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":55,"category":47,"text_question":56,"photo_question":49,"text_answer":57,"step_text_answer":19,"step_photo_answer":19,"views":58,"likes":59,"slug":60},538015,"Three times the sum of m an n","1. Identify the expression for the sum of \\( m \\) and \\( n \\): \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>m + n\u003C/math-field>\u003C/math-field>\u003Cbr />\n2. Multiply this sum by 3: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3(m+n)\u003C/math-field>\u003C/math-field>\u003Cbr />\n3. The final expression is the answer: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>3(m+n)\u003C/math-field>\u003C/math-field>",1237,247,"three-times-the-sum-of-m-an-n",{"id":62,"category":47,"text_question":63,"photo_question":49,"text_answer":64,"step_text_answer":19,"step_photo_answer":19,"views":65,"likes":66,"slug":67},538011,"mWhat is the image of point (-5, -2) when it is reflected across y = -x?","Solution:\u003Cbr />\n1. Reflection across the line \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>y = -x\u003C/math-field>\u003C/math-field>:\u003Cbr />\n - To find the image of a point \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>(x, y)\u003C/math-field>\u003C/math-field> after reflection across \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>y = -x\u003C/math-field>\u003C/math-field>, swap the coordinates and change their signs:\u003Cbr />\n - Original point: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>(-5, -2)\u003C/math-field>\u003C/math-field>\u003Cbr />\n - Swap and change signs: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>(-(-2), -(-5)) = (2, 5)\u003C/math-field>\u003C/math-field>\u003Cbr />\n2. The image of the point $(-5, -2)$ after reflection across the line \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>y = -x\u003C/math-field>\u003C/math-field> is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>(2, 5)\u003C/math-field>\u003C/math-field>.",1038,208,"mwhat-is-the-image-of-point-5-2-when-it-is-reflected-across-y-x",{"id":69,"category":47,"text_question":70,"photo_question":49,"text_answer":71,"step_text_answer":19,"step_photo_answer":19,"views":72,"likes":73,"slug":74},537995,"-x+2y=0\n3x+2y=8\nSolve","1. Let's solve the system of equations:\u003Cbr />\n\u003Cbr />\n The given equations are:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> -x + 2y = 0 \\quad ...(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> 3x + 2y = 8 \\quad ...(2) \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. First, solve equation (1) for \\( x \\):\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> -x + 2y = 0 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n Rearranging, we have:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> x = 2y \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Substitute \\( x = 2y \\) into equation (2):\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 3(2y) + 2y = 8 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n Simplify:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 6y + 2y = 8 \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> 8y = 8 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> y = 1 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Substitute \\( y = 1 \\) back into \\( x = 2y \\):\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> x = 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 = 2 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Therefore, the solution to the system is:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> x = 2, \\quad y = 1 \u003C/math-field>\u003C/math-field>",553,111,"x-2y-0-3x-2y-8-solve",{"id":76,"category":47,"text_question":77,"photo_question":49,"text_answer":78,"step_text_answer":19,"step_photo_answer":19,"views":79,"likes":80,"slug":81},537957,"5a/10b-10b2+8\nWhere a =-8 and b=-6","1. Substitute \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\( a = -8 \\)\u003C/math-field>\u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\( b = -6 \\)\u003C/math-field>\u003C/math-field> into the expression: \u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{5(-8)}{10(-6)} - 10(-6)^2 + 8 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>2. Simplify \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\( \\frac{5(-8)}{10(-6)} \\)\u003C/math-field>\u003C/math-field> :\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{5 \\times (-8)}{10 \\times (-6)} = \\frac{-40}{-60} = \\frac{2}{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>\\( 10(-6)^2 \\)\u003C/math-field>\u003C/math-field> :\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 10 \\times (-6)^2 = 10 \\times 36 = 360 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>4. Substitute these values back into the expression:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{2}{3} - 360 + 8 \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>5. Combine the terms carefully:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{2}{3} + 8 = \\frac{2}{3} + \\frac{24}{3} = \\frac{26}{3} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>6. Subtract 360:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{26}{3} - 360 = \\frac{26}{3} - \\frac{1080}{3} = \\frac{26 - 1080}{3} = \\frac{-1054}{3} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>7. Final result is:\u003Cbr>\u003Cbr>\u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> \\frac{-1054}{3} \u003C/math-field>\u003C/math-field> \u003Cbr>\u003Cbr>The answer is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\(-\\frac{1054}{3}\\)\u003C/math-field>\u003C/math-field>.",738,148,"5a-10b-10b2-8-where-a-8-and-b-6",{"id":83,"category":47,"text_question":84,"photo_question":49,"text_answer":85,"step_text_answer":19,"step_photo_answer":19,"views":86,"likes":87,"slug":88},537914,"Integration by parts for \\int \\ln x dx :","Solution:\u003Cbr />\n1. Identify the components of integration by parts:\u003Cbr />\n - Let \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>u = \\ln x\u003C/math-field>\u003C/math-field>, then \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>du = \\frac{1}{x} \\, dx\u003C/math-field>\u003C/math-field>.\u003Cbr />\n - Let \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>dv = dx\u003C/math-field>\u003C/math-field>, then \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>v = x\u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\n2. Apply the integration by parts formula \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\int u \\, dv = uv - \\int v \\, du\u003C/math-field>\u003C/math-field>:\u003Cbr />\n * The formula becomes:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\int \\ln x \\, dx = x \\ln x - \\int x \\left(\\frac{1}{x} \\right) \\, dx\u003C/math-field>\u003C/math-field>.\u003Cbr />\n\u003Cbr />\n3. Simplify the remaining integral:\u003Cbr />\n * \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\int x \\left(\\frac{1}{x} \\right) \\, dx = \\int 1 \\, dx = x + C\u003C/math-field>\u003C/math-field>, where \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>C\u003C/math-field>\u003C/math-field> is the integration constant.\u003Cbr />\n\u003Cbr />\n4. Substitute back to get the final result:\u003Cbr />\n * \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\int \\ln x \\, dx = x \\ln x - x + C\u003C/math-field>\u003C/math-field>.\u003Cbr />",258,52,"integration-by-parts-for-int-ln-x-dx",{"id":90,"category":47,"text_question":91,"photo_question":49,"text_answer":92,"step_text_answer":19,"step_photo_answer":19,"views":93,"likes":94,"slug":95},537845,"X-10=-25","To solve the equation \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> X - 10 = -25 \u003C/math-field>\u003C/math-field>, we need to isolate \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> X \u003C/math-field>\u003C/math-field> by adding 10 to both sides: \u003Cbr />\n\u003Cbr />\n1. Start with the original equation: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> X - 10 = -25 \u003C/math-field>\u003C/math-field>\u003Cbr />\n2. Add 10 to both sides: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> X - 10 + 10 = -25 + 10 \u003C/math-field>\u003C/math-field>\u003Cbr />\n3. Simplify both sides: \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> X = -15 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\nThe solution is \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> X = -15 \u003C/math-field>\u003C/math-field>",1297,259,"x-10-25",{"id":97,"category":47,"text_question":98,"photo_question":49,"text_answer":99,"step_text_answer":19,"step_photo_answer":19,"views":100,"likes":101,"slug":102},537797,"What is a rectangular prism measures 610cm by 514cm by 370 what is the surface area","1. Calculate the area of the top and bottom faces:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 2lw = 2(610 \\times 514) = 627,080 \\, \\text{cm}^2 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Calculate the area of the front and back faces:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 2lh = 2(610 \\times 370) = 451,400 \\, \\text{cm}^2 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Calculate the area of the left and right faces:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> 2wh = 2(514 \\times 370) = 380,240 \\, \\text{cm}^2 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Sum all the calculated areas:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> SA = 627,080 + 451,400 + 380,240 = 1,458,720 \\, \\text{cm}^2 \u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Final answer:\u003Cbr />\n\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only> SA = 1,458,720 \\, \\text{cm}^2 \u003C/math-field>\u003C/math-field>",1254,251,"what-is-a-rectangular-prism-measures-610cm-by-514cm-by-370-what-is-the-surface-area",{"id":104,"category":47,"text_question":105,"photo_question":49,"text_answer":106,"step_text_answer":19,"step_photo_answer":19,"views":107,"likes":108,"slug":109},537762,"X^2=81^1*1/2 \n Find x","Solution:\u003Cbr />\n1. Start with the equation:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>X^2 = 81^{\\frac{1}{2}}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. Recognize that \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>81^{\\frac{1}{2}}\u003C/math-field>\u003C/math-field> is the square root of 81:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>81^{\\frac{1}{2}} = \\sqrt{81}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Calculate the square root of 81:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>\\sqrt{81} = 9\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. Substitute back into the equation:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>X^2 = 9\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n5. Solve for \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>X\u003C/math-field>\u003C/math-field> by taking the square root of both sides:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>X = \\pm \\sqrt{9}\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n6. Simplify the result:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>X = \\pm 3\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n7. Thus, the solutions for \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>X\u003C/math-field>\u003C/math-field> are:\u003Cbr />\n \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>X = 3\u003C/math-field>\u003C/math-field> and \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>X = -3\u003C/math-field>\u003C/math-field>",411,82,"x-2-81-1-1-2-find-x",{"id":111,"category":47,"text_question":112,"photo_question":49,"text_answer":113,"step_text_answer":19,"step_photo_answer":19,"views":114,"likes":115,"slug":116},537681,"A square has a side length of s write a rule in function notation to represent the perimeter","Solution:\u003Cbr />\n1. Given:\u003Cbr />\n- A square with side length \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>s\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n2. The formula for the perimeter \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>P\u003C/math-field>\u003C/math-field> of a square is:\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>P = 4 \\cdot s\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n3. Express the formula as a function:\u003Cbr />\n- Let \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>P(s) = 4s\u003C/math-field>\u003C/math-field>\u003Cbr />\n\u003Cbr />\n4. The function notation for the perimeter of the square is:\u003Cbr />\n- \u003Cmath-field read-only default-mode=\"inline-math\" class=\"math-expression\">\u003Cmath-field read-only>P(s) = 4s\u003C/math-field>\u003C/math-field>",1119,224,"a-square-has-a-side-length-of-s-write-a-rule-in-function-notation-to-represent-the-perimeter",{"first":17,"last":5,"prev":19,"next":21},{"current_page":17,"from":17,"last_page":5,"links":119,"path":144,"per_page":145,"to":145,"total":13},[120,123,126,128,130,132,134,137,140,142],{"url":17,"label":121,"active":122},"1",true,{"url":21,"label":124,"active":125},"2",false,{"url":24,"label":127,"active":125},"3",{"url":27,"label":129,"active":125},"4",{"url":30,"label":131,"active":125},"5",{"url":33,"label":133,"active":125},"6",{"url":135,"label":136,"active":125},7,"7",{"url":138,"label":139,"active":125},8,"8",{"url":5,"label":141,"active":125},"9",{"url":21,"label":143,"active":125},"Next 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