{"id":6610,"date":"2020-10-01T17:30:19","date_gmt":"2020-10-01T17:30:19","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6610"},"modified":"2026-01-26T00:59:58","modified_gmt":"2026-01-26T00:59:58","slug":"1-9-solving-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/1-9-solving-formulas\/","title":{"raw":"1.10 Formulas","rendered":"1.10 Formulas"},"content":{"raw":"
\r\n

section 1.10 Learning Objectives<\/h3>\r\n1.10: Formulas<\/strong>\r\n
    \r\n \t
  • Evaluate a formula for given values<\/li>\r\n \t
  • Solve for a specified variable in a formula<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n\r\nConsider the following problem in the video:\r\n\r\nhttps:\/\/youtu.be\/jlxPgKQfhQs\r\n\r\nWe could have isolated the w<\/em> in the formula for perimeter before we solved the equation, and if we were going to use the formula many times, it could save a lot of time. The next example shows how to isolate a variable in a formula before substituting known dimensions or values into the formula.\r\n
    \r\n

    Example 1<\/h3>\r\nSolve for the variable, [latex]w[\/latex], <\/i>in the formula for the perimeter of a rectangle: <\/em>\r\n

    [latex]{p}=2\\left({l}\\right)+2\\left({w}\\right)[\/latex].<\/p>\r\n[reveal-answer q=\"967601\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"967601\"]\r\n\r\nFirst, isolate the term with [latex]w[\/latex]\u00a0by subtracting [latex]2l[\/latex]\u00a0from both sides of the equation.\r\n

    [latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,p\\,=\\,\\,\\,\\,2l+2w\\\\\\underline{\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,-2l\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,}\\\\p-2l=\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2w\\end{array}[\/latex]<\/i><\/p>\r\nNext, clear the coefficient of [latex]w[\/latex]\u00a0<\/i>by dividing both sides of the equation by 2.\r\n

    [latex]\\displaystyle \\begin{array}{l}\\underline{p-2l}=\\underline{2w}\\\\\\,\\,\\,\\,\\,\\,2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2\\\\ \\,\\,\\,\\frac{p-2l}{2}\\,\\,=\\,\\,w\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,w=\\frac{p-2l}{2}\\end{array}[\/latex]<\/p>\r\nYou can rewrite the equation so the isolated variable is on the left side.\r\n

    [latex]w=\\frac{p-2l}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Rates<\/h2>\r\nThere\u00a0is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would\u00a0want to know how long it would take you to reach your destination.\u00a0[latex]d=rt[\/latex] is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes.\r\n

    Distance, Rate, and Time<\/h2>\r\nIf you know two of the quantities in the relationship [latex]d=rt[\/latex], you can easily find the third using methods for solving linear equations. For example, if you\u00a0know that you\u00a0will be traveling on a road with a speed limit of\u00a0[latex]30\\frac{\\text{ miles }}{\\text{ hour }}[\/latex] for 2 hours, you can\u00a0find the distance you would travel by\u00a0multiplying rate times time or [latex]\\left(30\\frac{\\text{ miles }}{\\text{ hour }}\\right)\\left(2\\text{ hours }\\right)=60\\text{ miles }[\/latex].\r\n\r\nWe can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the\u00a0[latex]d=rt[\/latex] equation for t<\/em> using division:\r\n

    [latex]d=rt\\\\\\frac{d}{r}=t[\/latex]<\/p>\r\nLikewise, if we want to find rate, we can isolate r<\/em> using division:\r\n

    [latex]d=rt\\\\\\frac{d}{t}=r[\/latex]<\/p>\r\nIn the following example you will see how this formula is applied to answer a few questions about ultra marathon running.\r\n\r\n[caption id=\"attachment_3540\" align=\"alignleft\" width=\"211\"]\"Ann Ann Trason[\/caption]\r\n\r\nUltra marathon running (defined as anything longer than 26.2 miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River 50-mile Endurance Run which\u00a0begins in Sacramento, California, and ends in Auburn, California.[footnote]\"Ann Trason.\" Wikipedia. Accessed May 05, 2016. https:\/\/en.wikipedia.org\/wiki\/Ann_Trason.[\/footnote] In 1993 Trason finished the run with a time of 6:09:08. \u00a0The men's record for the same course was set in 1994\u00a0by Tom Johnson who finished the course with a time of 5:33:21.[footnote]\u00a0\"American River 50 Mile Endurance Run.\" Wikipedia. Accessed May 05, 2016. https:\/\/en.wikipedia.org\/wiki\/American_River_50_Mile_Endurance_Run.[\/footnote]\r\n\r\nIn the next examples we will use the [latex]d=rt[\/latex] formula to answer the following questions about the two runners.\r\n

      \r\n \t
    1. What was each runner's rate<\/strong> for their record-setting runs?<\/li>\r\n \t
    2. What was each runner's time<\/strong> for running one mile?<\/li>\r\n<\/ol>\r\nTo make answering the questions easier, we will round the two runners' times to 6 hours and 5.5 hours.\r\n
      \r\n

      Example 2<\/h3>\r\nWhat was each runner's rate for their record-setting runs?\r\n[reveal-answer q=\"55589\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"55589\"]\r\n\r\nRead and Understand:<\/strong>\u00a0We\u00a0are looking for rate and we know distance and time, so we can use the idea:\u00a0[latex]d=rt[\/latex]\r\n\r\nIf we divide both sides of the equation with t, we will have equation for the rate:\r\n

      [latex]\\frac{d}{t}=r[\/latex]<\/p>\r\nDefine and Translate:<\/strong> Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of what how all the terms are related to each other.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Runner<\/th>\r\nDistance =<\/th>\r\n(Rate )<\/th>\r\n(Time)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Trason<\/td>\r\n50 miles<\/td>\r\nr<\/td>\r\n6 hours<\/td>\r\n<\/tr>\r\n
      Johnson<\/td>\r\n50 miles<\/td>\r\nr<\/td>\r\n\u00a05.5 hours<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWrite and Solve:<\/strong>\r\n\r\nTrason's rate:<\/span>\r\n

      [latex]\\frac{d}{t}=r[\/latex]<\/p>\r\n

      [latex]\\begin{array}{c}\\frac{50\\text{ miles }}{6\\text{ hours }}=r\\end{array}[\/latex]<\/p>\r\nIf we divide 50 with 6 we get the approximate rate in terms of miles per hour (MPH)\r\n

      Trason's rate is [latex]\\frac{8.33\\text{ miles }}{\\text{ hour }} = 8.33[\/latex] MPH. (rounded to two decimal places)<\/p>\r\n \r\n\r\nJohnson's rate:<\/span>\r\n

      [latex]\\frac{d}{t}=r[\/latex]<\/p>\r\n

      [latex]\\begin{array}{c}\\frac{50\\text{ miles }}{5.5\\text{ hours }}=r\\end{array}[\/latex]<\/p>\r\nIf we divide 50 with 5.5 we get the approximate rate in terms of miles per hour (MPH)\r\n

      Johnson's rate is [latex]\\frac{9.1\\text{ miles }}{\\text{ hour }} = 9.1[\/latex] MPH. (rounded to two decimal places)<\/p>\r\nCheck and Interpret:<\/strong>\r\n\r\nWe can fill in our table with this information.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Runner<\/th>\r\nDistance =<\/th>\r\n(Rate )<\/th>\r\n(Time)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Trason<\/td>\r\n50 miles<\/td>\r\n8.33 [latex]\\frac{\\text{ miles }}{\\text{ hour }}[\/latex]<\/td>\r\n6 hours<\/td>\r\n<\/tr>\r\n
      Johnson<\/td>\r\n50 miles<\/td>\r\n9.1\u00a0[latex]\\frac{\\text{ miles }}{\\text{ hour }}[\/latex]<\/td>\r\n5.5 hours<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow we will tackle the second question where we are asked to find a time for each runner.\r\n
      \r\n

      Example 3<\/h3>\r\nWhat was each runner's time for running one mile?\r\n\r\n[reveal-answer q=\"757309\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"757309\"]\r\n\r\nHere is the table we created for reference:\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Runner<\/th>\r\nDistance =<\/th>\r\n(Rate )<\/th>\r\n(Time)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Trason<\/td>\r\n50 miles<\/td>\r\n8.33 [latex]\\frac{\\text{ miles }}{\\text{ hour }}[\/latex]<\/td>\r\n6 hours<\/td>\r\n<\/tr>\r\n
      Johnson<\/td>\r\n50 miles<\/td>\r\n9.1\u00a0[latex]\\frac{\\text{ miles }}{\\text{ hour }}[\/latex]<\/td>\r\n5.5 hours<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRead and Understand:<\/strong>\u00a0We are looking for time, and this time our distance has changed from 50 miles to 1 mile, so we can use [latex]d=rt[\/latex].\r\n\r\nIf we divide both sides of the equation with r, we will have the equation for the time:\r\n

      [latex]\\frac{d}{r}=t[\/latex]<\/p>\r\nDefine and Translate:<\/strong>\u00a0We can use the formula\u00a0[latex]d=rt[\/latex] to make our table. This time the unknown is t<\/em>, the distance is 1 mile, and we know each runner's rate. It may help to create a new table:\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Runner<\/th>\r\nDistance =<\/th>\r\n(Rate )<\/th>\r\n(Time)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Trason<\/td>\r\n1 mile<\/td>\r\n8.33 [latex]\\frac{\\text{ miles }}{\\text{ hour }}[\/latex]<\/td>\r\nt<\/em> hours<\/td>\r\n<\/tr>\r\n
      Johnson<\/td>\r\n1 mile<\/td>\r\n9.1\u00a0[latex]\\frac{\\text{ miles }}{\\text{ hour }}[\/latex]<\/td>\r\nt<\/em> hours<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWrite and Solve:<\/strong>\r\n\r\nTrason:<\/span>\r\n\r\nStart with the equation from above where we isolated the t.\r\n

      [latex]\\frac{d}{r}=t[\/latex]<\/p>\r\nNow substitute in the values for d=1, and r=8.33 from the table above.\r\n

      [latex]\\frac{1}{8.33}=t[\/latex] hours<\/p>\r\n

      [latex]0.12\\text{ hours }=t[\/latex].<\/p>\r\n0.12 hours is about 7.2 minutes, so Trason's time for running one mile was about 7.2 minutes. WOW! She did that for 6 hours!\r\n\r\nJohnson:<\/span>\r\n\r\nAgain, start with:\r\n

      [latex]\\frac{d}{r}=t[\/latex]<\/p>\r\nNow substitute in the values for d=1 and r=9.1 from the table above.\r\n

      [latex]\\begin{array}{c}\\frac{1}{9.1}=t\\text{ hours }\\\\\\\\0.11\\text{ hours }=t\\end{array}[\/latex].<\/p>\r\n0.11 hours is about 6.6 minutes, so Johnson's time for running one mile was about 6.6 minutes. WOW! He did that for 5.5 hours!\r\n\r\nCheck and Interpret:<\/strong>\r\n\r\nHave we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole 50-mile course. \u00a0Yes, we answered our question.\r\n\r\nTrason's mile time was [latex]7.2\\frac{\\text{minutes}}{\\text{mile}}[\/latex] and Johnsons' mile time was\u00a0[latex]6.6\\frac{\\text{minutes}}{\\text{mile}}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of answering many rate questions given distance and time.\r\n\r\nhttps:\/\/youtu.be\/3WLp5mY1FhU\r\n

      Area<\/h2>\r\nThe area of a triangle is given by [latex] A=\\frac{1}{2}bh[\/latex] where\r\n\r\n[latex]A[\/latex]\u00a0= area\r\n[latex]b[\/latex] = the length of the base\r\n[latex]h[\/latex] = the height of the triangle\r\n\r\nRemember that when two variables or a number and a variable are sitting next to each other without a mathematical operator between them, you can assume they are being multiplied. This can seem frustrating, but you can think of it like mathematical slang. Over the years, people who use math frequently have just made that shortcut enough that it has been adopted as convention.\r\n\r\nIn the next example we will use the formula for area of a triangle to find a missing dimension, as well as use substitution to solve for the base of a triangle given the area and height.\r\n
      \r\n

      Example 4<\/h3>\r\nFind the base,\u00a0[latex]b[\/latex],\u00a0of a triangle with an area,\u00a0[latex]A[\/latex],\u00a0of 20 square feet and a height,\u00a0[latex]h[\/latex],\u00a0of 8 feet.\r\n[reveal-answer q=\"698967\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"698967\"]\r\n\r\nUse the formula for the area of a triangle, [latex] {A}=\\frac{{1}}{{2}}{bh}[\/latex].<\/i>\r\n\r\nSubstitute the given lengths into the formula and solve for [latex]b[\/latex].<\/i>\r\n

      [latex]\\displaystyle \\begin{array}{l}\\,\\,A=\\frac{1}{2}bh\\\\\\\\20=\\frac{1}{2}b\\cdot 8\\\\\\\\20=\\frac{8}{2}b\\\\\\\\20=4b\\\\\\\\\\frac{20}{4}=\\frac{4b}{4}\\\\\\\\ \\,\\,\\,5=b\\end{array}[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\nThe base of the triangle measures 5 feet.[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can rewrite the formula in terms of b<\/em> or h<\/em> as we did with perimeter previously. This probably seems abstract, but it can help you develop your equation-solving skills, as well as help you get more comfortable with working with all kinds of variables, not just x<\/em>.\r\n
      \r\n

      Example 5<\/h3>\r\nUse the multiplication and division properties of equality to isolate the variable [latex]b[\/latex]\u00a0<\/em>in the Area formula [latex] A=\\frac{1}{2}bh[\/latex].\r\n\r\n[reveal-answer q=\"291790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"291790\"]\r\n

      [latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{h}=\\frac{bh}{h}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{h}=\\frac{b\\cancel{h}}{\\cancel{h}}\\end{array}[\/latex]<\/p>\r\nWrite the equation with the desired variable on the left-hand side as a matter of convention:\r\n

      [latex]b=\\frac{2A}{h}[\/latex]\r\n[\/hidden-answer]<\/p>\r\n\r\n\r\n

      \r\n\r\nUse the multiplication and division properties of equality to isolate the variable [latex]h[\/latex]\u00a0<\/em>in the Area formula [latex] A=\\frac{1}{2}bh[\/latex].\r\n[reveal-answer q=\"595790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"595790\"]\r\n

      [latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,A=\\frac{1}{2}bh\\\\\\\\\\left(2\\right)A=\\left(2\\right)\\frac{1}{2}bh\\\\\\\\\\,\\,\\,\\,\\,\\,2A=bh\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{b}=\\frac{bh}{b}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\frac{2A}{b}=\\frac{h\\cancel{b}}{\\cancel{b}}\\end{array}[\/latex]<\/p>\r\nWrite the equation with the desired variable on the left-hand side as a matter of convention:\r\n

      [latex]h=\\frac{2A}{b}[\/latex]\r\n[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nThe following video shows another example of finding the base of a triangle given area and height.\r\nhttps:\/\/youtu.be\/VQZQvJ3rXYg\r\n

      Temperature<\/span><\/h2>\r\nLet\u2019s look at another formula that includes parentheses and fractions, the formula for converting from the Fahrenheit temperature scale to the Celsius scale.\r\n

      [latex]C=\\left(F-32\\right)\\cdot \\frac{5}{9}[\/latex]<\/p>\r\nThis formula is telling us that to find the equivalent Celsius temperature, we will start with the Fahrenheit temperature, subtract 32 from it, and then multiply the result by [latex]\\frac{5}{9}[\/latex]. But what if we are starting with a Celsius temperature and want to convert to Fahrenheit? The example below will show how you can use the same formula to convert from Celsius to Fahrenheit.\r\n

      \r\n

      Example 6<\/h3>\r\nGiven a temperature of [latex]12^{\\circ}{C}[\/latex], find the equivalent in [latex]{}^{\\circ}{F}[\/latex].\r\n[reveal-answer q=\"594254\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"594254\"]\r\n\r\nSubstitute the given temperature in[latex]{}^{\\circ}{C}[\/latex] into the conversion formula:\r\n

      [latex]12=\\left(F-32\\right)\\cdot \\frac{5}{9}[\/latex]<\/p>\r\nIsolate the variable\u00a0[latex]F[\/latex] to obtain the equivalent temperature.\r\n

      [latex]\\begin{array}{r}12=\\left(F-32\\right)\\cdot \\frac{5}{9}\\\\\\\\\\left(\\frac{9}{5}\\right)12=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\\\left(\\frac{108}{5}\\right)=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\21.6=F-32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{+32\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,+32}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\\\53.6={}^{\\circ}{F}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video, we show another example of converting from celsius to fahrenheit.\r\nhttps:\/\/youtu.be\/DRydX8V-JwY\r\n\r\nAs with the other formulas we have worked with, in the examples above, we could have isolated the variable F first, then substituted in the given temperature in Celsius. The example below will show you how to solve the formula for F.\r\n

      \r\n

      Example 7<\/h3>\r\nThe formula below is used for converting from the Fahrenheit scale to the Celsius scale. Solve the formula for [latex]F[\/latex]. (This will change it into a formula that can be used to convert Celsius to Fahrenheit).\r\n\r\n[latex]C=\\left(F-32\\right)\\cdot \\frac{5}{9}[\/latex]\r\n\r\n[reveal-answer q=\"591790\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"591790\"]\r\n\r\nTo isolate the variable [latex]F[\/latex], it would be best to clear the fraction involving\u00a0[latex]F[\/latex] first. Multiply both sides of the equation by [latex] \\displaystyle \\frac{9}{5}[\/latex].\r\n

      [latex]\\begin{array}{l}\\\\\\,\\,\\,\\,\\left(\\frac{9}{5}\\right)C=\\left(F-32\\right)\\left(\\frac{5}{9}\\right)\\left(\\frac{9}{5}\\right)\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{9}{5}C=F-32\\end{array}[\/latex]<\/p>\r\nAdd 32 to both sides.\r\n

      [latex]\\begin{array}{l}\\frac{9}{5}\\,C+32=F-32+32\\\\\\\\\\frac{9}{5}\\,C+32=F\\end{array}[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]F=\\frac{9}{5}C+32[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n
      \r\n

      Think About It<\/h3>\r\nThe surface area of a cylinder is given by the following formula:\u00a0 [latex]s=2\\pi rh+2\\pi r^{2}[\/latex]. Use that formula to express height, [latex]h[\/latex], in terms of surface area and the radius.\r\n\r\nIn this example, the variable [latex]h[\/latex] is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate [latex]h[\/latex].\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"194805\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"194805\"]\r\nIsolate the term containing the variable, h<\/em>, by subtracting [latex]2\\pi r^{2}[\/latex]from both sides.\r\n

      [latex]\\begin{array}{r}S\\,\\,=2\\pi rh+2\\pi r^{2} \\\\ \\underline{-2\\pi r^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2\\pi r^{2}}\\\\S-2\\pi r^{2}\\,\\,\\,\\,=\\,\\,\\,\\,2\\pi rh\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nNext, isolate the variable\u00a0[latex]h[\/latex] by dividing both sides of the equation by [latex]2\\pi r[\/latex].\r\n

      [latex]\\begin{array}{r}\\frac{S-2\\pi r^{2}}{2\\pi r}=\\frac{2\\pi rh}{2\\pi r} \\\\\\\\ \\frac{S-2\\pi r^{2}}{2\\pi r}=h\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nYou can rewrite the equation so the isolated variable is on the left side.\r\n

      [latex]h=\\frac{S-2\\pi r^{2}}{2\\pi r}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>","rendered":"

      \n

      section 1.10 Learning Objectives<\/h3>\n

      1.10: Formulas<\/strong><\/p>\n

        \n
      • Evaluate a formula for given values<\/li>\n
      • Solve for a specified variable in a formula<\/li>\n<\/ul>\n<\/div>\n

         <\/p>\n

        Consider the following problem in the video:<\/p>\n