#set document(title: "2.1 Nonlinear Models", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 2.1#h(0.6em)Nonlinear Models In Chapter 1, we considered models described by linear functions. In this chapter, we begin our study of nonlinear models. === Solving Nonlinear Equations When studying nonlinear models, we will need to solve nonlinear equations. For example, in Perimeter and Area we used a graph to solve the quadratic equation #math.equation(block: true, alt: "18 x minus x squared equals 80")[$18 x − x^(2) = 80$] Here is another example. The figure shows a table and a graph for the function #math.equation(block: false, alt: "y equals 2 x squared minus 5")[$" " y = 2 x^(2) − 5$]. #figure(figph[parabola], alt: "parabola", caption: none) #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$" " x " "$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]]), [#math.equation(block: false, alt: "y")[$" " y " "$]], [#math.equation(block: false, alt: "13")[$13$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "13")[$13$]], )) You can see that there are two points on the graph for each #math.equation(block: false, alt: "y")[$y$]-value greater than #math.equation(block: false, alt: "minus 5")[$− 5$]. For example, the two points with #math.equation(block: false, alt: "y")[$y$]-coordinate #math.equation(block: false, alt: "7")[$7$] are shown. To solve the equation #math.equation(block: true, alt: "2 x squared minus 5 equals 7")[$2 x^(2) − 5 = 7$] we need only find the #math.equation(block: false, alt: "x")[$x$]-coordinates of these points. From the graph, the solutions appear to be about #math.equation(block: false, alt: "2.5")[$2.5$] and #math.equation(block: false, alt: "minus 2.5")[$− 2.5$]. How can we solve this equation algebraically? The opposite operation for squaring a number is taking a square root. So we can undo the operation of squaring by extracting square roots. We first solve for #math.equation(block: false, alt: "x squared")[$x^(2)$] to get #math.equation(block: true, alt: "2 x squared, equals 12; x squared, equals 6")[$2 x^(2) & = 12 \ x^(2) & = 6$] and then take square roots to find #math.equation(block: true, alt: "x equals plus or minus the square root of 6")[$x = ± sqrt(6)$] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Don't forget that every positive number has #emph[two] square roots. The symbol #math.equation(block: false, alt: "plus or minus")[$±$] (read “plus or minus”) is a shorthand notation used to indicate both square roots of #math.equation(block: false, alt: "6")[$6$]. ] The exact solutions are thus #math.equation(block: false, alt: "the square root of 6")[$sqrt(6)$] and #math.equation(block: false, alt: "minus the square root of 6")[$− sqrt(6)$]. We can also find decimal #emph[approximations] for the solutions using a calculator. Rounded to two decimal places, the approximate solutions are #math.equation(block: false, alt: "2.45")[$2.45$] and #math.equation(block: false, alt: "minus 2.45")[$− 2.45$]. Which statement is true? \_\_\_\_\_ #solutionbox[ A quadratic equation may include a linear or a constant term. ] Which statement is true? + A quadratic equation may include a linear or a constant term. + The solutions of a quadratic equation are always of the form #math.equation(block: false, alt: "plus or minus k")[$± k$] . + Your calculator gives exact decimal values for square roots of integers. + The coefficients of a quadratic equation are called parabolas. In general, we can solve equations of the form #math.equation(block: false, alt: "a x squared plus c equals 0")[$" " a x^(2) + c = 0 " "$] by isolating #math.equation(block: false, alt: "x squared")[$x^(2)$] on one side of the equation and then taking the square root of each side. This method for solving equations is called #strong[extraction of roots]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Extraction of Roots] To solve the equation #math.equation(block: true, alt: "a x squared plus c equals 0")[$a x^(2) + c = 0$] + Isolate #math.equation(block: false, alt: "x squared")[$x^(2)$]. + Take square roots of both sides. There are two solutions. ] #examplebox("Example 1")[][ Jax the cat falls off a tree branch 20 feet above the ground, and his height #math.equation(block: false, alt: "t")[$t$] seconds later is given by #math.equation(block: false, alt: "h equals 20 minus 16 t squared")[$" " h = 20 − 16 t^(2)$]. + What is Jax's height #math.equation(block: false, alt: "0.5")[$0.5$] second later? + How long does Jax have to get in position to land on its feet before he reaches the ground? #solutionbox[ + In this question, we are given a value of #math.equation(block: false, alt: "t")[$t$] and asked to find the corresponding value of #math.equation(block: false, alt: "h")[$h$]. To do this, we evaluate the formula for #math.equation(block: false, alt: "t equals 0.5")[$t = 0.5$]. We substitute #math.equation(block: false, alt: "0.5")[$0.5$] for #math.equation(block: false, alt: "t")[$t$] into the formula and simplify. #math.equation(block: true, alt: "h, equals 20 minus 16 open parenthesis 0.5 close parenthesis squared, Compute the power.; equals 20 minus 16 open parenthesis 0.25 close parenthesis, Multiply; then subtract.; equals 20 minus 4 equals 16")[$h & = 20 − 16 ( 0.5 )^(2) & & "Compute the power." \ & = 20 − 16 ( 0.25 ) & & "Multiply; then subtract." \ & = 20 − 4 = 16$] Jax is #math.equation(block: false, alt: "16")[$16$] feet above the ground after #math.equation(block: false, alt: "0.5")[$0.5$] second. + We would like to find the value of #math.equation(block: false, alt: "t")[$t$] when the height, #math.equation(block: false, alt: "h")[$h$], is known. We substitute #math.equation(block: false, alt: "h equals 0")[$h = 0$] into the equation to obtain #math.equation(block: true, alt: "0 equals 20 minus 16 t squared")[$0 = 20 − 16 t^(2)$] To solve this equation, we use extraction of roots. First we isolate #math.equation(block: false, alt: "t squared")[$t^(2)$] on one side of the equation. #math.equation(block: true, alt: "16 t squared, equals 20, Divide by 16.; t squared, equals the fraction 20 over 16 equals 1.25")[$16 t^(2) & = 20 & & "Divide by 16." \ t^(2) & = frac(20, 16) = 1.25$] Then we take the square root of both sides of the equation to find #figure(figph[half of parabola opening downward], alt: "half of parabola opening downward", caption: none) #math.equation(block: true, alt: "t equals plus or minus the square root of 1.25 approximately equals plus or minus 1.118")[$t = ± sqrt(1.25) ≈ ± 1.118$] Only the positive solution makes sense here, so Jax has approximately 1.12 seconds to get into position for landing. A graph of Jax's height after #math.equation(block: false, alt: "t")[$t$] seconds is shown at left. The points corresponding to parts (a) and (b) are labeled. ] ] Which solutions are approximations? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "x equals plus or minus 6.32455532")[$x = ± 6.32455532$] ] Which solutions are approximations? + #math.equation(block: false, alt: "x squared equals 40 ; plus or minus 6.32455532")[$x^(2) = 40 ; " " " " " " " " " " " " " " " " " " " " " " " " ± 6.32455532$] + #math.equation(block: false, alt: "t squared equals the fraction 81 over 64 ; plus or minus 1.125")[$t^(2) = display(frac(81, 64)) ; " " " " " " " " " " " " " " " " " " " " " " " " ± 1.125$] + #math.equation(block: false, alt: "w squared equals 50 ; plus or minus 5 the square root of 2")[$w^(2) = 50 ; " " " " " " " " " " " " " " " " " " " " " " " " ± 5 sqrt(2)$] + #math.equation(block: false, alt: "b squared equals open parenthesis 0.632 close parenthesis squared ; plus or minus 0.632")[$b^(2) = ( 0.632 )^(2) ; " " " " " " " " " " " " " " " " " " " " " " " " ± 0.632$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In part (a) of Example we #strong[evaluated] the expression #math.equation(block: false, alt: "20 minus 16 t squared")[$" " 20 − 16 t^(2) " "$] to find a value for #math.equation(block: false, alt: "h")[$h$], and in part (b) we #strong[solved] the equation #math.equation(block: false, alt: "0 equals 20 minus 16 t squared")[$" " 0 = 20 − 16 t^(2) " "$] to find a value for #math.equation(block: false, alt: "t")[$t$]. ] + Solve by extracting roots #math.equation(block: false, alt: "the fraction 3 x squared minus 8 over 5 equals 10")[$display(frac(3 x^(2) − 8, 5)) = 10$]. #linebreak() #math.equation(block: false, alt: "First, isolate x squared . x squared equals")[$"First, isolate " bold(italic(x))^(2) . #hide($0000000000000000$) x^(2) =$]\_\_\_\_\_ #linebreak() #math.equation(block: false, alt: "Take the square root of both sides. x equals plus or minus")[$"Take the square root of both sides." #hide($000$) x = ±$]\_\_\_\_\_ #linebreak() Note: Use "sqrt(2)" to get #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$], etc. + Give exact answers above; then give approximations rounded to two decimal places. #linebreak() #math.equation(block: false, alt: "x approximately equals")[$x ≈$]\_\_\_\_\_ #linebreak() Note: Enter a comma between solutions. #solutionbox[ #math.equation(block: true, alt: "x equals plus or minus the square root of the fraction 58 over 3 approximately equals plus or minus 4.40")[$x = ± sqrt(display(frac(58, 3))) ≈ ± 4.40$] ] + Solve by extracting roots #math.equation(block: false, alt: "the fraction 3 x squared minus 8 over 5 equals 10")[$" " " " " " " " display(frac(3 x^(2) − 8, 5)) = 10$]. + Give exact answers; then give approximations rounded to two decimal places. #solutionbox[ + We isolate #math.equation(block: false, alt: "x")[$x$] as follows: multiply both sides by 5, add 8 to both sides, then divide both sides by 3. This yields #math.equation(block: true, alt: "x squared equals the fraction 58 over 3")[$x^(2) = display(frac(58, 3))$] Finally we take square roots. + #math.equation(block: false, alt: "x equals plus or minus the square root of the fraction 58 over 3 , or approximately equals plus or minus 4.40")[$x = ± sqrt(display(frac(58, 3))) , "or" ≈ ± 4.40$] ] Use graphs to explain why a linear equation can have only one solution, but a quadratic equation may have two solutions. \_\_\_\_\_ Use graphs to explain why a linear equation can have only one solution, but a quadratic equation may have two solutions. === Solving Formulas We can use extraction of roots to solve many formulas involving the square of the variable. #examplebox("Example 2")[][ The formula #math.equation(block: true, alt: "V equals the fraction 1 over 3 π r squared h")[$V = display(frac(1, 3)) π r^(2) h$] gives the volume of a cone in terms of its height and radius. Solve the formula for #math.equation(block: false, alt: "r")[$r$] in terms of #math.equation(block: false, alt: "V")[$V$] and #math.equation(block: false, alt: "h")[$h$]. #solutionbox[ Because the variable we want is squared, we use extraction of roots. First, we multiply both sides by #math.equation(block: false, alt: "3")[$3$] to clear the fraction. #math.equation(block: true, alt: "3 V, equals 3 open parenthesis the fraction 1 over 3 π r squared h close parenthesis; 3 V, equals π r squared h, Divide both sides by π h .; the fraction 3 V over π h, equals r squared, Take square roots.; plus or minus the square root of the fraction 3 V over π h, equals r")[$3 V & = 3 ( frac(1, 3) π r^(2) h ) \ 3 V & = π r^(2) h & & "Divide both sides by " bold(italic(π)) bold(italic(h)) . \ frac(3 V, π h) & = r^(2) & & "Take square roots." \ ± sqrt(display(frac(3 V, π h))) & = r$] Because the radius of a cone must be a positive number, we use only the positive square root: #math.equation(block: false, alt: "r equals the square root of the fraction 3 V over π h")[$" " " " r = sqrt(display(frac(3 V, π h)))$]. ] ] Find a formula for the radius of a circle in terms of its area. #math.equation(block: false, alt: "r equals")[$r =$]\_\_\_\_\_ Note: use "sqrt(2)" to get #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$], and use "pi" to get #math.equation(block: false, alt: "π")[$π$]. Start with the formula for the area of a circle: #math.equation(block: false, alt: "A equals")[$A =$] Solve for #math.equation(block: false, alt: "r")[$r$] in terms of #math.equation(block: false, alt: "A")[$A$]. #solutionbox[ #math.equation(block: true, alt: "r equals the square root of A / π")[$r = sqrt(A / π)$] ] Find a formula for the radius of a circle in terms of its area. Hint: Start with the formula for the area of a circle, then solve for #math.equation(block: false, alt: "r")[$r$] in terms of #math.equation(block: false, alt: "A")[$A$]. #solutionbox[ #math.equation(block: true, alt: "r equals the square root of A / π")[$r = sqrt(A / π)$] ] === More Extraction of Roots Equations of the form #math.equation(block: true, alt: "a open parenthesis p x plus q close parenthesis squared plus r equals 0")[$bold(italic(a)) ( bold(italic(p)) bold(italic(x)) + bold(italic(q)) )^(2) + bold(italic(r)) = 0$] can also be solved by extraction of roots after isolating the squared expression, #math.equation(block: false, alt: "open parenthesis p x plus q close parenthesis squared")[$" " ( p x + q )^(2)$]. #examplebox("Example 3")[][ Solve the equation #math.equation(block: false, alt: "3 open parenthesis x minus 2 close parenthesis squared equals 48")[$" " 3 ( x − 2 )^(2) = 48$] #solutionbox[ First, we isolate the perfect square, #math.equation(block: false, alt: "open parenthesis x minus 2 close parenthesis squared")[$" " ( x − 2 )^(2)$]. #math.equation(block: true, alt: "3 open parenthesis x minus 2 close parenthesis squared, equals 48, Divide both sides by 3.; open parenthesis x minus 2 close parenthesis squared, equals 16, Take the square root of each side.; x minus 2, equals plus or minus the square root of 16 equals plus or minus 4")[$3 ( x − 2 )^(2) & = 48 & & "Divide both sides by " 3. \ ( x − 2 )^(2) & = 16 & & "Take the square root of each side." \ x − 2 & = ± sqrt(16) = ± 4$] This gives us two equations for #math.equation(block: false, alt: "x")[$x$], #math.equation(block: true, alt: "x minus 2, equals 4, or, x minus 2, equals minus 4, Solve each equation.; x, equals 6, or, x, equals minus 2")[$x − 2 & = 4 & & "or " & x − 2 & = − 4 & & "Solve each equation." \ x & = 6 & & "or " & x & = − 2$] The solutions are #math.equation(block: false, alt: "6")[$6$] and #math.equation(block: false, alt: "minus 2")[$− 2$]. ] ] Here is a general strategy for solving equations by extraction of roots. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Extraction of Roots] To solve the equation #math.equation(block: true, alt: "a open parenthesis p x plus q close parenthesis squared plus r equals 0")[$a ( p x + q )^(2) + r = 0$] + Isolate the squared expression, #math.equation(block: false, alt: "open parenthesis p x plus q close parenthesis squared")[$( p x + q )^(2)$]. + Take the square root of each side of the equation. Remember that a positive number has two square roots. + Solve each equation. There are two solutions. ] Solve #math.equation(block: false, alt: "2 open parenthesis 5 x plus 3 close parenthesis squared equals 38")[$2 ( 5 x + 3 )^(2) = 38$] by extracting roots. + Give your answers as exact values, separating the solutions with a comma. #linebreak() \_\_\_\_\_ #linebreak() Note: Use "sqrt(2)" to get #math.equation(block: false, alt: "the square root of 2")[$sqrt(2)$], and take care to use parentheses appropriately. #linebreak() Use the "Preview My Answers" button to see if you have entered valid syntax. + Find approximations for the solutions to two decimal places, separating the solutions with a comma. #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "x equals the fraction minus 3 plus or minus the square root of 19 over 5")[$x = display(frac(− 3 ± sqrt(19), 5))$] + #math.equation(block: false, alt: "x approximately equals minus 1.47")[$x ≈ − 1.47$] or #math.equation(block: false, alt: "x approximately equals 0.27")[$x ≈ 0.27$] ] Solve #math.equation(block: false, alt: "2 open parenthesis 5 x plus 3 close parenthesis squared equals 38")[$" " 2 ( 5 x + 3 )^(2) = 38 " "$] by extracting roots. + Give your answers as exact values. + Find approximations for the solutions to two decimal places. #solutionbox[ First we isolate the squared expression to get #math.equation(block: true, alt: "open parenthesis 5 x plus 3 close parenthesis squared, equals 19, Take the square root of each side.; 5 x plus 3, equals plus or minus the square root of 19")[$( 5 x + 3 )^(2) & = 19 & & "Take the square root of each side." \ 5 x + 3 & = ± sqrt(19)$] Finally, we subtract 3 and divide by 5 to solve for #math.equation(block: false, alt: "x")[$x$]. + #math.equation(block: false, alt: "x equals the fraction minus 3 plus or minus the square root of 19 over 5")[$x = display(frac(− 3 ± sqrt(19), 5))$] + #math.equation(block: false, alt: "x approximately equals minus 1.47")[$x ≈ − 1.47$] or #math.equation(block: false, alt: "x approximately equals 0.27")[$x ≈ 0.27$] ] Which of the following equations cannot be solved by extraction of roots? \_\_\_\_\_ #solutionbox[ The equation #math.equation(block: false, alt: "2 x squared plus 3 x equals 8")[$2 x^(2) + 3 x = 8$] cannot be solved by extraction of roots. ] Which of the following equations cannot be solved by extraction of roots? + #math.equation(block: false, alt: "2 x squared plus 3 equals 8")[$2 x^(2) + 3 = 8$] + #math.equation(block: false, alt: "2 x squared plus 3 x equals 8")[$2 x^(2) + 3 x = 8$] + #math.equation(block: false, alt: "2 open parenthesis x plus 3 close parenthesis squared equals 8")[$2 ( x + 3 )^(2) = 8$] + #math.equation(block: false, alt: "2 x squared minus 3 equals 0")[$2 x^(2) − 3 = 0$] === Compound Interest and Inflation Many savings institutions offer accounts on which the interest is #emph[compounded annually]. At the end of each year, the interest earned is added to the principal, and the interest for the next year is computed on this larger sum of money. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Compound Interest] If interest is compounded annually for #math.equation(block: false, alt: "n")[$n$] years, the amount, #math.equation(block: false, alt: "A")[$A$], of money in an account is given by #math.equation(block: true, alt: "A equals P open parenthesis 1 plus r close parenthesis to the power n")[$A = P ( 1 + r )^(n)$] where #math.equation(block: false, alt: "P")[$P$] is the principal and #math.equation(block: false, alt: "r")[$r$] is the interest rate, expressed as a decimal fraction. ] #examplebox("Example 4")[][ Carmella invests \$3000 in an account that pays an interest rate, #math.equation(block: false, alt: "r")[$r$], compounded annually. + Write an expression for the amount of money in Carmella’s account after two years. + What interest rate would be necessary for Carmella's account to grow to \$3500 in two years? #solutionbox[ + We use the formula above with #math.equation(block: false, alt: "P equals 3000")[$P = 3000$] and #math.equation(block: false, alt: "n equals 2")[$n = 2$]. Carmella’s account balance will be #math.equation(block: true, alt: "A equals 3000 open parenthesis 1 plus r close parenthesis squared")[$A = 3000 ( 1 + r )^(2)$] + We substitute #math.equation(block: false, alt: "3500")[$3500$] for #math.equation(block: false, alt: "A")[$A$] in the equation. #math.equation(block: true, alt: "3500 equals 3000 open parenthesis 1 plus r close parenthesis squared")[$3500 = 3000 ( 1 + r )^(2)$] We can solve this equation in #math.equation(block: false, alt: "r")[$r$] by extraction of roots. First, we isolate the perfect square. #math.equation(block: true, alt: "3500, equals 3000 open parenthesis 1 plus r close parenthesis squared, Divide both sides by 3000.; 1.16 bar, equals open parenthesis 1 plus r close parenthesis squared, Take the square root of both sides.")[$3500 & = 3000 ( 1 + r )^(2) & & "Divide both sides by 3000." \ 1.1 limits(6)^(―) & = ( 1 + r )^(2) & & "Take the square root of both sides."$] This give us two equations to solve. #math.equation(block: true, alt: "plus or minus 1.0801, approximately equals 1 plus r, Subtract 1 from both sides.; r approximately equals 0.0801 or r, approximately equals minus 2.0801")[$± 1.0801 & ≈ 1 + r & & "Subtract 1 from both sides." \ r ≈ 0.0801 " or " r & ≈ − 2.0801$] Because the interest rate must be a positive number, we discard the negative solution. Carmella needs an account with interest rate #math.equation(block: false, alt: "r approximately equals 0.0801")[$r ≈ 0.0801$], or just over 8%, to achieve an account balance of \$3500 in two years. ] ] What is the first step in solving the equation #math.equation(block: false, alt: "2 open parenthesis x plus 1 close parenthesis squared equals 17")[$2 ( x + 1 )^(2) = 17$]? \_\_\_\_\_ #solutionbox[ To solve #math.equation(block: false, alt: "2 open parenthesis x plus 1 close parenthesis squared equals 17")[$2 ( x + 1 )^(2) = 17$] by extraction of roots, we first divide both sides by 2. ] What is the first step in solving the equation #math.equation(block: false, alt: "2 open parenthesis x plus 1 close parenthesis squared equals 17")[$" " 2 ( x + 1 )^(2) = 17 " "$]? + Expand #math.equation(block: false, alt: "open parenthesis x plus 1 close parenthesis squared")[$( x + 1 )^(2)$]. + Get zero on one side. + Divide both sides by 2. + Take the square root of both sides. The formula for compound interest also applies to the effects of inflation. For instance, if there is a steady inflation rate of 4% per year, in two years an item that now costs \$100 will cost #math.equation(block: true, alt: "A, equals P open parenthesis 1 plus r close parenthesis squared; equals 100 open parenthesis 1 plus 0.04 close parenthesis squared equals $ 108.16")[$A & = P ( 1 + r )^(2) \ & = 100 ( 1 + 0.04 )^(2) = upright(\$) 108.16$] Two years ago, the average cost of dinner and a movie was \$42. This year the average cost is \$44.99. What was the annual rate of inflation over the past two years? #math.equation(block: false, alt: "r approximately equals")[$r ≈$]\_\_\_\_\_% #solutionbox[ #math.equation(block: false, alt: "r equals the square root of the fraction 44.99 over 42 minus 1")[$r = sqrt(display(frac(44.99, 42))) − 1$], or as a percent, the inflation rate was approximately #math.equation(block: false, alt: "3.5 %")[$3.5 upright(%)$]. ] Two years ago, the average cost of dinner and a movie was \$42. This year the average cost is \$44.99. What was the annual rate of inflation over the past two years? #solutionbox[ We solve the equation #math.equation(block: true, alt: "44.99 equals 42 open parenthesis 1 plus r close parenthesis squared")[$44.99 = 42 ( 1 + r )^(2)$] to find #math.equation(block: true, alt: "r equals the square root of the fraction 44.99 over 42 minus 1 approximately equals 0.035")[$r = sqrt(display(frac(44.99, 42))) − 1 ≈ 0.035$] The annual inflation rate was approximately #math.equation(block: false, alt: "3.5 %")[$3.5 upright(%)$]. ] === Other Nonlinear Equations Because squaring and taking square roots are opposite operations, we can solve the equation #math.equation(block: true, alt: "the square root of x equals 8.2")[$sqrt(x) = 8.2$] by squaring both sides to get #math.equation(block: true, alt: "open parenthesis the square root of x close parenthesis squared, equals 8.2 squared; x, equals 67.24")[$( sqrt(x) )^(2) & = 8.2^(2) \ x & = 67.24$] Similarly, we can solve #math.equation(block: true, alt: "x cubed equals 258")[$x^(3) = 258$] by taking the cube root of both sides, because cubing and taking cube roots are opposite operations. Rounding to three places, we find #math.equation(block: true, alt: "the cube root of x cubed, equals 258; x, approximately equals 6.366")[$root(3, x^(3)) & = 258 \ x & ≈ 6.366$] Delbert squared #math.equation(block: false, alt: "minus 6")[$− 6$] by entering #math.equation(block: false, alt: "minus 6 squared")[$− 6^(2)$] into his calculator, and got #math.equation(block: false, alt: "minus 36")[$− 36$]. What went wrong? \_\_\_\_\_ #solutionbox[ To square #math.equation(block: false, alt: "minus 6")[$− 6$], Delbert should have put parentheses around #math.equation(block: false, alt: "minus 6")[$− 6$]. ] Delbert squared #math.equation(block: false, alt: "minus 6")[$− 6$] by entering #math.equation(block: false, alt: "minus 6 squared")[$− 6^(2)$] into his calculator, and got #math.equation(block: false, alt: "minus 36")[$− 36$]. What went wrong? + Nothing; that is the right answer. + He should have put parentheses around #math.equation(block: false, alt: "6 squared")[$6^(2)$]. + He should have put parentheses around #math.equation(block: false, alt: "minus 6")[$− 6$]. + You cannot square a negative number on a calculator. The notion of undoing operations can help us solve a variety of simple nonlinear equations. The operation of taking a reciprocal is its own opposite, so we solve the equation #math.equation(block: true, alt: "the fraction 1 over x equals 50")[$frac(1, x) = 50$] by taking the reciprocal of both sides to get #math.equation(block: true, alt: "x equals the fraction 1 over 50 equals 0.02")[$x = frac(1, 50) = 0.02$] #examplebox("Example 5")[][ Solve #math.equation(block: false, alt: "the fraction 3 over x minus 2 equals 4")[$" " " " display(frac(3, x − 2)) = 4$] #solutionbox[ We begin by taking the reciprocal of both sides of the equation to get #math.equation(block: true, alt: "the fraction x minus 2 over 3 equals the fraction 1 over 4")[$frac(x − 2, 3) = frac(1, 4)$] We continue to undo the operations in reverse order. First, we multiply both sides by #math.equation(block: false, alt: "3")[$3$]. #math.equation(block: true, alt: "x minus 2, equals the fraction 3 over 4, Add 2 to both sides.; x, equals 2 plus the fraction 3 over 4 equals the fraction 11 over 4, the fraction 2 over 1 plus the fraction 3 over 4 equals the fraction 8 over 4 plus the fraction 3 over 4 equals the fraction 11 over 4")[$x − 2 & = frac(3, 4) & & "Add 2 to both sides." \ x & = 2 + frac(3, 4) = frac(11, 4) & & frac(2, 1) + frac(3, 4) = frac(8, 4) + frac(3, 4) = frac(11, 4)$] The solution is #math.equation(block: false, alt: "the fraction 11 over 4")[$display(frac(11, 4))$], or #math.equation(block: false, alt: "2.75")[$2.75$]. ] ] What is the reciprocal of #math.equation(block: false, alt: "the fraction 1 over the square root of x")[$display(frac(1, sqrt(x)))$]? \_\_\_\_\_ #solutionbox[ The reciprocal of #math.equation(block: false, alt: "the fraction 1 over the square root of x")[$display(frac(1, sqrt(x)))$] is #math.equation(block: false, alt: "the square root of x")[$sqrt(x)$]. ] What is the reciprocal of #math.equation(block: false, alt: "the fraction 1 over the square root of x")[$display(frac(1, sqrt(x)))$]? + #math.equation(block: false, alt: "the fraction 1 over x")[$display(frac(1, x))$] + #math.equation(block: false, alt: "the fraction 1 over x squared")[$display(frac(1, x^(2)))$] + #math.equation(block: false, alt: "x squared")[$x^(2)$] + #math.equation(block: false, alt: "the square root of x")[$sqrt(x)$] Solve #math.equation(block: false, alt: "2 the square root of x plus 4 equals 6")[$2 sqrt(x + 4) = 6$] #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "x equals 5")[$x = 5$] ] Solve #math.equation(block: false, alt: "2 the square root of x plus 4 equals 6")[$" " 2 sqrt(x + 4) = 6$] #solutionbox[ We divide both sides by 2 to get #math.equation(block: true, alt: "the square root of x plus 4, equals 3, Square both sides.; x plus 4, equals 9; x equals 5")[$sqrt(x + 4) & = 3 & & "Square both sides." \ x + 4 & = 9 \ x = 5$] ] What is the difference between a rational number and an irrational number? Give examples of each. (See the section The Real Number System in Appendix A: Algebra Skills Refresher to review rational and irrational numbers.) \_\_\_\_\_ What is the difference between a rational number and an irrational number? Give examples of each. (See the section The Real Number System in Appendix A: Algebra Skills Refresher to review rational and irrational numbers.) #notebox("Technology", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Using the Intersect Feature] We can use the #emph[intersect] feature on a graphing calculator to solve equations. ] #examplebox("Example 6")[][ Use a graphing calculator to solve #math.equation(block: false, alt: "the fraction 3 over x minus 2 equals 4")[$" " " " display(frac(3, x − 2)) = 4$] #solutionbox[ We would like to find the points on the graph of #math.equation(block: false, alt: "y equals the fraction 3 over x minus 2")[$" " " " y = display(frac(3, x − 2)) " " " "$] that have #math.equation(block: false, alt: "y")[$y$]-coordinate equal to #math.equation(block: false, alt: "4")[$4$]. We graph the two functions #math.equation(block: true, alt: "Y sub 1, equals 3 / open parenthesis X minus 2 close parenthesis; Y sub 2, equals 4")[$Y_(1) & = 3 / ( X − 2 ) \ Y_(2) & = 4$] in the window #math.equation(block: true, alt: "Xmin, equals minus 9.4, Xmax equals 9.4; Ymin, equals minus 10, Ymax equals 10")[$"Xmin" & = − 9.4 & & "Xmax" = 9.4 \ "Ymin" & = − 10 & & "Ymax" = 10$] The point where the two graphs intersect locates the solution of the equation. If we trace along the graph of #math.equation(block: false, alt: "Y sub 1")[$Y_(1)$], the closest we can get to the intersection point is #math.equation(block: false, alt: "open parenthesis 2.8 , 3.75 close parenthesis")[$( 2.8 , 3.75 )$], as shown in figure (a). We get a better approximation using the #emph[intersect] feature. #figure(figph[graphing calculator displays], alt: "graphing calculator displays", caption: none) Use the arrow keys to position the Trace bug as close to the intersection point as you can. Then press 2nd TRACE to see the Calculate menu. Press #math.equation(block: false, alt: "5")[$5$] for intersect; then respond to each of the calculator's questions, #emph[First curve?], #emph[Second curve?], and #emph[Guess?] by pressing ENTER. The calculator will then display the intersection point, #math.equation(block: false, alt: "x equals 2.75")[$x = 2.75$], #math.equation(block: false, alt: "y equals 4")[$y = 4$], as shown in figure (b). The solution of the original equation is #math.equation(block: false, alt: "x equals 2.75")[$x = 2.75$]. ] ] Use the intersect feature to solve the equation #math.equation(block: false, alt: "2 x squared minus 5 equals 7")[$2 x^(2) − 5 = 7$]. Round your answers to three decimal places. Separate solutions with a comma. \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "x equals plus or minus the square root of 6 approximately equals 2.449")[$x = ± sqrt(6) ≈ 2.449$] ] Use the intersect feature to solve the equation #math.equation(block: false, alt: "2 x squared minus 5 equals 7")[$" " 2 x^(2) − 5 = 7$]. Round your answers to three decimal places. #solutionbox[ #math.equation(block: true, alt: "x equals plus or minus the square root of 6 approximately equals 2.449")[$x = ± sqrt(6) ≈ 2.449$] ] === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Quadratic - Compound interest - Exact solution - Perfect square - Extraction of roots - Inflation - Area - Cube root - Isolate - Height - Perimeter - Reciprocal ==== CONCEPTS + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Extraction of Roots] To solve the equation #math.equation(block: true, alt: "a open parenthesis p x plus q close parenthesis squared plus r equals 0")[$a ( p x + q )^(2) + r = 0$]+ Isolate the squared expression, #math.equation(block: false, alt: "open parenthesis p x plus q close parenthesis squared")[$( p x + q )^(2)$]. + Take the square root of each side of the equation. Remember that a positive number has two square roots. + Solve each equation. There are two solutions. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Compound Interest] If interest is compounded annually for #math.equation(block: false, alt: "n")[$n$] years, the amount, #math.equation(block: false, alt: "A")[$A$], of money in an account is given by #math.equation(block: true, alt: "A equals P open parenthesis 1 plus r close parenthesis to the power n")[$A = P ( 1 + r )^(n)$]where #math.equation(block: false, alt: "P")[$P$] is the principal and #math.equation(block: false, alt: "r")[$r$] is the interest rate, expressed as a decimal fraction. ] + We can give exact answers to a simple nonlinear equation, or we can give decimal approximations. + Simple nonlinear equations can be solved by undoing the operations on the variable. ==== STUDY QUESTIONS + How many square roots does a positive number have? + What is the first step in solving the equation #math.equation(block: false, alt: "a open parenthesis p x plus q close parenthesis squared equals r")[$a ( p x + q )^(2) = r$] by extraction of roots? + Give the exact solutions of the equation #math.equation(block: false, alt: "x squared equals 10")[$x^(2) = 10$], and then give decimal approximations rounded to hundredths. + State a formula for the amount in an account on which #math.equation(block: false, alt: "5 %")[$5 upright(%)$] interest is compounded annually. + Give an example of two rectangles with the same perimeter but different areas. + The perimeter of a rectangle is #math.equation(block: false, alt: "50")[$50$] meters. Write an expression for the length of the rectangle in terms of its width. + What is the opposite operation for taking a reciprocal? + What is the reciprocal of #math.equation(block: false, alt: "the fraction 1 over the square root of x")[$display(frac(1, sqrt(x)))$] ? ==== SKILLS Practice each skill in the Homework problems listed. + Solve equations by extraction of roots: \#1–12, 31–42 + Solve formulas: \#13–16, 63–68 + Use the Pythagorean theorem: \#19–24 + Solve equations graphically: \#25–30 + Solve simple nonlinear equations: \#43–54 + Solve problems: \#55–62 === Homework 2.1 For Problems 1–6, solve by extracting roots. Give exact values for your answers. #math.equation(block: true, alt: "9 x squared equals 25")[$9 x^(2) = 25$] #math.equation(block: true, alt: "plus or minus the fraction 5 over 3")[$± display(frac(5, 3))$] #math.equation(block: true, alt: "4 x squared equals 9")[$4 x^(2) = 9$] #math.equation(block: true, alt: "4 x squared minus 24 equals 0")[$4 x^(2) − 24 = 0$] #math.equation(block: true, alt: "plus or minus the square root of 6")[$± sqrt(6)$] #math.equation(block: true, alt: "3 x squared minus 9 equals 0")[$3 x^(2) − 9 = 0$] #math.equation(block: true, alt: "the fraction 2 x squared over 3 equals 4")[$display(frac(2 x^(2), 3)) = 4$] #math.equation(block: true, alt: "plus or minus the square root of 6")[$± sqrt(6)$] #math.equation(block: true, alt: "the fraction 3 x squared over 5 equals 6")[$display(frac(3 x^(2), 5)) = 6$] For Problems 7–12, solve by extracting roots. Round your answers to two decimal places. #math.equation(block: true, alt: "2 x squared equals 14")[$2 x^(2) = 14$] #math.equation(block: true, alt: "plus or minus 2.65")[$± 2.65$] #math.equation(block: true, alt: "3 x squared equals 15")[$3 x^(2) = 15$] #math.equation(block: true, alt: "1.5 x squared equals 0.7 x squared plus 26.2")[$1.5 x^(2) = 0.7 x^(2) + 26.2$] #math.equation(block: true, alt: "plus or minus 5.72")[$± 5.72$] #math.equation(block: true, alt: "0.4 x squared equals 2 x squared minus 8.6")[$0.4 x^(2) = 2 x^(2) − 8.6$] #math.equation(block: true, alt: "5 x squared minus 97 equals 3.2 x squared minus 38")[$5 x^(2) − 97 = 3.2 x^(2) − 38$] #math.equation(block: true, alt: "plus or minus 5.73")[$± 5.73$] #math.equation(block: true, alt: "17 minus the fraction x squared over 4 equals 43 minus x squared")[$17 − display(frac(x^(2), 4)) = 43 − x^(2)$] For Problems 13–16, solve the formulas for the specified variable. #math.equation(block: false, alt: "F equals the fraction m v squared over r")[$F = display(frac(m v^(2), r))$], for #math.equation(block: false, alt: "v")[$v$] #math.equation(block: true, alt: "plus or minus the square root of the fraction F r over m")[$± sqrt(display(frac(F r, m)))$] #math.equation(block: false, alt: "A equals the fraction the square root of 3 over 4 s squared")[$A = display(frac(sqrt(3), 4)) s^(2)$], for #math.equation(block: false, alt: "s")[$s$] #math.equation(block: false, alt: "s equals the fraction 1 over 2 g t squared")[$s = display(frac(1, 2)) g t^(2)$], for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: true, alt: "plus or minus the square root of the fraction 2 s over g")[$± sqrt(display(frac(2 s, g)))$] #math.equation(block: false, alt: "S equals 4 π r squared")[$S = 4 π r^(2)$], for #math.equation(block: false, alt: "r")[$r$] For Problems 17 and 18, refer to the geometric formulas in Geometry formulas. A conical coffee filter is #math.equation(block: false, alt: "8.4")[$8.4$] centimeters tall. + Write a formula for the filter's volume in terms of its widest radius (at the top of the filter). + Complete the table of values for the volume equation. If you double the radius of the filter, by what factor does the volume increase?#figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "V")[$V$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + If the volume of the filter is #math.equation(block: false, alt: "302.4")[$302.4$] cubic centimeters, what is its radius? + Use your calculator to graph the volume equation. Locate the point on the graph that corresponds to the filter in part (c). + #math.equation(block: false, alt: "V equals 2.8 π r squared approximately equals 8.8 r squared")[$V = 2.8 π r^(2) ≈ 8.8 r^(2)$] + #figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "V")[$V$]], [#math.equation(block: false, alt: "8.8")[$8.8$]], [#math.equation(block: false, alt: "35.2")[$35.2$]], [#math.equation(block: false, alt: "79.2")[$79.2$]], [#math.equation(block: false, alt: "140.7")[$140.7$]], [#math.equation(block: false, alt: "219.9")[$219.9$]], [#math.equation(block: false, alt: "316.7")[$316.7$]], [#math.equation(block: false, alt: "431.0")[$431.0$]], [#math.equation(block: false, alt: "563.0")[$563.0$]], )) #linebreak() The volume increases by a factor of #math.equation(block: false, alt: "4")[$4$]. + #math.equation(block: false, alt: "5.86")[$5.86$] cm + #figure(figph[GC graph], alt: "GC graph", caption: none) A large bottle of shampoo is #math.equation(block: false, alt: "20")[$20$] centimeters tall and cylindrical in shape. + Write a formula for the volume of the bottle in terms of its radius. + Complete the table of values for the volume equation. If you halve the radius of the bottle, by what factor does the volume decrease?#figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "V")[$V$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + What radius should the bottle have if it must hold #math.equation(block: false, alt: "240")[$240$] milliliters of shampoo? (One milliliter is equal to 1 cubic centimeter.) + Use your calculator to graph the volume equation. Locate the point on the graph that corresponds to the bottle in part (c). For Problems 19–24, + Make a sketch of the situation described, and label a right triangle. + Use the Pythagorean theorem to solve each problem. (See Algebra Skills Refresher Facts from Geometry to review the Pythagorean theorem.) The size of a TV screen is the length of its diagonal. If the width of a #math.equation(block: false, alt: "35")[$35$]-inch TV screen is #math.equation(block: false, alt: "28")[$28$] inches, what is its height? #figure(figph[tv screen], alt: "tv screen", caption: none) #math.equation(block: false, alt: "21")[$21$] in. How high on a building will a #math.equation(block: false, alt: "25")[$25$]-foot ladder reach if its foot is #math.equation(block: false, alt: "15")[$15$] feet away from the base of the wall? #figure(figph[ladder against building], alt: "ladder against building", caption: none) If a #math.equation(block: false, alt: "30")[$30$]-meter pine tree casts a shadow of #math.equation(block: false, alt: "30")[$30$] meters, how far is the tip of the shadow from the top of the tree? #figure(figph[tree and shadow], alt: "tree and shadow", caption: none) #math.equation(block: false, alt: "the square root of 1800 approximately equals 42.4")[$sqrt(1800) ≈ 42.4$] m A baseball diamond is a square whose sides are #math.equation(block: false, alt: "90")[$90$] feet in length. Find the straight-line distance from home plate to second base. #figure(figph[baseball diamond], alt: "baseball diamond", caption: none) What size square can be inscribed in a circle of radius #math.equation(block: false, alt: "8")[$8$] inches? #figure(figph[square in circle], alt: "square in circle", caption: none) #math.equation(block: false, alt: "the square root of 128")[$sqrt(128)$] in. by #math.equation(block: false, alt: "the square root of 128")[$sqrt(128)$] in. #math.equation(block: false, alt: "approximately equals 11.3")[$≈ 11.3$] in. #math.equation(block: false, alt: "times 11.3")[$× 11.3$] in. What size rectangle can be inscribed in a circle of radius 30 feet if the length of the rectangle must be 3 times its width? #figure(figph[rectangle in circle], alt: "rectangle in circle", caption: none) For Problems 25–30, + Use a calculator or computer to graph the function in the suggested window. + Use your graph to find two solutions for the given equation. (See Graphs of Functions to review graphical solution of equations.) + Check your solutions algebraically, using mental arithmetic. + #math.equation(block: false, alt: "y equals the fraction 1 over 4 x squared")[$y = display(frac(1, 4)) x^(2)$] #math.equation(block: true, alt: "Xmin, equals minus 15, Xmax equals 15; Ymin, equals minus 10, Ymax equals 40")[$"Xmin" & = − 15 & & "Xmax" = 15 \ "Ymin" & = − 10 & & "Ymax" = 40$] + #math.equation(block: false, alt: "the fraction 1 over 4 x squared equals 36")[$display(frac(1, 4)) x^(2) = 36$] + #figure(figph[GC graph], alt: "GC graph", caption: none) + #math.equation(block: false, alt: "x equals plus or minus 12")[$x = ± 12$] + #math.equation(block: false, alt: "y equals 8 x squared")[$y = 8 x^(2)$] #math.equation(block: true, alt: "Xmin, equals minus 15, Xmax equals 15; Ymin, equals minus 50, Ymax equals 450")[$"Xmin" & = − 15 & & "Xmax" = 15 \ "Ymin" & = − 50 & & "Ymax" = 450$] + #math.equation(block: false, alt: "8 x squared equals 392")[$8 x^(2) = 392$] + #math.equation(block: false, alt: "y equals open parenthesis x minus 5 close parenthesis squared")[$y = ( x − 5 )^(2)$] #math.equation(block: true, alt: "Xmin, equals minus 5, Xmax equals 15; Ymin, equals minus 5, Ymax equals 25")[$"Xmin" & = − 5 & & "Xmax" = 15 \ "Ymin" & = − 5 & & "Ymax" = 25$] + #math.equation(block: false, alt: "open parenthesis x minus 5 close parenthesis squared equals 16")[$( x − 5 )^(2) = 16$] + #figure(figph[GC graph], alt: "GC graph", caption: none) + #math.equation(block: false, alt: "x equals 1")[$x = 1$] or #math.equation(block: false, alt: "x equals 9")[$x = 9$] + #math.equation(block: false, alt: "y equals open parenthesis x plus 2 close parenthesis squared")[$y = ( x + 2 )^(2)$] #math.equation(block: true, alt: "Xmin, equals minus 10, Xmax equals 10; Ymin, equals minus 2, Ymax equals 12")[$"Xmin" & = − 10 & & "Xmax" = 10 \ "Ymin" & = − 2 & & "Ymax" = 12$] + #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis squared equals 9")[$( x + 2 )^(2) = 9$] + #math.equation(block: false, alt: "y equals 3 open parenthesis x minus 4 close parenthesis squared")[$y = 3 ( x − 4 )^(2)$] #math.equation(block: true, alt: "Xmin, equals minus 5, Xmax equals 15; Ymin, equals minus 20, Ymax equals 130")[$"Xmin" & = − 5 & & "Xmax" = 15 \ "Ymin" & = − 20 & & "Ymax" = 130$] + #math.equation(block: false, alt: "3 open parenthesis x minus 4 close parenthesis squared equals 108")[$3 ( x − 4 )^(2) = 108$] + #figure(figph[GC graph], alt: "GC graph", caption: none) + #math.equation(block: false, alt: "x equals 10")[$x = 10$] or #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$] + #math.equation(block: false, alt: "y equals the fraction 1 over 2 open parenthesis x plus 3 close parenthesis squared")[$y = display(frac(1, 2)) ( x + 3 )^(2)$] #math.equation(block: true, alt: "Xmin, equals minus 15, Xmax equals 5; Ymin, equals minus 5, Ymax equals 15")[$"Xmin" & = − 15 & & "Xmax" = 5 \ "Ymin" & = − 5 & & "Ymax" = 15$] + #math.equation(block: false, alt: "the fraction 1 over 2 open parenthesis x plus 3 close parenthesis squared equals 8")[$display(frac(1, 2)) ( x + 3 )^(2) = 8$] For Problems 31–42, solve by extraction of roots. #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis squared equals 9")[$( x − 2 )^(2) = 9$] #math.equation(block: true, alt: "5 , minus 1")[$5 , − 1$] #math.equation(block: true, alt: "open parenthesis x plus 3 close parenthesis squared equals 4")[$( x + 3 )^(2) = 4$] #math.equation(block: true, alt: "open parenthesis 2 x minus 1 close parenthesis squared equals 16")[$( 2 x − 1 )^(2) = 16$] #math.equation(block: true, alt: "the fraction 5 over 2 , the fraction minus 3 over 2")[$display(frac(5, 2)) , display(frac(− 3, 2))$] #math.equation(block: true, alt: "open parenthesis 3 x plus 1 close parenthesis squared equals 25")[$( 3 x + 1 )^(2) = 25$] #math.equation(block: true, alt: "4 open parenthesis x plus 2 close parenthesis squared equals 12")[$4 ( x + 2 )^(2) = 12$] #math.equation(block: true, alt: "minus 2 plus or minus the square root of 3")[$− 2 ± sqrt(3)$] #math.equation(block: true, alt: "6 open parenthesis x minus 5 close parenthesis squared equals 42")[$6 ( x − 5 )^(2) = 42$] #math.equation(block: true, alt: "open parenthesis x minus the fraction 1 over 2 close parenthesis squared equals the fraction 3 over 4")[$attach(( x − display(frac(1, 2)) ), t: 2) = display(frac(3, 4))$] #math.equation(block: true, alt: "the fraction 1 over 2 plus or minus the fraction the square root of 3 over 2")[$display(frac(1, 2)) ± display(frac(sqrt(3), 2))$] #math.equation(block: true, alt: "open parenthesis x minus the fraction 2 over 3 close parenthesis squared equals the fraction 5 over 9")[$attach(( x − display(frac(2, 3)) ), t: 2) = display(frac(5, 9))$] #math.equation(block: true, alt: "81 open parenthesis x plus the fraction 1 over 3 close parenthesis squared equals 1")[$81 attach(( x + display(frac(1, 3)) ), t: 2) = 1$] #math.equation(block: true, alt: "the fraction minus 2 over 9 , the fraction minus 4 over 9")[$display(frac(− 2, 9)) , display(frac(− 4, 9))$] #math.equation(block: true, alt: "16 open parenthesis x plus the fraction 1 over 2 close parenthesis squared equals 1")[$16 attach(( x + display(frac(1, 2)) ), t: 2) = 1$] #math.equation(block: true, alt: "3 open parenthesis 8 x minus 7 close parenthesis squared equals 24")[$3 ( 8 x − 7 )^(2) = 24$] #math.equation(block: true, alt: "the fraction 7 over 8 plus or minus the fraction the square root of 8 over 8")[$display(frac(7, 8)) ± display(frac(sqrt(8), 8))$] #math.equation(block: true, alt: "2 open parenthesis 5 x minus 12 close parenthesis squared equals 48")[$2 ( 5 x − 12 )^(2) = 48$] For Problems 43–54, + Solve algebraically. + Use the #strong[intersect] feature on a graphing calculator to solve. #math.equation(block: true, alt: "4 x cubed minus 12 equals 852")[$4 x^(3) − 12 = 852$] #math.equation(block: true, alt: "6")[$6$] #math.equation(block: true, alt: "the fraction 8 x cubed plus 6 over 3 equals 74")[$display(frac(8 x^(3) + 6, 3)) = 74$] #math.equation(block: true, alt: "5 the square root of x minus 9 equals 31")[$5 sqrt(x) − 9 = 31$] #math.equation(block: true, alt: "64")[$64$] #math.equation(block: true, alt: "25 minus 2 the square root of x equals 1")[$25 − 2 sqrt(x) = 1$] #math.equation(block: true, alt: "the fraction 1 over 2 x minus 3 equals the fraction 3 over 4")[$display(frac(1, 2 x − 3)) = display(frac(3, 4))$] #math.equation(block: true, alt: "the fraction 13 over 6")[$display(frac(13, 6))$] #math.equation(block: true, alt: "the fraction 15 over x plus 16 equals 3")[$display(frac(15, x + 16)) = 3$] #math.equation(block: true, alt: "8 minus 6 the cube root of x equals minus 4")[$8 − 6 root(3, x) = − 4$] #math.equation(block: true, alt: "8")[$8$] #math.equation(block: true, alt: "the fraction 4 the cube root of x over 5 plus 3 equals 7")[$display(frac(4 root(3, x), 5)) + 3 = 7$] #math.equation(block: true, alt: "the square root of 3 x minus 2 plus 3 equals 8")[$sqrt(3 x − 2) + 3 = 8$] #math.equation(block: true, alt: "9")[$9$] #math.equation(block: true, alt: "6 the square root of 1 minus 2 x equals 30")[$6 sqrt(1 − 2 x) = 30$] #math.equation(block: true, alt: "the fraction 2 over the square root of 4 x minus 2 equals 8")[$display(frac(2, sqrt(4 x − 2))) = 8$] #math.equation(block: true, alt: "the fraction 33 over 64")[$display(frac(33, 64))$] #math.equation(block: true, alt: "the fraction 1 over the square root of x plus 2 equals the fraction 3 over 4")[$display(frac(1, sqrt(x + 2))) = display(frac(3, 4))$] Cyril plans to invest \$#math.equation(block: false, alt: "5000")[$5000$] in a money market account that pays interest compounded annually. + Write a formula for the balance, #math.equation(block: false, alt: "B")[$B$], in Cyril's account after two years as a function of the interest rate, #math.equation(block: false, alt: "r")[$r$]. + If Cyril would like to have \$#math.equation(block: false, alt: "6250")[$6250$] in two years, what interest rate must the account pay? + Use your calculator to graph the formula for Cyril's account balance. Locate the point on the graph that corresponds to the amount in part (b). + #math.equation(block: false, alt: "B equals 5000 open parenthesis 1 plus r close parenthesis squared")[$B = 5000 ( 1 + r )^(2)$] + #math.equation(block: false, alt: "11.8 %")[$11.8 upright(%)$] + #figure(figph[GC graph], alt: "GC graph", caption: none) You plan to deposit your savings of \$#math.equation(block: false, alt: "1600")[$1600$] in an account that compounds interest annually. + Write a formula for the amount in your savings account after two years as a function of the interest rate, #math.equation(block: false, alt: "r")[$r$]. + To the nearest tenth of a percent, what interest rate will you require if you want your \$#math.equation(block: false, alt: "1600")[$1600$] to grow to \$#math.equation(block: false, alt: "2000")[$2000$] in two years? + Use your calculator to graph the formula for the account balance. Locate the point on the graph that corresponds to the amount in part (b). Carol's living expenses two years ago were \$#math.equation(block: false, alt: "1200")[$1200$] per month. This year, the same items cost Carol \$#math.equation(block: false, alt: "1400")[$1400$] per month. What was the annual inflation rate for the past two years? #math.equation(block: true, alt: "8 %")[$8 upright(%)$] Two years ago, the average price of a house in the suburbs was \$#math.equation(block: false, alt: "188 , 600")[$188 , 600$]. This year, the average price is \$#math.equation(block: false, alt: "203 , 700")[$203 , 700$]. What was the annual percent increase in the cost of a house? A machinist wants to make a metal section of pipe that is #math.equation(block: false, alt: "80")[$80$] millimeters long and has an interior volume of #math.equation(block: false, alt: "9000")[$9000$] cubic millimeters. If the pipe is #math.equation(block: false, alt: "2")[$2$] millimeters thick, its interior volume is given by the formula #math.equation(block: true, alt: "V equals π open parenthesis r minus 2 close parenthesis squared h")[$V = π ( r − 2 )^(2) h$] where #math.equation(block: false, alt: "h")[$h$] is the length of the pipe and #math.equation(block: false, alt: "r")[$r$] is its radius. What should the radius of the pipe be? #math.equation(block: false, alt: "7.98")[$7.98$] mm A storage box for sweaters is constructed from a square sheet of corrugated cardboard measuring #math.equation(block: false, alt: "x")[$x$] inches on a side. The volume of the box, in cubic inches, is #math.equation(block: true, alt: "V equals 10 open parenthesis x minus 20 close parenthesis squared")[$V = 10 ( x − 20 )^(2)$] If the box should have a volume of #math.equation(block: false, alt: "1960")[$1960$] cubic inches, what size cardboard square is needed? The area of an equilateral triangle is given by the formula #math.equation(block: false, alt: "A equals the fraction the square root of 3 over 4 s squared")[$A = display(frac(sqrt(3), 4)) s^(2)$], where #math.equation(block: false, alt: "s")[$s$] is the length of the side. + Find the areas of equilateral triangles with sides of length #math.equation(block: false, alt: "2")[$2$] centimeters, #math.equation(block: false, alt: "4")[$4$] centimeters, and #math.equation(block: false, alt: "10")[$10$] centimeters. First give exact values, then approximations to hundredths. + Graph the area equation in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 14.1; Ymin, equals 0, Ymax equals 60")[$"Xmin" & = 0 & & "Xmax" = 14.1 \ "Ymin" & = 0 & & "Ymax" = 60$] Use the TRACE or #strong[value] feature to verify your answers to part (a). + Trace along the curve to the point #math.equation(block: false, alt: "open parenthesis 5.1 , 11.26266 close parenthesis")[$( 5.1 , 11.26266 )$]. What do the coordinates of this point represent? + Use your graph to estimate the side of an equilateral triangle whose area is #math.equation(block: false, alt: "20")[$20$] square centimeters. + Write and solve an equation to answer part (d). + If the area of an equilateral triangle is #math.equation(block: false, alt: "100 the square root of 3")[$100 sqrt(3)$] square centimeters, what is the length of its side? + #math.equation(block: false, alt: "the square root of 3 approximately equals 1.73")[$sqrt(3) ≈ 1.73$] sq cm, #math.equation(block: false, alt: "4 the square root of 3 approximately equals 6.93")[$4 sqrt(3) ≈ 6.93$] sq cm, #math.equation(block: false, alt: "25 the square root of 3 approximately equals 43.3")[$25 sqrt(3) ≈ 43.3$] sq cm + #figure(figph[GC graph], alt: "GC graph", caption: none) + An equilateral triangle with side #math.equation(block: false, alt: "5.1")[$5.1$] cm has area #math.equation(block: false, alt: "11.263 cm squared")[$11.263 attach(" cm", t: 2)$]. + #math.equation(block: false, alt: "side approximately equals 6.8")[$"side" ≈ 6.8$] cm + #math.equation(block: false, alt: "the fraction the square root of 3 over 4 s squared equals 20")[$display(frac(sqrt(3), 4)) s^(2) = 20$]; #math.equation(block: false, alt: "s approximately equals 6.8")[$s ≈ 6.8$] + #math.equation(block: false, alt: "approximately equals 20")[$≈ 20$] cm The area of the ring in the figure is given by the formula #math.equation(block: false, alt: "A equals π R squared minus π r squared")[$A = π R^(2) − π r^(2)$], where #math.equation(block: false, alt: "R")[$R$] is the radius of the outer circle and #math.equation(block: false, alt: "r")[$r$] is the radius of the inner circle. #figure(figph[annulus], alt: "annulus", caption: none) + Suppose the inner radius of the ring is kept fixed at #math.equation(block: false, alt: "r equals 4")[$r = 4$] centimeters, but the radius of the outer circle, #math.equation(block: false, alt: "R")[$R$], is allowed to vary. Find the area of the ring when the outer radius is #math.equation(block: false, alt: "6")[$6$] centimeters, #math.equation(block: false, alt: "8")[$8$] centimeters, and #math.equation(block: false, alt: "12")[$12$] centimeters. First give exact values, then approximations to hundredths. + Graph the area equation, with r = 4, in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 14.1; Ymin, equals 0, Ymax equals 400")[$"Xmin" & = 0 & & "Xmax" = 14.1 \ "Ymin" & = 0 & & "Ymax" = 400$] Use the TRACE feature to verify your answers to part (a). + #strong[Trace] along the curve to the point #math.equation(block: false, alt: "open parenthesis 9.75 , 248.38217 close parenthesis")[$( 9.75 , 248.38217 )$]. What do the coordinates of this point represent? + Use your graph to estimate the outer radius of the ring when its area is #math.equation(block: false, alt: "100")[$100$] square centimeters. + Write and solve an equation to answer part (d). + If the area of the ring is #math.equation(block: false, alt: "9 π")[$9 π$] square centimeters, what is the radius of the outer circle? For Problems 63–68, solve for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$]. #math.equation(block: true, alt: "the fraction a x squared over b equals c")[$display(frac(a x^(2), b)) = c$] #math.equation(block: true, alt: "plus or minus the square root of the fraction b c over a")[$± sqrt(display(frac(b c, a)))$] #math.equation(block: true, alt: "the fraction b x squared over c minus a equals 0")[$display(frac(b x^(2), c)) − a = 0$] #math.equation(block: true, alt: "open parenthesis x minus a close parenthesis squared equals 16")[$( x − a )^(2) = 16$] #math.equation(block: true, alt: "a plus or minus 4")[$a ± 4$] #math.equation(block: true, alt: "open parenthesis x plus a close parenthesis squared equals 36")[$( x + a )^(2) = 36$] #math.equation(block: true, alt: "open parenthesis a x plus b close parenthesis squared equals 9")[$( a x + b )^(2) = 9$] #math.equation(block: true, alt: "the fraction minus b plus or minus 3 over a")[$display(frac(− b ± 3, a))$] #math.equation(block: true, alt: "open parenthesis a x minus b close parenthesis squared equals 25")[$( a x − b )^(2) = 25$] You have #math.equation(block: false, alt: "36")[$36$] feet of rope and you want to enclose a rectangular display area against one wall of an exhibit hall. The area enclosed depends on the dimensions of the rectangle you make. Because the wall makes one side of the rectangle, the length of the rope accounts for only three sides. Thus #math.equation(block: true, alt: "Base plus 2 open parenthesis Height close parenthesis equals 36")[$"Base" + 2 ( "Height" ) = 36$] #figure(figph[rectangle against wall], alt: "rectangle against wall", caption: none) + Complete the table showing the base and the area of the rectangle for the given heights.#figure(table( columns: 7, align: left, inset: 6pt, table.header([Height], [Base], [Area], [], [Height], [Base], [Area]), [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "34")[$34$]], [#math.equation(block: false, alt: "34")[$34$]], [], [#math.equation(block: false, alt: "10")[$10$]], [], [], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "64")[$64$]], [], [#math.equation(block: false, alt: "11")[$11$]], [], [], [#math.equation(block: false, alt: "3")[$3$]], [], [], [], [#math.equation(block: false, alt: "12")[$12$]], [], [], [#math.equation(block: false, alt: "4")[$4$]], [], [], [], [#math.equation(block: false, alt: "13")[$13$]], [], [], [#math.equation(block: false, alt: "5")[$5$]], [], [], [], [#math.equation(block: false, alt: "14")[$14$]], [], [], [#math.equation(block: false, alt: "6")[$6$]], [], [], [], [#math.equation(block: false, alt: "15")[$15$]], [], [], [#math.equation(block: false, alt: "7")[$7$]], [], [], [], [#math.equation(block: false, alt: "16")[$16$]], [], [], [#math.equation(block: false, alt: "8")[$8$]], [], [], [], [#math.equation(block: false, alt: "17")[$17$]], [], [], [#math.equation(block: false, alt: "9")[$9$]], [], [], [], [#math.equation(block: false, alt: "18")[$18$]], [], [], )) + Make a graph with #emph[Height] on the horizontal axis and #emph[Area] on the vertical axis. Draw a smooth curve through your data points. + What is the area of the largest rectangle you can enclose in this way? What are its dimensions? On your graph, label the point that corresponds to this rectangle with the letter #math.equation(block: false, alt: "M")[$M$]. + Let #math.equation(block: false, alt: "x")[$x$] stand for the height of a rectangle and write algebraic expressions for the base and the area of the rectangle. + Enter your algebraic expression for the area in your calculator, then use the #strong[Table] feature to verify the entries in your table in part (a). + Graph your formula for area on your graphing calculator. Use your table of values and your handdrawn graph to help you choose appropriate WINDOW settings. + Use the #strong[intersect] command to find the height of the rectangle whose area is #math.equation(block: false, alt: "149.5")[$149.5$] square feet. + #figure(table( columns: 7, align: left, inset: 6pt, table.header([Height], [Base], [Area], [], [Height], [Base], [Area]), [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "34")[$34$]], [#math.equation(block: false, alt: "34")[$34$]], [], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "160")[$160$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "64")[$64$]], [], [#math.equation(block: false, alt: "11")[$11$]], [#math.equation(block: false, alt: "14")[$14$]], [#math.equation(block: false, alt: "154")[$154$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "90")[$90$]], [], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "144")[$144$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "28")[$28$]], [#math.equation(block: false, alt: "112")[$112$]], [], [#math.equation(block: false, alt: "13")[$13$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "130")[$130$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "26")[$26$]], [#math.equation(block: false, alt: "130")[$130$]], [], [#math.equation(block: false, alt: "14")[$14$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "112")[$112$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "144")[$144$]], [], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "22")[$22$]], [#math.equation(block: false, alt: "154")[$154$]], [], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "64")[$64$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "160")[$160$]], [], [#math.equation(block: false, alt: "17")[$17$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "34")[$34$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "162")[$162$]], [], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], )) + #figure(figph[points and curve], alt: "points and curve", caption: none) + #math.equation(block: false, alt: "162")[$162$] sq ft, with base #math.equation(block: false, alt: "18")[$18$] ft, height #math.equation(block: false, alt: "9")[$9$] ft + Base: #math.equation(block: false, alt: "36 minus 2 x")[$36 − 2 x$]; area: #math.equation(block: false, alt: "x open parenthesis 36 minus 2 x close parenthesis")[$x ( 36 − 2 x )$] + See (a) + #math.equation(block: false, alt: "6.5")[$6.5$] ft or #math.equation(block: false, alt: "11.5")[$11.5$] ft We are going to make an open box from a square piece of cardboard by cutting #math.equation(block: false, alt: "3")[$3$]-inch squares from each corner and then turning up the edges as shown in the figure. #figure(figph[rectangle box], alt: "rectangle box", caption: none) + Complete the table showing the side of the original sheet of cardboard, the dimensions of the box created from it, and the volume of the box.#figure(table( columns: 5, align: left, inset: 6pt, table.header([Side], [Length #linebreak() of box #linebreak()], [Width #linebreak() of box #linebreak()], [Height #linebreak() of box #linebreak()], [Volume #linebreak() of box #linebreak()]), [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "9")[$9$]], [], [], [], [], [#math.equation(block: false, alt: "10")[$10$]], [], [], [], [], [#math.equation(block: false, alt: "11")[$11$]], [], [], [], [], [#math.equation(block: false, alt: "12")[$12$]], [], [], [], [], [#math.equation(block: false, alt: "13")[$13$]], [], [], [], [], [#math.equation(block: false, alt: "14")[$14$]], [], [], [], [], [#math.equation(block: false, alt: "15")[$15$]], [], [], [], [], )) #linebreak() Explain why the side of the cardboard square cannot be smaller than #math.equation(block: false, alt: "6")[$6$] inches. What happens if the cardboard is exactly #math.equation(block: false, alt: "6")[$6$] inches on a side? + Make a graph with #emph[Side] on the horizontal axis and #emph[Volume] on the vertical axis. Draw a smooth curve through your data points. (Use your table to help you decide on appropriate scales for the axes.) + Let #math.equation(block: false, alt: "x")[$x$] represent the side of the original sheet of cardboard. Write algebraic expressions for the dimensions of the box and for its volume. + Enter your expression for the volume of the box in your calculator; then use the #strong[Table] feature to verify the values in your table in part (a). + Graph your formula for volume on your graphing calculator. Use your table of values and your handdrawn graph to help you choose appropriate WINDOW settings. + Use the #strong[intersect] command to find out how large a square of cardboard you need to make a box with volume #math.equation(block: false, alt: "126.75")[$126.75$] cubic inches. + Does your graph have a highest point? What happens to the volume of the box as you increase #math.equation(block: false, alt: "x")[$x$]? The jump height, #math.equation(block: false, alt: "J")[$J$], in meters, achieved by a pole vaulter is given approximately by #math.equation(block: true, alt: "J equals v squared / open parenthesis 2 g close parenthesis")[$J = v^(2) / ( 2 g )$] , where v is the vaulter's speed in meters per second at the end of his run, and #math.equation(block: false, alt: "g equals 9.8")[$g = 9.8$] is the gravitational acceleration. (Source: Alexander, 1992) + Fill in the table of values for jump heights achieved with values of #math.equation(block: false, alt: "v")[$v$] from #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "11")[$11$] meters per second.#figure(table( columns: 13, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "11")[$11$]]), [#math.equation(block: false, alt: "J")[$J$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Graph the jump height versus final speed. (Use the table values to help you choose a window for the graph.) + The jump height should be added to the height of the vaulter's center of gravity (at about hip level) to give the maximum height, #math.equation(block: false, alt: "H")[$H$], he can clear. For a typical pole vaulter, his center of gravity at the end of the run is #math.equation(block: false, alt: "0.9")[$0.9$] meters from the ground. Complete the table of values for maximum heights, #math.equation(block: false, alt: "H")[$H$], and graph #math.equation(block: false, alt: "H")[$H$] on your graph of #math.equation(block: false, alt: "J")[$J$].#figure(table( columns: 13, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "11")[$11$]]), [#math.equation(block: false, alt: "H")[$H$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + A good pole vaulter can reach a final speed of #math.equation(block: false, alt: "9.5")[$9.5$] meters per second. What height will he clear? + In 2016, the world record in pole vaulting, established by Renaud Lavillenie in 2014, was 6.16 meters. What was the vaulter's speed at the end of his run? + #figure(table( columns: 13, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "11")[$11$]]), [#math.equation(block: false, alt: "J")[$J$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.05")[$0.05$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.46")[$0.46$]], [#math.equation(block: false, alt: "0.82")[$0.82$]], [#math.equation(block: false, alt: "1.28")[$1.28$]], [#math.equation(block: false, alt: "1.84")[$1.84$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "3.27")[$3.27$]], [#math.equation(block: false, alt: "4.13")[$4.13$]], [#math.equation(block: false, alt: "5.1")[$5.1$]], [#math.equation(block: false, alt: "6.17")[$6.17$]], )) + #figure(figph[two curves], alt: "two curves", caption: none) + #figure(table( columns: 13, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "11")[$11$]]), [#math.equation(block: false, alt: "H")[$H$]], [#math.equation(block: false, alt: "0.9")[$0.9$]], [#math.equation(block: false, alt: "0.95")[$0.95$]], [#math.equation(block: false, alt: "1.1")[$1.1$]], [#math.equation(block: false, alt: "1.36")[$1.36$]], [#math.equation(block: false, alt: "1.72")[$1.72$]], [#math.equation(block: false, alt: "2.18")[$2.18$]], [#math.equation(block: false, alt: "2.74")[$2.74$]], [#math.equation(block: false, alt: "3.4")[$3.4$]], [#math.equation(block: false, alt: "4.17")[$4.17$]], [#math.equation(block: false, alt: "5.03")[$5.03$]], [#math.equation(block: false, alt: "6.0")[$6.0$]], [#math.equation(block: false, alt: "7.07")[$7.07$]], )) + #math.equation(block: false, alt: "5.5")[$5.5$] meters + #math.equation(block: false, alt: "10.15")[$10.15$] meters per second To be launched into space, a satellite must travel fast enough to escape Earth's gravity. This escape velocity, #math.equation(block: false, alt: "v")[$v$], satisfies the equation #math.equation(block: true, alt: "the fraction 1 over 2 m v squared equals the fraction G M m over R")[$frac(1, 2) m v^(2) = frac(G M m, R)$] where #math.equation(block: false, alt: "m")[$m$] is the mass of the satellite, #math.equation(block: false, alt: "M")[$M$] is the mass of the Earth, #math.equation(block: false, alt: "R")[$R$] is the radius of the Earth, and #math.equation(block: false, alt: "G")[$G$] is the universal gravitational constant. + Solve the equation for #math.equation(block: false, alt: "v")[$v$] in terms of the other variables. + The equation #math.equation(block: true, alt: "m g equals the fraction G M m over R squared")[$m g = frac(G M m, R^(2))$] gives the force of gravity at the Earth's surface. We can use this equation to simplify the expression for #math.equation(block: false, alt: "v")[$v$]: First, multiply both sides of the equation by #math.equation(block: false, alt: "the fraction R over m")[$frac(R, m)$]. You now have an expression for #math.equation(block: false, alt: "the fraction G M over R")[$frac(G M, R)$]. Substitute this new expression into your formula for #math.equation(block: false, alt: "v")[$v$]. + The radius of the Earth is about #math.equation(block: false, alt: "6400")[$6400$] km, and #math.equation(block: false, alt: "g equals 0.0098")[$g = 0.0098$]. Calculate the escape velocity from Earth in kilometers per second. Convert your answer to miles per hour. (One kilometer is #math.equation(block: false, alt: "0.621")[$0.621$] miles.) + The radius of the moon is #math.equation(block: false, alt: "1740")[$1740$] km, and the value of #math.equation(block: false, alt: "g")[$g$] at the moon's surface is #math.equation(block: false, alt: "0.0016")[$0.0016$]. Calculate the escape velocity from the moon in kilometers per second and convert to miles per hour.