MATH MADE EASY - MathMaster

and N

$7, 9$ ?","Solution:\u003Cbr />\n1. Identify the coordinates of points L and N.\u003Cbr />\n * Point L: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>

$x_1, y_1$ =

$1, 3$ \u003C/math-field>\u003C/math-field>\u003Cbr />\n * Point N: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>

$x_2, y_2$ =

$7, 9$ \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Use the distance formula between two points \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>

$x_1, y_1$ \u003C/math-field>\u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>

$x_2, y_2$ \u003C/math-field>\u003C/math-field>:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>d = \\sqrt{

$x_2 - x_1$ ^2 +

$y_2 - y_1$ ^2}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Substitute the given coordinates into the distance formula:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>d = \\sqrt{

$7 - 1$ ^2 +

$9 - 3$ ^2}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Simplify the expression:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>d = \\sqrt{6^2 + 6^2}\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>d = \\sqrt{36 + 36}\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>d = \\sqrt{72}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Simplify further if possible:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>d = \\sqrt{36 \\cdot 2} = \\sqrt{36} \\cdot \\sqrt{2} = 6\\sqrt{2}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nResult:\u003Cbr />\n- The distance between L and N is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>6\\sqrt{2}\u003C/math-field>\u003C/math-field>.",1009,202,"what-is-the-distance-between-line-segment-ln-with-endpoints-l-1-3-and-n-7-9",{"id":64,"category":49,"text_question":65,"photo_question":51,"text_answer":66,"step_text_answer":8,"step_photo_answer":8,"views":67,"likes":68,"slug":69},538008,"A circle is circumscribed and inscribed in a square of area 64cm. Calculate the area of the circular ring bounded by these two circles.","1. Given that the area of the square is 64 cm\$^2\$, so the side length \$ s \$ of the square is given by:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> s^2 = 64 \\implies s = 8 \\, \\text{cm} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>2. The circumscribed circle has radius \$ R \$ equal to half the diagonal of the square, which is equal to the side length times the root of 2, \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> R = \\frac{8\\sqrt{2}}{2} = 4\\sqrt{2} \\, \\text{cm} \u003C/math-field>\u003C/math-field> \u003Cdiv>\u003Cbr>\u003Cbr>\u003C/div>2. The area of the circumscribed circle:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A_{\\text{circumscribed}} = \\pi (4\\sqrt{2})^2 = 32\\pi \\, \\text{cm}^2 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cspan style=\"font-size: 14.304px;\">2. The inscribed circle has radius \$ r \$ equal to half the side length of the square,\u003Cbr style=\"font-size: 14.304px;\">\u003Cbr style=\"font-size: 14.304px;\">\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>r=\\frac{8}{2}=4\\text{cm}\u003C/math-field>\u003C/math-field>\u003Cspan style=\"font-size: 14.304px;\">\u003Cdiv style=\"font-size: 14.304px;\">\u003Cbr>\u003Cbr>\u003C/div>\u003Cspan style=\"font-size: 14.304px;\">2. The area of the inscribed circle:\u003Cbr style=\"font-size: 14.304px;\">\u003Cbr style=\"font-size: 14.304px;\">\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>A_{\\text{inscribed}}=\\pi(4)^2=16\\pi\\,\\text{cm}^2\u003C/math-field>\u003C/math-field>\u003Cspan style=\"font-size: 14.304px;\">\u003Cbr style=\"font-size: 14.304px;\">\u003Cbr style=\"font-size: 14.304px;\">3. The area of the ring:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A_{\\text{ring}} = A_{\\text{circumscribed}} - A_{\\text{inscribed}} = (32\\pi - 16\\pi) = 16\\pi \\, \\text{cm}^2 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cdiv>\u003Cbr>\u003Cbr>\u003C/div>\u003Cdiv>\u003Cbr>\u003Cbr>\u003C/div>\u003Cdiv>Answer: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>16\\pi\\operatorname{cm}^2\u003C/math-field>\u003C/math-field> \u003C/div>",1296,259,"a-circle-is-circumscribed-and-inscribed-in-a-square-of-area-64cm-calculate-the-area-of-the-circular-ring-bounded-by-these-two-circles",{"id":71,"category":49,"text_question":72,"photo_question":51,"text_answer":73,"step_text_answer":8,"step_photo_answer":8,"views":74,"likes":75,"slug":76},537996,"Y=2x+1\n Y=2x+9\n Solve","1. Set the two equations equal to each other since they both represent \$ Y \$:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 2x + 1 = 2x + 9 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Subtract \$ 2x \$ from both sides:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 1 = 9 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. The above statement is a contradiction, meaning no value of \$ x \$ satisfies it.\u003Cbr />\n\u003Cbr />\n4. Since \$ 2x + 1 \$ and \$ 2x + 9 \$ are equations of lines with the same slope but different y-intercepts, these lines are parallel and do not intersect.\u003Cbr />\n\u003Cbr />\n5. Therefore, there are no solutions to the system.",730,146,"y-2x-1-y-2x-9-solve",{"id":78,"category":49,"text_question":79,"photo_question":51,"text_answer":80,"step_text_answer":8,"step_photo_answer":8,"views":81,"likes":82,"slug":83},537939,"From 2010/11 to 2012/13 what is the percentage change in the total value of open tuition loans?","Solution:\u003Cbr />\n1. Let the total value of outstanding tuition fee loans in 2010/11 be \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V_{2010/11}\u003C/math-field>\u003C/math-field> and in 2012/13 be \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>V_{2012/13}\u003C/math-field>\u003C/math-field>.\u003Cbr />\n2. The percentage change formula is given by:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\text{Percentage Change} = \\frac{V_{2012/13} - V_{2010/11}}{V_{2010/11}} \\times 100\\%\u003C/math-field>\u003C/math-field>\u003Cbr />\n3. Substitute the provided or known values into the formula if available.\u003Cbr />\n\u003Cbr />\n* Note: As specific numerical values for 2010/11 and 2012/13 are not provided, numbers need to be inputted to complete this calculation.",1486,297,"from-2010-11-to-2012-13-what-is-the-percentage-change-in-the-total-value-of-open-tuition-loans",{"id":85,"category":49,"text_question":86,"photo_question":51,"text_answer":87,"step_text_answer":8,"step_photo_answer":8,"views":88,"likes":89,"slug":90},537922,"Find three consecutive integers whose sum is 84","1. Represent the consecutive integers as $x$, $x+1$, and $x+2$.\u003Cbr />\n\u003Cbr />\n2. Write the equation for their sum: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x + (x+1) + (x+2) = 84\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Simplify and solve for $x$:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3x + 3 = 84\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3x = 81\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x = 27\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. The integers are $x = 27$, $x + 1 = 28$, $x + 2 = 29$.\u003Cbr />\n\u003Cbr />\n5. Therefore, the three consecutive integers are 27, 28, and 29.",1301,260,"find-three-consecutive-integers-whose-sum-is-84",{"id":92,"category":49,"text_question":93,"photo_question":51,"text_answer":94,"step_text_answer":8,"step_photo_answer":8,"views":95,"likes":96,"slug":97},537859,"A music store buys instruments and then sells them for 30% more than they originally paid. if the store buys a guitar for $45 then what will the store sell it for?","1. Calculate 30% of the original price:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 0.30 \\times 45 = 13.5 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>2. Add this amount to the original price:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 45 + 13.5 = 58.5 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>Hence, the store will sell the guitar for \u003Cstrong>$58.50\u003C/strong>.",1419,284,"a-music-store-buys-instruments-and-then-sells-them-for-30-more-than-they-originally-paid-if-the-store-buys-a-guitar-for-45-then-what-will-the-store-sell-it-for",{"id":99,"category":49,"text_question":100,"photo_question":51,"text_answer":101,"step_text_answer":8,"step_photo_answer":8,"views":102,"likes":103,"slug":104},537749,"Consider X, a variable whose distribution is normal with a mean of 5 and a variance of 2. We obtain the variable Z by standardizing X. If the value of z is -1 for an individual, what is the value of x for that same individual? Show your approach. (1 point).","1. Utiliser la formule de standardisation :\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> Z = \\frac{X - \\mu}{\\sigma} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Remplacer les valeurs données dans la formule :\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> -1 = \\frac{X - 5}{\\sqrt{2}} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Résoudre l'équation pour \$X\$ :\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> -1 \\cdot \\sqrt{2} = X - 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> X = 5 - \\sqrt{2} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. La valeur de \$x\$ pour laquelle \$z = -1\$ est :\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> x = 5 - \\sqrt{2} \u003C/math-field>\u003C/math-field>",448,90,"consider-x-a-variable-whose-distribution-is-normal-with-a-mean-of-5-and-a-variance-of-2-we-obtain-the-variable-z-by-standardizing-x-if-the-value-of-z-is-1-for-an-individual-what-is-the-value-of-x",{"id":106,"category":49,"text_question":107,"photo_question":51,"text_answer":108,"step_text_answer":8,"step_photo_answer":8,"views":109,"likes":110,"slug":111},537710,"By solving the combined operation 3 - 2 ∙ (2 ∙ 3 - 2 ∙ 4), we obtain:","1. Resolver la operación dentro del paréntesis:\u003Cbr />\n \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2 \\cdot 3 - 2 \\cdot 4\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n Esto se convierte en:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>6 - 8 = -2\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Sustituir el resultado en la expresión completa:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3 - 2 \\cdot (-2)\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Resolver la multiplicación:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2 \\cdot (-2) = -4\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Finalmente, resolver la resta:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3 - (-4) = 3 + 4 = 7\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nPor lo tanto, la respuesta es:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>7\u003C/math-field>\u003C/math-field>",343,69,"by-solving-the-combined-operation-3-2-2-3-2-4-we-obtain",{"id":113,"category":49,"text_question":114,"photo_question":51,"text_answer":115,"step_text_answer":8,"step_photo_answer":8,"views":116,"likes":117,"slug":118},537630,"3.\tDue to the force of gravity, a 15.0 kg wooden crate slides down a steel ramp with an incline of 25.0°. Compute the acceleration of the crate down the incline. \nThe coefficient of kinetic friction is 0.300.","1. Calculate the gravitational force component down the ramp:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> F_{\\text{gravity}} = 15.0 \\cdot 9.81 \\cdot \\sin(25.0^\\circ) = 62.08 \\, \\text{N} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Calculate the normal force:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> F_{\\text{normal}} = 15.0 \\cdot 9.81 \\cdot \\cos(25.0^\\circ) = 133.39 \\, \\text{N} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Find the frictional force:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> F_{\\text{friction}} = 0.300 \\cdot 133.39 = 40.02 \\, \\text{N} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Determine the net force down the ramp:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> F_{\\text{net}} = 62.08 - 40.02 = 22.06 \\, \\text{N} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Calculate the acceleration:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> a = \\frac{22.06}{15.0} = 1.47 \\, \\text{m/s}^2 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nThus, the acceleration of the crate down the incline is approximately \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 1.47 \\, \\text{m/s}^2 \u003C/math-field>\u003C/math-field>. (Please note that the original solution provided had an incorrect calculation. The correct acceleration should be \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 1.47 \\, \\text{m/s}^2 \u003C/math-field>\u003C/math-field>.)",1327,265,"3-due-to-the-force-of-gravity-a-15-0-kg-wooden-crate-slides-down-a-steel-ramp-with-an-incline-of-25-0-compute-the-acceleration-of-the-crate-down-the-incline-the-coefficient-of-kinetic-friction-i",{"first":6,"last":120,"prev":8,"next":10},11,{"current_page":6,"from":6,"last_page":120,"links":122,"path":153,"per_page":147,"to":147,"total":36},[123,126,129,131,133,135,137,140,143,146,149,151],{"url":6,"label":124,"active":125},"1",true,{"url":10,"label":127,"active":128},"2",false,{"url":13,"label":130,"active":128},"3",{"url":16,"label":132,"active":128},"4",{"url":19,"label":134,"active":128},"5",{"url":22,"label":136,"active":128},"6",{"url":138,"label":139,"active":128},7,"7",{"url":141,"label":142,"active":128},8,"8",{"url":144,"label":145,"active":128},9,"9",{"url":147,"label":148,"active":128},10,"10",{"url":120,"label":150,"active":128},"11",{"url":10,"label":152,"active":128},"Next 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