynomial after addition is:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x^3 + 2x^2 + 2x + 1\u003C/math-field>\u003C/math-field>",739,148,"add-the-polynomials-g-x-x3-2x2-3x-1-4x2-x-2",{"id":51,"category":36,"text_question":52,"photo_question":38,"text_answer":53,"step_text_answer":8,"step_photo_answer":8,"views":54,"likes":55,"slug":56},538085,"R=3m. Calculate the volume of the sphere. Round to the nearest tenth if necessary","1. The formula for the volume of a sphere is \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. Substitute the given radius \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> R = 3 \\, \\text{m} \u003C/math-field>\u003C/math-field> into the formula:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> V = \\frac{4}{3} \\pi 3^3 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>3. Calculate \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 3^3 = 27 \u003C/math-field>\u003C/math-field> .\u003Cbr>\u003Cbr>4. Thus, the volume becomes:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> V = \\frac{4}{3} \\pi \\times 27 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>5. Simplify the expression:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> V = \\frac{4 \\times 27}{3} \\pi = 36 \\pi \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>6. Use the approximation \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\pi \\approx 3.1416 \u003C/math-field>\u003C/math-field> :\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> V \\approx 36 \\times 3.1416 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>7. Calculate the approximate volume:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V\\approx113.0973\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>8. Round to the nearest tenth:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> V \\approx 113.1 \\, \\text{m}^3 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>Therefore, the volume of the sphere is approximately \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 113.1 \\, \\text{m}^3 \u003C/math-field>\u003C/math-field> .",1203,241,"r-3m-calculate-the-volume-of-the-sphere-round-to-the-nearest-tenth-if-necessary",{"id":58,"category":36,"text_question":59,"photo_question":38,"text_answer":60,"step_text_answer":8,"step_photo_answer":8,"views":61,"likes":62,"slug":63},538084,"Width of 12 in. Calculate the volume of the sphere. Round to the nearest tenth if necessary","1. Identify the radius of the sphere. Given the width is 12 inches, the diameter is 12 inches. Therefore, the radius is half of the diameter:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> r = \\frac{12}{2} = 6 \\, \\text{in} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Use the formula for the volume of a sphere:\u003Cbr />\n \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 />\n\u003Cbr />\n3. Substitute the radius into the formula:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> V = \\frac{4}{3} \\pi 6^3 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Calculate:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> V = \\frac{4}{3} \\pi \\times 216 = \\frac{864}{3} \\pi = 288 \\pi \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Approximate using \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\pi \\approx 3.1416 \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 288 \\times 3.1416 = 904.8 \\, \\text{in}^3 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n6. The volume of the sphere, rounded to the nearest tenth, is approximately:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 904.8 \\, \\text{in}^3 \u003C/math-field>\u003C/math-field>",278,56,"width-of-12-in-calculate-the-volume-of-the-sphere-round-to-the-nearest-tenth-if-necessary",{"id":65,"category":36,"text_question":66,"photo_question":38,"text_answer":67,"step_text_answer":8,"step_photo_answer":8,"views":68,"likes":69,"slug":70},538083,"Calculate the volume tothenearesttenthofacubiccentimeter 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":72,"category":36,"text_question":73,"photo_question":38,"text_answer":74,"step_text_answer":8,"step_photo_answer":8,"views":75,"likes":76,"slug":77},538082,"Find the length of each base edge tothenearesttenthofameter of the 24m tall glass square pyramids of the Muttart Conservatory in Alberta, Canada, if each contains 5280m^3 of space","1. Volume V of a square pyramid is given by the formula:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V = \\frac{1}{3} B h\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>where B is the area of the base and h is the height of the pyramid.\u003Cbr>\u003Cbr>2. Given that the height h = 24 m and the volume V = 5280 m^3.\u003Cbr>\u003Cbr>3. The base is square, so if the side length of the base is s, then:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>B = s^2\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>4. Substituting into the volume formula:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5280 = \\frac{1}{3} s^2 \\times 24\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>5. Simplify and solve for s^2:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5280 = 8 s^2\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>s^2 = \\frac{5280}{8} = 660\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>6. Solve for s:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>s = \\sqrt{660} \\approx 25.7\u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>7. To find the length of each base edge to the nearest tenth of a meter, compute:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>s \\approx 25.7 \\, \\text{m}\u003C/math-field>\u003C/math-field>",418,84,"find-the-length-of-each-base-edge-to-the-nearest-tenth-of-a-meter-of-the-24m-tall-glass-square-pyramids-of-the-muttart-conservatory-in-alberta-canada-if-each-contains-5280m-3-of-space",{"id":79,"category":36,"text_question":80,"photo_question":38,"text_answer":81,"step_text_answer":8,"step_photo_answer":8,"views":82,"likes":83,"slug":84},538081,"An observer is 150 meters away\n distance of a hot air balloon online\n straight line at ground level. From your position,\n measures an elevation angle of 40° up to\n the base of the balloon. At what height is\n find the hot air balloon?","Solution:\u003Cbr />\n1. Dado:\u003Cbr />\n- Distancia horizontal desde el observador hasta la base del globo: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>d = 150 \\ m\u003C/math-field>\u003C/math-field>\u003Cbr />\n- Ángulo de elevación: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\theta = 40^{\\circ}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Usamos la función tangente para encontrar la altura \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>h\u003C/math-field>\u003C/math-field> del globo aerostático. La tangente de un ángulo en un triángulo rectángulo es la razón entre la altura y la distancia horizontal:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\tantheta = \\frac{h}{d}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Sustituimos los valores conocidos en la ecuación:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\tan40circ = \\frac{h}{150}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Resolvemos para \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>h\u003C/math-field>\u003C/math-field>:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>h = 150 \\times \\tan40circ\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Calculamos el valor numérico:\u003Cbr />\n* Usando una calculadora, \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\tan40circ \\approx 0.8391\u003C/math-field>\u003C/math-field>\u003Cbr />\n* Entonces: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>h \\approx 150 \\times 0.8391 = 125.865 \\ m\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nLa altura del globo aerostático es aproximadamente \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>125.865 \\ m\u003C/math-field>\u003C/math-field>.",667,133,"an-observer-is-150-meters-away-distance-of-a-hot-air-balloon-online-straight-line-at-ground-level-from-your-position-measures-an-elevation-angle-of-40-up-to-the-base-of-the-balloon-at-what-hei",{"id":86,"category":36,"text_question":87,"photo_question":38,"text_answer":88,"step_text_answer":8,"step_photo_answer":8,"views":89,"likes":90,"slug":91},538080,"A plane ticket has gone up 18%, now costing 4,720.Howmuchdiditcostbeforetheincrease?","\u003Cmathfieldreadonlydefaultmode=\"inlinemath\"class=\"mathexpression\">\u003Cmathfieldreadonly>textSolution:\u003C/mathfield>\u003C/mathfield>\u003Cbr/>\n1.Definevariables:\u003Cbr/>\nLet\u003Cmathfieldreadonlydefaultmode=\"inlinemath\"class=\"mathexpression\">\u003Cmathfieldreadonly>P\u003C/mathfield>\u003C/mathfield>betheoriginalpriceoftheplaneticket.\u003Cbr/>\n\u003Cmathfieldreadonlydefaultmode=\"inlinemath\"class=\"mathexpression\">\u003Cmathfieldreadonly>P\u003C/mathfield>\u003C/mathfield>increasedby18\\frac{x}{15}=\\frac{4}{3}\u003C/mathfield>\n\u003Cbr>\n\u003C/div>\n\n\u003Cdiv>\n\n\u003Cmathfieldstyle=\"fontsize:16px;padding:8px;borderradius:8px;border:1pxsolidrgba(0,0,0,.3);boxshadow:000rgba(0,0,0,.2)\n\"readonly>x=20\u003C/math-field>\n    \u003Cbr>\n  \u003C/div>",467,93,"15-75-of",{"id":149,"category":36,"text_question":150,"photo_question":38,"text_answer":151,"step_text_answer":8,"step_photo_answer":8,"views":152,"likes":90,"slug":153},538067,"Naria Wants to build a perimeter wall fence around 400 sqm lot. The frontage of the lot is 20 meters. The wall height is 1.2 meters below the ground and 3.8 meters above the ground .\nThe cost of constructing the wall is 750 per square meter. Additionally,she plans to install a 5-meter-wide gate that costs ₱50,000.\nHow much Maria spend in total for the wall and the gate\na. Php 281,250.00\nb. Php 106,250.00\nc. Php 331,250.00\nd. Php 218,250.00","1. Determine the dimensions of the lot. Given the frontage is 20 meters, calculate the other side using the area:\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Area} = 20 \\times x = 400 \u003C/math-field>\u003C/math-field>\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> x = \\frac{400}{20} = 20 \\, \\text{meters} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Calculate the perimeter of the lot:\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Perimeter} = 2 \\times (20 + 20) = 80 \\, \\text{meters} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Calculate the total height of the wall:\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Total height} = 1.2 + 3.8 = 5 \\, \\text{meters} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Calculate the wall area excluding the gate:\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Wall area} = (\\text{Perimeter} - \\text{Gate width}) \\times \\text{Total height} \u003C/math-field>\u003C/math-field>\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Wall area} = (80 - 5) \\times 5 = 75 \\times 5 = 375 \\, \\text{sqm} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Calculate the cost of the wall:\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Wall cost} = 375 \\times 750 = 281,250 \\, \\text{PHP} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n6. Calculate total cost including the gate:\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Total cost} = \\text{Wall cost} + \\text{Gate cost} \u003C/math-field>\u003C/math-field>\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Total cost} = 281,250 + 50,000 = 331,250 \\, \\text{PHP} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n7. Therefore, the total cost is: \u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Php} \\, 331,250.00 \u003C/math-field>\u003C/math-field>",727,"naria-wants-to-build-a-perimeter-wall-fence-around-400-sqm-lot-the-frontage-of-the-lot-is-20-meters-the-wall-height-is-1-2-meters-below-the-ground-and-3-8-meters-above-the-ground-the-cost-of-const",{"id":155,"category":36,"text_question":156,"photo_question":38,"text_answer":157,"step_text_answer":8,"step_photo_answer":8,"views":158,"likes":76,"slug":159},538066,"2. A rectangular lot has a\n30 meters.\nallocated along the frontage. What is the gross area\nfrontage of 18 meters and a depth of\nHowever, a public right of way of 3 meters is\nReve\nof\nthe\nlot?\na. 520 sqm\nb. 540 sqm\nC. 560 sqm\nd. 580 sqm\nReV","1. Calculate the area of the rectangular lot without the right of way:\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\text{Area}_{\\text{total}} = \\text{frontage} \\times \\text{depth} = 18 \\, \\text{m} \\times 30 \\, \\text{m} = 540 \\, \\text{sqm} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. No need to subtract any area because the problem states that the right of way has already been accounted for in the dimensions provided.\u003Cbr />\n\u003Cbr />\n3. Therefore, the gross area of the lot remains:\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 540 \\, \\text{sqm} \u003C/math-field>\u003C/math-field> \u003Cbr />\n   \u003Cbr />\nThus, the answer is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>540 \\, \\text{sqm}\u003C/math-field>\u003C/math-field>.",420,"2-a-rectangular-lot-has-a-30-meters-allocated-along-the-frontage-what-is-the-gross-area-frontage-of-18-meters-and-a-depth-of-however-a-public-right-of-way-of-3-meters-is-reve-of-the-lot-a-520-sq",{"id":161,"category":36,"text_question":162,"photo_question":38,"text_answer":163,"step_text_answer":8,"step_photo_answer":8,"views":164,"likes":165,"slug":166},538065,"A triangular lot has a base of 15 meters and heights of 10 meters . What is the total area of the lot ?","1. The formula for the area of a triangle is given by:\u003Cbr />\n\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = \\frac{1}{2} \\times \\text{base} \\times \\text{height} \u003C/math-field>\u003C/math-field> \u003Cbr />\n\u003Cbr />\n2. Substitute the given values into the formula:\u003Cbr />\n\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = \\frac{1}{2} \\times 15 \\times 10 \u003C/math-field>\u003C/math-field> \u003Cbr />\n\u003Cbr />\n3. Calculate the area:\u003Cbr />\n\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = \\frac{1}{2} \\times 150 \u003C/math-field>\u003C/math-field> \u003Cbr />\n\u003Cbr />\n4. Simplify the expression:\u003Cbr />\n\u003Cbr />\n   \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = 75 \u003C/math-field>\u003C/math-field> \u003Cbr />\n\u003Cbr />\nTherefore, the total area of the lot is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>75 \\, \\text{square meters}\u003C/math-field>\u003C/math-field>.",425,85,"a-triangular-lot-has-a-base-of-15-meters-and-heights-of-10-meters-what-is-the-total-area-of-the-lot",{"id":168,"category":36,"text_question":169,"photo_question":38,"text_answer":170,"step_text_answer":8,"step_photo_answer":8,"views":171,"likes":172,"slug":173},538064,"4x+(x-3)=2x-(3x-4)+5","Solution:\u003Cbr />\n1. Simplify both sides of the equation.\u003Cbr />\n- Left side: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>4x + (x - 3) = 4x + x - 3 = 5x - 3\u003C/math-field>\u003C/math-field>\u003Cbr />\n- Right side: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2x - (3x - 4) + 5 = 2x - 3x + 4 + 5 = -x + 9\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. The equation is now: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5x - 3 = -x + 9\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Add \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x\u003C/math-field>\u003C/math-field> to both sides to eliminate the \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>-x\u003C/math-field>\u003C/math-field> from the right side and simplify:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5x + x - 3 = 9\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Simplify: \u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>6x - 3 = 9\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Add 3 to both sides to isolate terms with \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x\u003C/math-field>\u003C/math-field>:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>6x = 9 + 3\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n6. Simplify:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>6x = 12\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n7. Divide both sides by 6 to solve for \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x\u003C/math-field>\u003C/math-field>:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x = \\frac{12}{6}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n8. Simplify the fraction:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x = 2\u003C/math-field>\u003C/math-field>",1332,266,"4x-x-3-2x-3x-4-5",{"first":6,"last":175,"prev":8,"next":10},187,{"current_page":6,"from":6,"last_page":175,"links":177,"path":211,"per_page":212,"to":212,"total":213},[178,181,184,186,188,190,192,195,198,201,204,207,209],{"url":6,"label":179,"active":180},"1",true,{"url":10,"label":182,"active":183},"2",false,{"url":13,"label":185,"active":183},"3",{"url":16,"label":187,"active":183},"4",{"url":19,"label":189,"active":183},"5",{"url":22,"label":191,"active":183},"6",{"url":193,"label":194,"active":183},7,"7",{"url":196,"label":197,"active":183},8,"8",{"url":199,"label":200,"active":183},9,"9",{"url":202,"label":203,"active":183},10,"10",{"url":205,"label":206,"active":183},186,"186",{"url":175,"label":208,"active":183},"187",{"url":10,"label":210,"active":183},"Next »","https://api.math-master.org/api/question",20,3737,{"data":215},{"questions":216},[217,221,225,229,233,237,241,245,249,253,257,261,265,269,273,277,281,285,289,293],{"id":218,"category":36,"text_question":219,"slug":220},532021,"Solution of the equation y'' - y' -6y = 0","solution-of-the-equation-y-39-39-y-39-6y-0",{"id":222,"category":36,"text_question":223,"slug":224},532048,"CASE 6-1: PREPARE A PRODUCTION PLAN: WHAT PROBLEMS ARRIVE? Midwest Plastics Company has conducted profit planning for several years. The president stated (with justification) that inventory control and planning had not been satisfactory, which was mainly due to poor planning of production and inventory budgets. Please analyze and provide recommendations, in detail, on the issue regarding the 20B profit plan, which is now being prepared. Their analysis and recommendations will be presented to the executive committee. Despite the seasonality factor, the sales department has been successful in developing a sales plan, on a monthly basis, for each year. The following sales data is available for 20B. 1. Sales plan summary for 20B: 2. Finished goods inventory, as of January 1, 20B, is 96,000 units. 3. Work-in-process inventory will remain constant. 4. Actual annual sales in 20A, including the estimate for December, were 350,000 units. 5. The average finished goods inventory during 20A was 70,000 units. IT IS REQUESTED. 1. Prepare the annual production budget, assuming that management policy is to budget ending finished goods inventory at a standard quantity, based on the ratio of historical sales of 20A to inventory turnover. 2. Prepare a schedule showing sales, production, and inventory levels for each month, assuming: 1) stable inventory, 2) stable production, and 3) recommended inventory-production levels. In developing your recommendations, assume that the following policies have been established: a) The president has set the policy that a maximum inventory of 85,000 units and a minimum inventory of 75,000 units should be used, except in abnormal circumstances. b) A stable level of production is definitely preferred, except that during the holiday season in July and August, production may be reduced by 25 percent. Likewise, a variation in production of 7.5 percent above and below the average level is acceptable. 3. What are the main problems faced by the company in production planning? Make your general recommendations.","case-6-1-prepare-a-production-plan-what-problems-arrive-midwest-plastics-company-has-conducted-profit-planning-for-several-years-the-president-stated-with-justification-that-inventory-control-an",{"id":226,"category":36,"text_question":227,"slug":228},532054,"A software company incurs a cost of50 per license sold plus 5,000infixedcosts.Howmanylicensesshouldyouselltominimizetotalcosts?","asoftwarecompanyincursacostof50perlicensesoldplus5000infixedcostshowmanylicensesshouldyouselltominimizetotalcosts","id":230,"category":36,"textquestion":231,"slug":232,532055,"Acollegebelievesthat2248, the monthly fixed costs amount to 125,000andthevariablecostformakingaguavarollis28.\n Determine:\n a) The equation of the total income from the production of guava rolls.","a-regional-candy-factory-sells-a-guava-roll-at-a-price-of-48-the-monthly-fixed-costs-amount-to-125-000-and-the-variable-cost-for-making-a-guava-roll-is-28-determine-a-the-equation-of-the-tota",{"id":345,"category":36,"text_question":346,"slug":347},534145,"The average number of babies born at a hospital is 6 per hour. What is the probability that three babies are born during a particular 1 hour period?","the-average-number-of-babies-born-at-a-hospital-is-6-per-hour-what-is-the-probability-that-three-babies-are-born-during-a-particular-1-hour-period",{"id":349,"category":36,"text_question":350,"slug":351},534253,"If X1 and X2 are independent standard normal variables, find PX12+X22>2.41","if-x1-and-x2-are-independent-standard-normal-variables-find-p-x1-2-x2-2-2-41",{"id":353,"category":36,"text_question":354,"slug":355},534270,"Convert 5/9 to a decimal","convert-5-9-to-a-decimal",{"id":357,"category":36,"text_question":358,"slug":359},534345,"On+January+10+2023+the+CONSTRUCTORA+DEL+ORIENTE+SAC+company+acquires+land+to+develop+a+real estate+project%2C+which+prev%C3% A9+enable+50+lots+for+commercial+use+valued+in+S%2F+50%2C000.00+each+one%2C+the+company+has+as+a+business+model+generate+ cash+flow+through%C3%A9s+of+the+rental%2C+so+47%2C+of+the+50+enabled+lots+are+planned to lease+47%2C+and+ the+rest+will be%C3%A1n+used+by+the+company+for+management%C3%B3n+and+land+control","on-january-10-2023-the-constructora-del-oriente-sac-company-acquires-land-to-develop-a-real-estate-project-2c-which-prev-c3-a9-enable-50-lots-for-commercial-use-valued-in-s-2f-50-2c000-00-each-one-2c",{"id":361,"category":36,"text_question":362,"slug":363},534384,"A,B,C and D are the corners of a rectangular building. Find the lengths the diagonals if AB measures 38' - 9\" and AD measures 56' - 3\"","a-b-c-and-d-are-the-corners-of-a-rectangular-building-find-the-lengths-the-diagonals-if-ab-measures-38-9-and-ad-measures-56-3",{"id":365,"category":36,"text_question":366,"slug":367},534473,"A 20,000 kg school bus is moving at 30 km per hour on a straight road. At that moment, it applies the brakes until it comes to a complete stop after 15 seconds. 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