Loading [MathJax]/jax/output/HTML-CSS/jax.js
^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},{"id":216,"category":36,"slug":217,"text_question":218,"photo_question":8,"text_answer":219,"step_text_answer":8,"step_photo_answer":8,"views":220,"likes":221,"expert":222},537082,"an-atom-a-has-8-electrons-in-its-right-shell-and-is-isoelectronically-with-the-trivalent-cation-of-an-atom-b-when-the-quantum-numbers-of-the-last-electron-of-the-electron-distribution-are-determin","An atom (A) has 8 electrons in its right shell and is isoelectronically with the trivalent cation of an atom (B) when the quantum numbers of the last electron of the electron distribution are determined, when the atom (B) becomes negatively ionized.","1. Determina la configuración electrónica del átomo (A) para encontrar que tiene 8 electrones en la última capa:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>1s^2 2s^2 2p^6\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Determina la configuración electrónica del mismo átomo (A) pero como ion trivalente para que sea iso-electrónico:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>A^{3-} \\Rightarrow 1s^2 2s^2 2p^6 3s^2 3p^6\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Identifica la configuración electrónica restante del átomo (B) original antes de ser ionizado negativamente, quitando los 3 electrones añadidos previamente:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>B \\Rightarrow 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Para el último electrón de este átomo (considerando ionización):\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3d^3 \\Rightarrow l=2,\\ m=1,\\ s=\\frac{1}{2}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Utiliza los números cuánticos del último electrón del ion negativamente cargado:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>n=3,\\ l=2,\\ m=1,\\ s=\\frac{1}{2}\u003C/math-field>\u003C/math-field> \u003Cbr />\n\u003Cbr />\nConclusión:\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{3}{2},2,1,\\frac{1}{2}\u003C/math-field>\u003C/math-field>",1322,264,{"id":19,"name":223,"photo":224,"biography":225,"created_at":8,"updated_at":8,"rating":226,"total_answer":227},"Rasheed","https://api.math-master.org/img/experts/5/5.webp","Born with an insatiable curiosity, I began my journey with mathematics during my school years. From solving basic arithmetic problems to unraveling complex algebraic equations, I developed a deep passion for the subject. Upon entering university, my learnings diving into advanced topics like abstract algebra, numerical analysis and differential equations.\nIn retrospect, my journey through the realm of mathematics has been an exhilarating adventure of discovery, learning, and growth.\n",4.7,105,{"data":229},{"questions":230},[231,235,239,243,247,251,255,259,263,267,271,275,279,283,287,291,295,299,303,307],{"id":232,"category":36,"text_question":233,"slug":234},531999,"How to find the value of x and y which satisfy both equations x-2y=24 and 8x-y=117","how-to-find-the-value-of-x-and-y-which-satisfy-both-equations-x-2y-24-and-8x-y-117",{"id":236,"category":36,"text_question":237,"slug":238},532052,"-6n+5=-13","6n-5-13",{"id":240,"category":36,"text_question":241,"slug":242},532072,"Since one of the three integers whose product is (-60) is (+4), write the values that two integers can take.","since-one-of-the-three-integers-whose-product-is-60-is-4-write-the-values-that-two-integers-can-take",{"id":244,"category":36,"text_question":245,"slug":246},532311,"2x-y=5 x-y=4","2x-y-5-x-y-4",{"id":248,"category":36,"text_question":249,"slug":250},532312,"Kayla has8,836.00 in her savings account. The bank gives Kayla 5%of the amount of money in account as a customer bonus. What amount of money does the bank give Kayla? Justify your answer on a 6th grade level.","kayla-has-8-836-00-in-her-savings-account-the-bank-gives-kayla-5-of-the-amount-of-money-in-account-as-a-customer-bonus-what-amount-of-money-does-the-bank-give-kayla-justify-your-answer-on-a-6th-gr",{"id":252,"category":36,"text_question":253,"slug":254},534057,"Identify a pattern in the list of numbers.Then use this pattern to find the next number. 37,31,25,19,13","identify-a-pattern-in-the-list-of-numbers-then-use-this-pattern-to-find-the-next-number-37-31-25-19-13",{"id":256,"category":36,"text_question":257,"slug":258},534097,"Pedro had 80% of the amount needed to buy a game. Of this amount, you spent 15% on a watch and therefore, you will need to add another R640.00topurchasethisgame.Isthevalueofthegame?","pedrohad80oftheamountneededtobuyagameofthisamountyouspent15onawatchandthereforeyouwillneedtoaddanotherr64000topurchasethisgameisthevalueofthegame","id":260,"category":36,"textquestion":261,"slug":262,534121,"Inagrocerystore,whenyoutakeout3peppersand4carrots,thereare26peppersand46carrotsleft.\nHowmanypeppersandcarrotswerethereinitially?","inagrocerystorewhenyoutakeout3peppersand4carrotsthereare26peppersand46carrotslefthowmanypeppersandcarrotswerethereinitially","id":264,"category":36,"textquestion":265,"slug":266,534129,"DeterminethegeneralequationofthestraightlinethatpassesthroughthepointP(2;3)andisparalleltothestraightlinewiththeequation5x2y1=0:","determinethegeneralequationofthestraightlinethatpassesthroughthepointp23andisparalleltothestraightlinewiththeequation5x2y10","id":268,"category":36,"textquestion":269,"slug":270,534199,"Apersondecidestoinvestmoneyinfixedincomesecuritiestoredeemitattheendof3years.Inthisway,youmakemonthlydepositsofR300.00 in the 1st year, R400.00inthe2ndyearandR500.00 in the 3rd year. Calculate the amount, knowing that compound interest is 0.6% per month for the entire period.\n The answer is 15,828.60","a-person-decides-to-invest-money-in-fixed-income-securities-to-redeem-it-at-the-end-of-3-years-in-this-way-you-make-monthly-deposits-of-r-300-00-in-the-1st-year-r-400-00-in-the-2nd-year-and-r-500-0",{"id":272,"category":36,"text_question":273,"slug":274},534294,"TEST 123123+1236ttttt","test-123123-1236ttttt",{"id":276,"category":36,"text_question":277,"slug":278},534347,"Buffalo Company makes and sells shampoo. Each unit requires 1.40laborcosts,materialcostsperunitare0.90 and other variable costs are 0.30.Itsellsshampoofor4.45 to retailers. Fixed costs are 15,000.Itsold25,000unitsinthecurrentmonth.\n\nWhatistheBreakEvenpointinunits?\nWhatistheBreakEvenpointindollars?\nWhatisthecontributionmarginofBuffaloCompany?","buffalocompanymakesandsellsshampooeachunitrequires140laborcostsmaterialcostsperunitare090andothervariablecostsare030itsellsshampoofor445toretailersfixedcost","id":280,"category":36,"textquestion":281,"slug":282,534359,"Expressthetrigonometricformofthecomplexz=1+i.","expressthetrigonometricformofthecomplexz1i","id":284,"category":36,"textquestion":285,"slug":286,534368,"Inmeasuringtheinternalradiusofacircularsewerthemeasurementis2104259 is taken out for 10 years with an annual interest rate of 9.4%, compounded quarterly. What quarterly payment is required to pay the loan off in 10 years? \n\nEnter to the nearest cent twodecimals. Do not use signsorcommasintheanswer.","17aloanfor104259istakenoutfor10yearswithanannualinterestrateof94compoundedquarterlywhatquarterlypaymentisrequiredtopaytheloanoffin10yearsentertothenearest","id":372,"category":36,"textquestion":373,"slug":374,534421,"Kayadeposits25,000intoanaccountthatearns3sicons":406},{"bxl:facebook-circle":407,"bxl:instagram":411,"mdi:web":413,"la:apple":415,"ph:google-logo-bold":418,"ph:google-logo":421},{"left":408,"top":408,"width":409,"height":409,"rotate":408,"vFlip":183,"hFlip":183,"body":410},0,24,"\u003Cpath fill=\"currentColor\" d=\"M12.001 2.002c-5.522 0-9.999 4.477-9.999 9.999c0 4.99 3.656 9.126 8.437 9.879v-6.988h-2.54v-2.891h2.54V9.798c0-2.508 1.493-3.891 3.776-3.891c1.094 0 2.24.195 2.24.195v2.459h-1.264c-1.24 0-1.628.772-1.628 1.563v1.875h2.771l-.443 2.891h-2.328v6.988C18.344 21.129 22 16.992 22 12.001c0-5.522-4.477-9.999-9.999-9.999\"/>",{"left":408,"top":408,"width":409,"height":409,"rotate":408,"vFlip":183,"hFlip":183,"body":412},"\u003Cpath fill=\"currentColor\" d=\"M11.999 7.377a4.623 4.623 0 1 0 0 9.248a4.623 4.623 0 0 0 0-9.248m0 7.627a3.004 3.004 0 1 1 0-6.008a3.004 3.004 0 0 1 0 6.008\"/>\u003Ccircle cx=\"16.806\" cy=\"7.207\" r=\"1.078\" fill=\"currentColor\"/>\u003Cpath fill=\"currentColor\" d=\"M20.533 6.111A4.6 4.6 0 0 0 17.9 3.479a6.6 6.6 0 0 0-2.186-.42c-.963-.042-1.268-.054-3.71-.054s-2.755 0-3.71.054a6.6 6.6 0 0 0-2.184.42a4.6 4.6 0 0 0-2.633 2.632a6.6 6.6 0 0 0-.419 2.186c-.043.962-.056 1.267-.056 3.71s0 2.753.056 3.71c.015.748.156 1.486.419 2.187a4.6 4.6 0 0 0 2.634 2.632a6.6 6.6 0 0 0 2.185.45c.963.042 1.268.055 3.71.055s2.755 0 3.71-.055a6.6 6.6 0 0 0 2.186-.419a4.6 4.6 0 0 0 2.633-2.633c.263-.7.404-1.438.419-2.186c.043-.962.056-1.267.056-3.71s0-2.753-.056-3.71a6.6 6.6 0 0 0-.421-2.217m-1.218 9.532a5 5 0 0 1-.311 1.688a3 3 0 0 1-1.712 1.711a5 5 0 0 1-1.67.311c-.95.044-1.218.055-3.654.055c-2.438 0-2.687 0-3.655-.055a5 5 0 0 1-1.669-.311a3 3 0 0 1-1.719-1.711a5.1 5.1 0 0 1-.311-1.669c-.043-.95-.053-1.218-.053-3.654s0-2.686.053-3.655a5 5 0 0 1 .311-1.687c.305-.789.93-1.41 1.719-1.712a5 5 0 0 1 1.669-.311c.951-.043 1.218-.055 3.655-.055s2.687 0 3.654.055a5 5 0 0 1 1.67.311a3 3 0 0 1 1.712 1.712a5.1 5.1 0 0 1 .311 1.669c.043.951.054 1.218.054 3.655s0 2.698-.043 3.654z\"/>",{"left":408,"top":408,"width":409,"height":409,"rotate":408,"vFlip":183,"hFlip":183,"body":414},"\u003Cpath fill=\"currentColor\" d=\"M16.36 14c.08-.66.14-1.32.14-2s-.06-1.34-.14-2h3.38c.16.64.26 1.31.26 2s-.1 1.36-.26 2m-5.15 5.56c.6-1.11 1.06-2.31 1.38-3.56h2.95a8.03 8.03 0 0 1-4.33 3.56M14.34 14H9.66c-.1-.66-.16-1.32-.16-2s.06-1.35.16-2h4.68c.09.65.16 1.32.16 2s-.07 1.34-.16 2M12 19.96c-.83-1.2-1.5-2.53-1.91-3.96h3.82c-.41 1.43-1.08 2.76-1.91 3.96M8 8H5.08A7.92 7.92 0 0 1 9.4 4.44C8.8 5.55 8.35 6.75 8 8m-2.92 8H8c.35 1.25.8 2.45 1.4 3.56A8 8 0 0 1 5.08 16m-.82-2C4.1 13.36 4 12.69 4 12s.1-1.36.26-2h3.38c-.08.66-.14 1.32-.14 2s.06 1.34.14 2M12 4.03c.83 1.2 1.5 2.54 1.91 3.97h-3.82c.41-1.43 1.08-2.77 1.91-3.97M18.92 8h-2.95a15.7 15.7 0 0 0-1.38-3.56c1.84.63 3.37 1.9 4.33 3.56M12 2C6.47 2 2 6.5 2 12a10 10 0 0 0 10 10a10 10 0 0 0 10-10A10 10 0 0 0 12 2\"/>",{"left":408,"top":408,"width":416,"height":416,"rotate":408,"vFlip":183,"hFlip":183,"body":417},32,"\u003Cpath fill=\"currentColor\" d=\"M20.844 2c-1.64 0-3.297.852-4.407 2.156v.032c-.789.98-1.644 2.527-1.375 4.312c-.128-.05-.136-.035-.28-.094c-.692-.281-1.548-.594-2.563-.594c-3.98 0-7 3.606-7 8.344c0 3.067 1.031 5.942 2.406 8.094c.688 1.078 1.469 1.965 2.281 2.625S11.57 28 12.531 28s1.68-.324 2.219-.563c.54-.238.957-.437 1.75-.437c.715 0 1.078.195 1.625.438c.547.242 1.293.562 2.281.562c1.07 0 1.98-.523 2.719-1.188s1.36-1.519 1.875-2.343c.516-.824.922-1.633 1.219-2.282c.148-.324.258-.593.343-.812s.13-.281.188-.531l.188-.813l-.75-.343a5.3 5.3 0 0 1-1.5-1.063c-.625-.637-1.157-1.508-1.157-2.844A4.08 4.08 0 0 1 24.563 13c.265-.309.542-.563.75-.719c.105-.078.187-.117.25-.156c.062-.04.05-.027.156-.094l.843-.531l-.562-.844c-1.633-2.511-4.246-2.844-5.281-2.844c-.48 0-.82.168-1.25.25c.242-.226.554-.367.75-.624c.004-.004-.004-.028 0-.032q.018-.016.031-.031h.031a6.16 6.16 0 0 0 1.563-4.438L21.78 2zm-1.188 2.313c-.172.66-.453 1.289-.906 1.78l-.063.063c-.382.516-.972.899-1.562 1.125c.164-.652.45-1.312.844-1.812c.008-.012.023-.02.031-.032c.438-.5 1.043-.875 1.656-1.125zm-7.437 5.5c.558 0 1.172.21 1.812.468s1.239.594 2.094.594c.852 0 1.496-.336 2.25-.594s1.559-.469 2.344-.469c.523 0 1.816.333 2.906 1.344c-.191.172-.36.297-.563.531a6.2 6.2 0 0 0-1.53 4.094c0 1.906.831 3.34 1.718 4.25c.55.563.89.696 1.313.938c-.055.125-.086.222-.157.375a19 19 0 0 1-1.093 2.062c-.454.727-1.004 1.434-1.532 1.907c-.527.472-1 .687-1.375.687c-.566 0-.898-.156-1.468-.406S17.581 25 16.5 25c-1.137 0-1.977.336-2.563.594c-.585.258-.89.406-1.406.406c-.246 0-.777-.2-1.375-.688c-.597-.488-1.254-1.23-1.844-2.156c-1.183-1.851-2.093-4.394-2.093-7c0-3.941 2.199-6.343 5-6.343\"/>",{"left":408,"top":408,"width":419,"height":419,"rotate":408,"vFlip":183,"hFlip":183,"body":420},256,"\u003Cpath fill=\"currentColor\" d=\"M228 128a100 100 0 1 1-22.86-63.64a12 12 0 0 1-18.51 15.28A76 76 0 1 0 203.05 140H128a12 12 0 0 1 0-24h88a12 12 0 0 1 12 12\"/>",{"left":408,"top":408,"width":419,"height":419,"rotate":408,"vFlip":183,"hFlip":183,"body":422},"\u003Cpath fill=\"currentColor\" d=\"M224 128a96 96 0 1 1-21.95-61.09a8 8 0 1 1-12.33 10.18A80 80 0 1 0 207.6 136H128a8 8 0 0 1 0-16h88a8 8 0 0 1 8 8\"/>",{"oVhJaef6Ht":8,"t96FybqVTi":8,"5lK7LS5al0":8,"jWdRXNxOQ0":8,"5oSQ2a90xd":8,"2QISyIzlyM":8,"HGsO2Ckakl":8},"/general/an-atom-a-has-8-electrons-in-its-right-shell-and-is-isoelectronically-with-the-trivalent-cation-of-an-atom-b-when-the-quantum-numbers-of-the-last-electron-of-the-electron-distribution-are-determin"] AppleWebKit/537.36 KHTML,likeGecko Chrome/64.0.3282.39 Safari/537.36",refreshOnResize:false}},app:{baseURL:"/",buildAssetsDir:"/_nuxt/",cdnURL:"https://gcdn.fx2.io/math-master.org/"}}