MATH MADE EASY - MathMaster

+ 4y = 16\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>4 + 4y = 16\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>4y = 16 - 4\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>4y = 12\u003C/math-field>\u003C/math-field>\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>y = \\frac{12}{4} = 3\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nThe solution to the system of equations is:\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>\u003Cbr />\n* \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>y = 3\u003C/math-field>\u003C/math-field>",671,134,"3x-4y-6-2x-4y-16",{"id":55,"category":47,"text_question":56,"photo_question":49,"text_answer":57,"step_text_answer":8,"step_photo_answer":8,"views":58,"likes":59,"slug":60},538057,"The product of two consecutive even integers is 8 less than five times their sum. Find the two integers","Let the two consecutive even integers be \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x\u003C/math-field>\u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x+2\u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\nTheir product is given by:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x

$x+2$ \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nThe sum of the integers is:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x +

$x+2$ = 2x + 2\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nFive times their sum is:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5

$2x + 2$ = 10x + 10\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nAccording to the problem, the product is 8 less than five times their sum:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x

$x+2$ = 10x + 10 - 8\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nSimplifying, we have:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x

$x+2$ = 10x + 2\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nThis expands to:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x^2 + 2x = 10x + 2\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nRearranging, we get:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x^2 + 2x - 10x - 2 = 0\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nSimplifying, we have:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x^2 - 8x - 2 = 0\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nSolve this quadratic equation using the quadratic formula \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}\u003C/math-field>\u003C/math-field>, where \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>a = 1\u003C/math-field>\u003C/math-field>, \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>b = -8\u003C/math-field>\u003C/math-field>, and \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>c = -2\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 = \\frac{-

$-8$ \\pm \\sqrt{

$-8$ ^2 - 4

$1$

$-2$ }}{2

$1$ } \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 = \\frac{8 \\pm \\sqrt{64 + 8}}{2} \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 = \\frac{8 \\pm \\sqrt{72}}{2} \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 = \\frac{8 \\pm \\sqrt{36 \\times 2}}{2} \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 = \\frac{8 \\pm 6\\sqrt{2}}{2} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nContinuing to simplify gives us:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> x = 4 \\pm 3\\sqrt{2} \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nSince the solutions involve square roots, there are no integer solutions to the problem.",570,114,"the-product-of-two-consecutive-even-integers-is-8-less-than-five-times-their-sum-find-the-two-integers",{"id":62,"category":47,"text_question":63,"photo_question":49,"text_answer":64,"step_text_answer":8,"step_photo_answer":8,"views":65,"likes":66,"slug":67},538046," A cylinder makes six turns in 2 seconds, calculate: a) its angular velocity in rad/s; b) its period and c) its frequency.","a) Angular velocity

$\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\omega\u003C/math-field>\u003C/math-field>$ is given by the formula:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\omega = \\frac{\\Delta \\theta}{\\Delta t}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nHere, the cylinder makes 6 turns, and each turn is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>2\\pi\u003C/math-field>\u003C/math-field> radians. Therefore, in 6 turns, the angle in radians is:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\Delta \\theta = 6 \\times 2\\pi = 12\\pi \\text{ radians}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nThe time period

$\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\Delta t\u003C/math-field>\u003C/math-field>$ is 2 seconds, so the angular velocity is:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\omega = \\frac{12\\pi}{2} = 6\\pi \\text{ rad/s}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nb) The period

$\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>T\u003C/math-field>\u003C/math-field>$ is the time it takes to complete one full rotation

$1 turn$ . Since the cylinder makes 6 turns in 2 seconds:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>T = \\frac{\\Delta t}{\\text{number of turns}} = \\frac{2}{6} = \\frac{1}{3} \\text{ s}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nc) The frequency

$\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>f\u003C/math-field>\u003C/math-field>$ is the reciprocal of the period:\u003Cbr />\n\u003Cbr />\n\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>f = \\frac{1}{T} = \\frac{1}{\\frac{1}{3}} = 3 \\text{ Hz}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nThus, the answers are:\u003Cbr />\n\u003Cbr />\na) \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\omega = 6\\pi \\text{ rad/s}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nb) \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>T = \\frac{1}{3} \\text{ s}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nc) \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>f = 3 \\text{ Hz}\u003C/math-field>\u003C/math-field>",250,50,"a-cylinder-makes-six-turns-in-2-seconds-calculate-a-its-angular-velocity-in-rad-s-b-its-period-and-c-its-frequency",{"id":69,"category":47,"text_question":70,"photo_question":49,"text_answer":71,"step_text_answer":8,"step_photo_answer":8,"views":72,"likes":73,"slug":74},538019,"Perform the indicated operation \n3x over X-5 minus 2x over X-2","Solution:\u003Cbr />\n1. Given expressions:\u003Cbr />\n \u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{3x}{x-5}\u003C/math-field>\u003C/math-field>\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{2x}{x-2}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Find the common denominator.\u003Cbr />\n- The denominators are \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>

$x-5$ \u003C/math-field>\u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>

$x-2$ \u003C/math-field>\u003C/math-field>.\u003Cbr />\n- The common denominator is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>

$x-5$

$x-2$ \u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\n3. Rewrite each fraction with the common denominator:\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{3x}{x-5} = \\frac{3x

$x-2$ }{

$x-5$

$x-2$ } = \\frac{3x^2 - 6x}{

$x-5$

$x-2$ }\u003C/math-field>\u003C/math-field>\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{2x}{x-2} = \\frac{2x

$x-5$ }{

$x-5$

$x-2$ } = \\frac{2x^2 - 10x}{

$x-5$

$x-2$ }\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Subtract the fractions:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{3x^2 - 6x}{

$x-5$

$x-2$ } - \\frac{2x^2 - 10x}{

$x-5$

$x-2$ } = \\frac{

$3x^2 - 6x$ -

$2x^2 - 10x$ }{

$x-5$

$x-2$ }\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Simplify the numerator:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3x^2 - 6x -

$2x^2 - 10x$ = 3x^2 - 6x - 2x^2 + 10x = x^2 + 4x\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n6. Final expression:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\frac{x^2 + 4x}{

$x-5$

$x-2$ }\u003C/math-field>\u003C/math-field>",1464,293,"perform-the-indicated-operation-3x-over-x-5-minus-2x-over-x-2",{"id":76,"category":47,"text_question":77,"photo_question":49,"text_answer":78,"step_text_answer":8,"step_photo_answer":8,"views":79,"likes":80,"slug":81},537911,"Task 4.

$1.5 points$ There are several students in the class and some pairs are friends

$the friendship relationship is symmetrical$ . Every student has at least one friend. The whole class came to MatFyz, where each student chooses exactly one of the lectures: either mathematics or computer science. Can you prove that

$in each such class$ the students can divide themselves into two lectures so that each studentZhad at least one friend who attended a different lecture than the student himself Z.","1. Consider the students and friends as a graph where each student is a vertex, and an undirected edge exists between two vertices if both students are friends. \u003Cbr />\n2. The problem requires proving that this graph is bipartite. \u003Cbr />\n3. A graph is bipartite if and only if it has no odd-length cycles.\u003Cbr />\n4. Start with any vertex and perform BFS or DFS to try to color the graph using two colors, such that no two adjacent vertices share the same color. \u003Cbr />\n5. If successful, you have a bipartite graph, meaning that nodes can be divided into two groups, each corresponding to one lecture.\u003Cbr />\n6. If at any point you find a vertex with the same color during the graph traversal across an edge, there exists an odd-length cycle.\u003Cbr />\n7. The presence of such a cycle would contradict the criteria as friendship is symmetrical, ensuring each cycle detected can be colored alternately, validating bipartiteness.\u003Cbr />\n8. Since each student has at least one friend, and that ensures the graph has no isolated points, every path will connect properly into two sets.\u003Cbr />\n9. Thus, the students can always be divided into two lectures maintaining the condition.\u003Cbr />\n\u003Cbr />\nAnswer: The students can always be divided in a manner where each student has at least one friend attending a different lecture.",1006,201,"task-4-1-5-points-there-are-several-students-in-the-class-and-some-pairs-are-friends-the-friendship-relationship-is-symmetrical-every-student-has-at-least-one-friend-the-whole-class-came-to-matfy",{"id":83,"category":47,"text_question":84,"photo_question":49,"text_answer":85,"step_text_answer":8,"step_photo_answer":8,"views":86,"likes":87,"slug":88},537902,"i need 3 addition or subtraction problems with 2 or 3 digit number that equal 5","1. Solve \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>12 - 7\u003C/math-field>\u003C/math-field>:\u003Cbr />\n\u003Cbr />\n - Calculate: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>12 - 7 = 5\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n - Answer: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Solve \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>102 - 97\u003C/math-field>\u003C/math-field>:\u003Cbr />\n\u003Cbr />\n - Calculate: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>102 - 97 = 5\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n - Answer: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Solve \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>55 - 50\u003C/math-field>\u003C/math-field>:\u003Cbr />\n\u003Cbr />\n - Calculate: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>55 - 50 = 5\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n - Answer: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>5\u003C/math-field>\u003C/math-field>",783,157,"i-need-3-addition-or-subtraction-problems-with-2-or-3-digit-number-that-equal-5",{"id":90,"category":47,"text_question":91,"photo_question":49,"text_answer":92,"step_text_answer":8,"step_photo_answer":8,"views":93,"likes":94,"slug":95},537843,"64=-8m","1. Start with the equation: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>64 = -8m\u003C/math-field>\u003C/math-field>\u003Cbr />\n2. Divide both sides by -8 to solve for \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>m\u003C/math-field>\u003C/math-field>: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>m = \\frac{64}{-8}\u003C/math-field>\u003C/math-field> \u003Cbr />\n3. Simplify the right side: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>m = -8\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nAnswer: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>m = -8\u003C/math-field>\u003C/math-field>",455,91,"64-8m",{"id":97,"category":47,"text_question":98,"photo_question":49,"text_answer":99,"step_text_answer":8,"step_photo_answer":8,"views":51,"likes":52,"slug":100},537810,"Alice spent

$15 more than half as much as Mrs. Bai spent on her shopping trip. If they spent a total of$ 375, how much did each spend?","1.A is the amount Alice spent, and B is the amount Mrs. Bai spent.\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = \\frac{1}{2} B + 15 \u003C/math-field>\u003C/math-field> \u003Cdiv>\u003Cbr>\u003Cbr>\u003C/div>\u003Cdiv>They spent a total of

$375, so:\u003C/div>\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A + B = 375 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>2. Substitute A in the second equation:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\left(\\frac{1}{2} B + 15\\right) + B = 375 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>3. Simplify and solve for B:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{1}{2} B + B + 15 = 375 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{3}{2} B + 15 = 375 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{3}{2} B = 375 - 15 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{3}{2} B = 360 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> B = \\frac{360 \\times 2}{3} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> B = 240 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>4. Substitute B = 240$ back into the equation for A\u003Cspan style=\"font-size: 0.894rem;\">:\u003Cdiv>\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = \\frac{1}{2}

$240$ + 15 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = 120 + 15 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> A = 135 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>\u003C/div>\u003Cdiv>\u003Cbr>\u003Cbr>\u003C/div>\u003Cdiv>Alice spent

$135 and Mrs.Bai spent$ 240\u003C/div>","alice-spent-15-more-than-half-as-much-as-mrs-bai-spent-on-her-shopping-trip-if-they-spent-a-total-of-375-how-much-did-each-spend",{"id":102,"category":47,"text_question":103,"photo_question":49,"text_answer":104,"step_text_answer":8,"step_photo_answer":8,"views":105,"likes":106,"slug":107},537745,"I need to know why mathematical calculations such as the functions f

$x, y$ are relevant in geomensura.","1. **Understanding the Purpose:**\u003Cbr />\n - Geomensura involves measuring and projecting geographical data.\u003Cbr />\n - Functions \$ f(x, y) \$ can represent fields like elevation over a 2D area.\u003Cbr />\n\u003Cbr />\n2. **Modeling Geographical Features:**\u003Cbr />\n - By inputting two independent variables (i.e., \$ x \$ and \$ y \$), functions can output a third variable (e.g., elevation),\u003Cbr />\n allowing for 3D modeling of the terrain: \u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> f(x, y) \\Rightarrow \\text{Elevation at coordinates } (x, y) \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. **Calculations in Practice:**\u003Cbr />\n - Functions allow calculational tasks such as finding contour lines or slope of a terrain,\u003Cbr />\n which is essential for safe and efficient surveying practices.\u003Cbr />\n\u003Cbr />\n4. **Conclusion:**\u003Cbr />\n - The use of functions in modeling geographical features provides a necessary\u003Cbr />\n mathematical foundation for the practical tasks involved in geomensura.\u003Cbr />\n\u003Cbr />\n**Answer:**\u003Cbr />\n - Functions \$ f(x, y) \$ are used to model geographical features essential for geomensura's analysis and projections.",584,117,"i-need-to-know-why-mathematical-calculations-such-as-the-functions-f-x-y-are-relevant-in-geomensura",{"id":109,"category":47,"text_question":110,"photo_question":49,"text_answer":111,"step_text_answer":8,"step_photo_answer":8,"views":112,"likes":113,"slug":114},537744,"In the rectilinear path of a road described by the equation L: 3x +4y -5 = 0, a public telephone is located at Q. If a person is at point P(12;6), calculate the distance in meters to make the call.","1. Reemplace \$ x \$ y \$ y \$ en la ecuación del plano \$ 3x + 4y - 5 = 0 \$ con las coordenadas del punto \$ P(12, 6) \$:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> d = \\frac{|3(12) + 4(6) - 5|}{\\sqrt{3^2 + 4^2}} \u003C/math-field>\u003C/math-field>\u003Cbr />\n2. Simplifique la expresión para numerador:\u003Cbr />\n - Calcule \$ 3(12) + 4(6) - 5 \$:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 36 + 24 - 5 = 55 \u003C/math-field>\u003C/math-field>\u003Cbr />\n3. Simplifique la expresión para the denominador:\u003Cbr />\n - Calcule \$ \\sqrt{3^2 + 4^2} \$:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\sqrt{9 + 16} = \\sqrt{25} = 5 \u003C/math-field>\u003C/math-field>\u003Cbr />\n4. Calcule el resultado final:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{|55|}{5} = \\frac{55}{5} = 11 \u003C/math-field>\u003C/math-field>\u003Cbr />\n5. La distancia desde el punto \$ P \$ a la línea es \$ d = 11 \$ metros",1498,300,"in-the-rectilinear-path-of-a-road-described-by-the-equation-l-3x-4y-5-0-a-public-telephone-is-located-at-q-if-a-person-is-at-point-p-12-6-calculate-the-distance-in-meters-to-make-the-call",{"first":6,"last":116,"prev":8,"next":10},9,{"current_page":6,"from":6,"last_page":116,"links":118,"path":143,"per_page":34,"to":34,"total":42},[119,122,125,127,129,131,133,136,139,141],{"url":6,"label":120,"active":121},"1",true,{"url":10,"label":123,"active":124},"2",false,{"url":13,"label":126,"active":124},"3",{"url":16,"label":128,"active":124},"4",{"url":19,"label":130,"active":124},"5",{"url":22,"label":132,"active":124},"6",{"url":134,"label":135,"active":124},7,"7",{"url":137,"label":138,"active":124},8,"8",{"url":116,"label":140,"active":124},"9",{"url":10,"label":142,"active":124},"Next 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