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== 3.6#h(0.6em)Chapter Summary and Review


    
      
=== Key Concepts


      + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Direct and Inverse Variation]

- #strong[#math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$]] if the ratio #math.equation(block: false, alt: "the fraction y over x")[$display(frac(y, x))$] is constant, that is, if #math.equation(block: false, alt: "y equals k x .")[$y = k x .$]
- #strong[#math.equation(block: false, alt: "y")[$y$] varies directly with a power of #math.equation(block: false, alt: "x")[$x$]] if the ratio #math.equation(block: false, alt: "the fraction y over x to the power n")[$display(frac(y, x^(n)))$] is constant, that is, if #math.equation(block: false, alt: "y equals k x to the power n")[$y = k x^(n)$] .
- #strong[#math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x")[$x$]] if the product #math.equation(block: false, alt: "x y")[$x y$] is constant, that is, if #math.equation(block: false, alt: "y equals the fraction k over x")[$y = display(frac(k, x))$] .
- #strong[#math.equation(block: false, alt: "y")[$y$] varies inversely with a power of #math.equation(block: false, alt: "x")[$x$]] if the product #math.equation(block: false, alt: "x to the power n y")[$x^(n) y$] is constant, that is, if #math.equation(block: false, alt: "y equals the fraction k over x to the power n")[$y = display(frac(k, x^(n)))$] .
]
+ The graph of a direct variation passes through the origin. The graph of an inverse variation has a vertical asymptote at the origin.
+ If #math.equation(block: false, alt: "y equals k x to the power n")[$y = k x^(n)$], we say that #math.equation(block: false, alt: "y")[$y$] #strong[scales] as #math.equation(block: false, alt: "x to the power n")[$x^(n)$].
+ #math.equation(block: false, alt: "n")[$n$]th roots: #math.equation(block: false, alt: "s")[$s$] is called an #math.equation(block: false, alt: "n")[$n$]th root of #math.equation(block: false, alt: "b")[$b$] if #math.equation(block: false, alt: "s to the power n equals b")[$s^(n) = b$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Exponential Notation]

The absolute value has the following properties:

#math.equation(block: true, alt: "a to the power minus n equals the fraction 1 over x to the power n, a not equal to 0; a to the power 0 equals 1, a not equal to 0; a to the power 1 / n equals the nth root of a, n an integer, n greater than 2; a to the power m / n equals open parenthesis a to the power 1 / n close parenthesis to the power m equals open parenthesis a to the power m close parenthesis to the power 1 / n ,, a greater than 0 , n not equal to 0")[$& a^(− n) = frac(1, x^(n)) & #hide($00000$) & a ≠ 0 \  & a^(0) = 1 &  & a ≠ 0 \  & a^(1 / n) = root(n, a) &  & n " an integer, " n > 2 \  & a^(m / n) = ( a^(1 / n) )^(m) = ( a^(m) )^(1 / n) , &  & a > 0 , " " n ≠ 0$]
]
+ In particular, a negative exponent denotes a reciprocal, and a fractional exponent denotes a root.
+ #math.equation(block: false, alt: "a to the power m / n equals the nth root of a to the power m equals open parenthesis the nth root of a close parenthesis to the power m")[$a^(m / n) = root(n, a^(m)) = attach(( root(n, a) ), t: m)$]
+ To compute #math.equation(block: false, alt: "a to the power m / n")[$a^(m / n)$], we can compute the #math.equation(block: false, alt: "n")[$n$]th root first, or the #math.equation(block: false, alt: "m")[$m$]th power, whichever is easier.
+ We cannot write down an exact decimal equivalent for an irrational number, but we can approximate an irrational number to as many decimal places as we like.
+ The laws of exponents are valid for all exponents #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "n")[$n$], and for #math.equation(block: false, alt: "b not equal to 0")[$b ≠ 0$].#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Laws of Exponents]

+ #math.equation(block: false, alt: "a to the power m times a to the power n equals a to the power m plus n")[$a^(m) ⋅ a^(n) = a^(m + n)$]
+ #math.equation(block: false, alt: "the fraction a to the power m over a to the power n equals a to the power m minus n")[$display(frac(a^(m), a^(n))) = a^(m − n)$]
+ #math.equation(block: false, alt: "open parenthesis a to the power m close parenthesis to the power n equals a to the power m n")[$attach(( a^(m) ), t: n) = a^(m n)$]
+ #math.equation(block: false, alt: "open parenthesis a b close parenthesis to the power n equals a to the power n b to the power n")[$attach(( a b ), t: n) = a^(n) b^(n)$]
+ #math.equation(block: false, alt: "open parenthesis the fraction a over b close parenthesis to the power n equals the fraction a to the power n over b to the power n")[$attach(( display(frac(a, b)) ), t: n) = display(frac(a^(n), b^(n)))$]
]
+ A function of the form #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals k x to the power p")[$f ( x ) = k x^(p)$], where #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "p")[$p$] are constants, is called a #strong[power function].
+ An #strong[allometric equation] is a power function of the form #math.equation(block: false, alt: "variable equals k open parenthesis mass close parenthesis to the power p")[$"variable" = k ( "mass" )^(p)$].
+ We can solve the equation #math.equation(block: false, alt: "x to the power n equals b")[$x^(n) = b$] by raising both sides to the #math.equation(block: false, alt: "the fraction 1 over n")[$display(frac(1, n))$] power
+ We can solve the equation #math.equation(block: false, alt: "x to the power 1 / n equals b")[$x^(1 / n) = b$] by raising both sides to the #math.equation(block: false, alt: "n")[$n$]th power.
+ To solve the equation #math.equation(block: false, alt: "x to the power m / n equals k")[$x^(m / n) = k$], we raise both sides to the power #math.equation(block: false, alt: "n / m")[$n / m$].
+ The graphs of power functions #math.equation(block: false, alt: "y equals x to the power m / n")[$y = x^(m / n)$], where #math.equation(block: false, alt: "m / n")[$m / n$] is positive are all increasing for #math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$]. If #math.equation(block: false, alt: "m / n greater than 1")[$m / n > 1$], the graph is concave up. If #math.equation(block: false, alt: "0 less than m / n less than 1")[$0 < m / n < 1$], the graph is concave down.
+ The notation #math.equation(block: false, alt: "z equals f open parenthesis x , y close parenthesis")[$z = f ( x , y )$] indicates that #math.equation(block: false, alt: "z")[$z$] is a function of two variables, #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$].
+ We can use a table with rows and columns to display the output values for a function of two variables.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Joint Variation]

- We say that #math.equation(block: false, alt: "z")[$z$] #strong[varies jointly] with #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] if
            		#math.equation(block: true, alt: "z equals k x y , k not equal to 0")[$z = k x y , " " " " " " k ≠ 0$]
- We say that #math.equation(block: false, alt: "z")[$z$] varies directly with #math.equation(block: false, alt: "x")[$x$] and inversely with #math.equation(block: false, alt: "y")[$y$] if
            		#math.equation(block: true, alt: "z equals k the fraction x over y , k not equal to 0 , y not equal to 0")[$z = k display(frac(x, y)) , " " " " " " k ≠ 0 , " " " " y ≠ 0$]
]
+ We can represent a function of two variables graphically by showing a set of graphs for several fixed values of one of the variables.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Roots of Real Numbers]

- Every positive number has two real-valued roots, one positive and one negative, if the index is even.
- A negative number has no real-valued root if the index is even.
- Every real number, positive, negative, or zero, has exactly one real-valued root if the index is odd.
]


    
    
      
=== Chapter 3 Review Problems


      
        The distance s a pebble falls through a thick liquid varies directly with the square of the length of time #math.equation(block: false, alt: "t")[$t$] it falls.


        + If the pebble falls #math.equation(block: false, alt: "28")[$28$] centimeters in #math.equation(block: false, alt: "4")[$4$] seconds, express the distance it will fall as a function of time.
+ Find the distance the pebble will fall in #math.equation(block: false, alt: "6")[$6$] seconds.


        
          + #math.equation(block: false, alt: "d equals 1.75 t squared")[$d = 1.75 t^(2)$]
+ #math.equation(block: false, alt: "63")[$63$] cm


        
      
      
        The volume, #math.equation(block: false, alt: "V")[$V$], of a gas varies directly with the temperature, #math.equation(block: false, alt: "T")[$T$], and inversely with the pressure, #math.equation(block: false, alt: "P")[$P$], of the gas.


        + If #math.equation(block: false, alt: "V equals 40")[$V = 40$] when #math.equation(block: false, alt: "T equals 300")[$T = 300$] and #math.equation(block: false, alt: "P equals 30")[$P = 30$], express the volume of the gas as a function of the temperature and pressure of the gas.
+ Find the volume when #math.equation(block: false, alt: "T equals 320")[$T = 320$] and #math.equation(block: false, alt: "P equals 40")[$P = 40$].


      
      
        The demand for bottled water is inversely proportional to the price per bottle. If Droplets can sell #math.equation(block: false, alt: "600")[$600$] bottles at \$#math.equation(block: false, alt: "8")[$8$] each, how many bottles can the company sell at \$#math.equation(block: false, alt: "10")[$10$] each?


        
          #math.equation(block: false, alt: "480")[$480$] bottles


        
      
      
        The intensity of illumination from a light source varies inversely with the square of the distance from the source. If a reading lamp has an intensity of #math.equation(block: false, alt: "100")[$100$] lumens at a distance of #math.equation(block: false, alt: "3")[$3$] feet, what is its intensity #math.equation(block: false, alt: "8")[$8$] feet away?


      
      
        A person's weight, #math.equation(block: false, alt: "w")[$w$], varies inversely with the square of his or her distance, #math.equation(block: false, alt: "r")[$r$], from the center of the Earth.


        + Express #math.equation(block: false, alt: "w")[$w$] as a function of #math.equation(block: false, alt: "r")[$r$]. Let #math.equation(block: false, alt: "k")[$k$] stand for the constant of variation.
+ Make a rough graph of your function.
+ How far from the center of the Earth must Neil be in order to weigh one-third of his weight on the surface? The radius of the Earth is about #math.equation(block: false, alt: "3960")[$3960$] miles.


        
          + #math.equation(block: false, alt: "w equals the fraction k over r squared")[$w = display(frac(k, r^(2)))$]
+ #figure(figph[inverse square in first quadrant], alt: "inverse square in first quadrant", caption: none)
+ #math.equation(block: false, alt: "3960 the square root of 3 approximately equals 6860")[$3960 sqrt(3) ≈ 6860$] miles


        
      
      
        The period, #math.equation(block: false, alt: "T")[$T$], of a pendulum varies directly with the square root of its length, #math.equation(block: false, alt: "L")[$L$].


        + Express #math.equation(block: false, alt: "T")[$T$] as a function of #math.equation(block: false, alt: "L")[$L$]. Let #math.equation(block: false, alt: "k")[$k$] stand for the constant of variation.
+ Make a rough graph of your function.
+ If a certain pendulum is replaced by a new one four-fifths as long as the old one, what happens to the period?


      
      In Problems 7–10, #math.equation(block: false, alt: "y")[$y$] varies directly or inversely with a power of #math.equation(block: false, alt: "x")[$x$]. Find the power of #math.equation(block: false, alt: "x")[$x$] and the constant of variation, #math.equation(block: false, alt: "k")[$k$]. Write a formula for each function of the form #math.equation(block: false, alt: "y equals k x to the power n")[$y = k x^(n)$] or #math.equation(block: false, alt: "y equals the fraction k over x to the power n")[$y = display(frac(k, x^(n)))$].


      
        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4.8")[$4.8$]],
  [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "30.0")[$30.0$]],
  [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "76.8")[$76.8$]],
  [#math.equation(block: false, alt: "11")[$11$]], [#math.equation(block: false, alt: "145.2")[$145.2$]],
))


        
          #math.equation(block: true, alt: "y equals 1.2 x squared")[$y = 1.2 x^(2)$]


        
      
      
        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "1.4")[$1.4$]], [#math.equation(block: false, alt: "75.6")[$75.6$]],
  [#math.equation(block: false, alt: "2.3")[$2.3$]], [#math.equation(block: false, alt: "124.2")[$124.2$]],
  [#math.equation(block: false, alt: "5.9")[$5.9$]], [#math.equation(block: false, alt: "318.6")[$318.6$]],
  [#math.equation(block: false, alt: "8.3")[$8.3$]], [#math.equation(block: false, alt: "448.2")[$448.2$]],
))


      
      
        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "40.0")[$40.0$]],
  [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "10.0")[$10.0$]],
  [#math.equation(block: false, alt: "4.0")[$4.0$]], [#math.equation(block: false, alt: "5.0")[$5.0$]],
  [#math.equation(block: false, alt: "8.0")[$8.0$]], [#math.equation(block: false, alt: "2.5")[$2.5$]],
))


        
          #math.equation(block: true, alt: "y equals the fraction 20 over x")[$y = display(frac(20, x))$]


        
      
      
        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]),
  [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "320.0")[$320.0$]],
  [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "115.2")[$115.2$]],
  [#math.equation(block: false, alt: "4.0")[$4.0$]], [#math.equation(block: false, alt: "45.0")[$45.0$]],
  [#math.equation(block: false, alt: "6.0")[$6.0$]], [#math.equation(block: false, alt: "20.0")[$20.0$]],
))


      
      For Problems 11–16, write without negative exponents and simplify.


      
        + #math.equation(block: false, alt: "open parenthesis minus 3 close parenthesis to the power minus 4")[$( − 3 )^(− 4)$]
+ #math.equation(block: false, alt: "4 to the power minus 3")[$4^(− 3)$]


        
          + #math.equation(block: false, alt: "the fraction 1 over 81")[$display(frac(1, 81))$]
+ #math.equation(block: false, alt: "the fraction 1 over 64")[$display(frac(1, 64))$]


        
      
      
        + #math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power minus 2")[$attach(( display(frac(1, 3)) ), t: − 2)$]
+ #math.equation(block: false, alt: "the fraction 3 over 5 to the power minus 2")[$display(frac(3, 5^(− 2)))$]


      
      
        + #math.equation(block: false, alt: "open parenthesis 3 m close parenthesis to the power minus 5")[$( 3 m )^(− 5)$]
+ #math.equation(block: false, alt: "minus 7 y to the power minus 8")[$− 7 y^(− 8)$]


        
          + #math.equation(block: false, alt: "the fraction 1 over 243 m to the power 5")[$display(frac(1, 243 m^(5)))$]
+ #math.equation(block: false, alt: "the fraction minus 7 over y to the power 8")[$display(frac(− 7, y^(8)))$]


        
      
      
        + #math.equation(block: false, alt: "a to the power minus 1 plus a to the power minus 2")[$a^(− 1) + a^(− 2)$]
+ #math.equation(block: false, alt: "the fraction 3 q to the power minus 9 over r to the power minus 2")[$display(frac(3 q^(− 9), r^(− 2)))$]


      
      
        + #math.equation(block: false, alt: "6 c to the power minus 7 times open parenthesis 3 close parenthesis to the power minus 1 c to the power 4")[$6 c^(− 7) ⋅ ( 3 )^(− 1) c^(4)$]
+ #math.equation(block: false, alt: "the fraction 11 z to the power minus 7 over 3 to the power minus 2 z to the power minus 5")[$display(frac(11 z^(− 7), 3^(− 2) z^(− 5)))$]


        
          + #math.equation(block: false, alt: "the fraction 2 over c cubed")[$display(frac(2, c^(3)))$]
+ #math.equation(block: false, alt: "the fraction 99 over z squared")[$display(frac(99, z^(2)))$]


        
      
      
        + #math.equation(block: false, alt: "open parenthesis 2 d to the power minus 2 k cubed close parenthesis to the power minus 4")[$attach(( 2 d^(− 2) k^(3) ), t: − 4)$]
+ #math.equation(block: false, alt: "the fraction 2 w cubed open parenthesis w to the power minus 2 close parenthesis to the power minus 3 over 5 w to the power minus 5")[$display(frac(2 w^(3) ( w^(− 2) )^(− 3), 5 w^(− 5)))$]


      
      For Problems 17–20, write each power in radical form.


      
        + #math.equation(block: false, alt: "25 m to the power 1 / 2")[$25 m^(1 / 2)$]
+ #math.equation(block: false, alt: "8 n to the power minus 1 / 3")[$8 n^(− 1 / 3)$]


        
          + #math.equation(block: false, alt: "25 the square root of m")[$25 sqrt(m)$]
+ #math.equation(block: false, alt: "the fraction 8 over the cube root of n")[$display(frac(8, root(3, n)))$]


        
      
      
        + #math.equation(block: false, alt: "open parenthesis 13 d close parenthesis to the power 2 / 3")[$( 13 d )^(2 / 3)$]
+ #math.equation(block: false, alt: "6 x to the power 2 / 5 y to the power 3 / 5")[$6 x^(2 / 5) y^(3 / 5)$]


      
      
        + #math.equation(block: false, alt: "open parenthesis 3 q close parenthesis to the power minus 3 / 4")[$( 3 q )^(− 3 / 4)$]
+ #math.equation(block: false, alt: "7 open parenthesis u v close parenthesis to the power 3 / 2")[$7 ( u v )^(3 / 2)$]


        
          + #math.equation(block: false, alt: "the fraction 1 over the 4th root of 27 q cubed")[$display(frac(1, root(4, 27 q^(3))))$]
+ #math.equation(block: false, alt: "7 the square root of u cubed v cubed")[$7 sqrt(u^(3) v^(3))$]


        
      
      
        + #math.equation(block: false, alt: "open parenthesis a squared plus b squared close parenthesis to the power 0.5")[$( a^(2) + b^(2) )^(0.5)$]
+ #math.equation(block: false, alt: "open parenthesis 16 minus x squared close parenthesis to the power 0.25")[$( 16 − x^(2) )^(0.25)$]


      
      For Problems 21–24, write each radical as a power with a fractional exponent.


      
        + #math.equation(block: false, alt: "2 the cube root of x squared")[$2 root(3, x^(2))$]
+ #math.equation(block: false, alt: "the fraction 1 over 4 the 4th root of x")[$display(frac(1, 4)) root(4, x)$]


        
          + #math.equation(block: false, alt: "2 x to the power 2 / 3")[$2 x^(2 / 3)$]
+ #math.equation(block: false, alt: "the fraction 1 over 4 x to the power 1 / 4")[$display(frac(1, 4)) x^(1 / 4)$]


        
      
      
        + #math.equation(block: false, alt: "z squared the square root of z")[$z^(2) sqrt(z)$]
+ #math.equation(block: false, alt: "z the cube root of z")[$z root(3, z)$]


      
      
        + #math.equation(block: false, alt: "the fraction 6 over the 4th root of b cubed")[$display(frac(6, root(4, b^(3))))$]
+ #math.equation(block: false, alt: "the fraction minus 1 over 3 the cube root of b")[$display(frac(− 1, 3 root(3, b)))$]


        
          + #math.equation(block: false, alt: "6 b to the power minus 3 / 4")[$6 b^(− 3 / 4)$]
+ #math.equation(block: false, alt: "the fraction minus 1 over 3 b to the power minus 1 / 3")[$display(frac(− 1, 3)) b^(− 1 / 3)$]


        
      
      
        + #math.equation(block: false, alt: "the fraction minus 4 over open parenthesis the 4th root of a close parenthesis squared")[$display(frac(− 4, ( root(4, a) )^(2)))$]
+ #math.equation(block: false, alt: "the fraction 2 over open parenthesis the square root of a close parenthesis cubed")[$display(frac(2, ( sqrt(a) )^(3)))$]


      
      For Problems 25–28, sketch graphs by hand for each function on the domain #math.equation(block: false, alt: "open parenthesis 0 , ∞ close parenthesis")[$( 0 , upright(∞) )$].


      
        #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x squared")[$x^(2)$]. The constant of variation is #math.equation(block: false, alt: "k equals 0.25")[$k = 0.25$].


        
          #figure(figph[inverse square], alt: "inverse square", caption: none)


        
      
      
        #math.equation(block: false, alt: "y")[$y$] varies directly with #math.equation(block: false, alt: "x")[$x$]. The constant of variation is #math.equation(block: false, alt: "k equals 1.5")[$k = 1.5$].


      
      
        #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x")[$x$]. The constant of variation is #math.equation(block: false, alt: "k equals 2")[$k = 2$].


        
          #figure(figph[inverse square], alt: "inverse square", caption: none)


        
      
      
        #math.equation(block: false, alt: "y")[$y$] varies inversely with #math.equation(block: false, alt: "x squared")[$x^(2)$]. The constant of variation is #math.equation(block: false, alt: "k equals 4")[$k = 4$].


      
      For Problems 29–30, write each function in the form #math.equation(block: false, alt: "y equals k x to the power p")[$y = k x^(p)$].


      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 2 over 3 x to the power 4")[$f ( x ) = display(frac(2, 3 x^(4)))$]


        
          #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 2 over 3 x to the power minus 4")[$f ( x ) = display(frac(2, 3)) x^(− 4)$]


        
      
      
        #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals the fraction 8 x to the power 7 over 29")[$g ( x ) = display(frac(8 x^(7), 29))$]


      
      For Problems 31–34,


      + Evaluate each function for the given values.
+ Graph the function.


      
        #math.equation(block: true, alt: "Q open parenthesis x close parenthesis equals 4 x to the power 5 / 2")[$Q ( x ) = 4 x^(5 / 2)$]


        #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "100")[$100$]]),
  [#math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$],
))


        
          + #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "100")[$100$]]),
  [#math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$]], [#math.equation(block: false, alt: "4096")[$4096$]], [#math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$]], [#math.equation(block: false, alt: "4 the square root of 3 to the power 5 approximately equals 62.35")[$4 sqrt(3^(5)) ≈ 62.35$]], [#math.equation(block: false, alt: "400 , 000")[$400 , 000$]],
))
+ #figure(figph[increasing concave up], alt: "increasing concave up", caption: none)


        
      
      
        #math.equation(block: true, alt: "T open parenthesis w close parenthesis equals minus 3 w to the power 2 / 3")[$T ( w ) = − 3 w^(2 / 3)$]


        #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "w")[$w$]], [#math.equation(block: false, alt: "27")[$27$]], [#math.equation(block: false, alt: "the fraction 1 over 8")[$display(frac(1, 8))$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "1000")[$1000$]]),
  [#math.equation(block: false, alt: "T open parenthesis w close parenthesis")[$T ( w )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$],
))


      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x to the power 0.3")[$f ( x ) = x^(0.3)$]


        #figure(table(
  columns: 9, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "100")[$100$]]),
  [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$],
))


        
          + #figure(table(
  columns: 9, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "100")[$100$]]),
  [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1.62")[$1.62$]], [#math.equation(block: false, alt: "2.00")[$2.00$]], [#math.equation(block: false, alt: "2.46")[$2.46$]], [#math.equation(block: false, alt: "3.23")[$3.23$]], [#math.equation(block: false, alt: "3.58")[$3.58$]], [#math.equation(block: false, alt: "3.98")[$3.98$]],
))
+ #figure(figph[increasing concave up], alt: "increasing concave up", caption: none)


        
      
      
        #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals minus x to the power minus 0.7")[$g ( x ) = − x^(− 0.7)$]


        #figure(table(
  columns: 9, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]),
  [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$],
))


      
      
        According to the theory of relativity, the mass of an object traveling at velocity #math.equation(block: false, alt: "v")[$v$] is given by the function


        #math.equation(block: true, alt: "m equals the fraction M over the square root of 1 minus the fraction v squared over c squared")[$m = display(frac(M, sqrt(1 − display(frac(v^(2), c^(2))))))$]
        where #math.equation(block: false, alt: "M")[$M$] is the mass of the object at rest and #math.equation(block: false, alt: "c")[$c$] is the speed of light. Find the mass of a man traveling at a velocity of #math.equation(block: false, alt: "0.7 c")[$0.7 c$] if his rest mass is #math.equation(block: false, alt: "80")[$80$] kilograms.


        
          #math.equation(block: false, alt: "112")[$112$] kg


        
      
      
        The cylinder of smallest surface area for a given volume has a radius and height both equal to #math.equation(block: false, alt: "the cube root of the fraction V over π")[$root(3, display(frac(V, π)))$]. Find the dimensions of the tin can of smallest surface area with volume #math.equation(block: false, alt: "60")[$60$] cubic inches.


      
      
        Membership in the Wildlife Society has grown according to the function


        #math.equation(block: true, alt: "M open parenthesis t close parenthesis equals 30 t to the power 3 / 4")[$M ( t ) = 30 t^(3 / 4)$]
        where #math.equation(block: false, alt: "t")[$t$] is the number of years since its founding in #math.equation(block: false, alt: "1970")[$1970$].


        + Sketch a graph of the function #math.equation(block: false, alt: "M open parenthesis t close parenthesis")[$M ( t )$].
+ What was the society's membership in #math.equation(block: false, alt: "1990")[$1990$]?
+ In what year will the membership be #math.equation(block: false, alt: "810")[$810$] people?


        
          + #figure(figph[increasing concave down], alt: "increasing concave down", caption: none)
+ 283
+ 2051


        
      
      
        The heron population in Saltmarsh Refuge is estimated by conservationists at


        #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 360 t to the power minus 2 / 3")[$P ( t ) = 360 t^(− 2 / 3)$]
        where #math.equation(block: false, alt: "t")[$t$] is the number of years since the refuge was established in #math.equation(block: false, alt: "1990")[$1990$].


        + Sketch a graph of the function #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$].
+ How many heron were there in #math.equation(block: false, alt: "1995")[$1995$]?
+ In what year will there be only #math.equation(block: false, alt: "40")[$40$] heron left?


      
      
        Manufacturers of ships (and other complex products) find that the average cost of producing a ship decreases as more of those ships are produced. This relationship is called the #strong[experience curve], given by the equation


        #math.equation(block: true, alt: "C equals a x to the power minus b")[$C = a x^(− b)$]
        where #math.equation(block: false, alt: "C")[$C$] is the average cost per ship in millions of dollars and #math.equation(block: false, alt: "x")[$x$] is the number of ships produced. The value of the constant #math.equation(block: false, alt: "b")[$b$] depends on the complexity of the ship. (Source: Storch, Hammon, and Bunch, 1988)


        + What is the significance of the constant of proportionality #math.equation(block: false, alt: "a")[$a$]? Hint: What is the value of #math.equation(block: false, alt: "C")[$C$] if only one ship is built?
+ For one kind of ship, #math.equation(block: false, alt: "b equals the fraction 1 over 8")[$b = display(frac(1, 8))$], and the cost of producing the first ship is \$#math.equation(block: false, alt: "12")[$12$] million. Write the equation for #math.equation(block: false, alt: "C")[$C$] as a function of #math.equation(block: false, alt: "x")[$x$] using radical notation.
+ Compute the cost per ship when #math.equation(block: false, alt: "2")[$2$] ships have been built. By what percent does the cost per ship decrease? By what percent does the cost per ship decrease from building #math.equation(block: false, alt: "2")[$2$] ships to building #math.equation(block: false, alt: "4")[$4$] ships?
+ By what percent does the average cost decrease from building #math.equation(block: false, alt: "n")[$n$] ships to building #math.equation(block: false, alt: "2 n")[$2 n$] ships? (In the shipbuilding industry, the average cost per ship usually decreases by #math.equation(block: false, alt: "5")[$5$] to #math.equation(block: false, alt: "10 %")[$10 upright(%)$] each time the number of ships doubles.)


        
          + It is the cost of producing the first ship.
+ #math.equation(block: false, alt: "C equals the fraction 12 over the 8th root of x")[$C = display(frac(12, root(8, x)))$] million
+ About \$#math.equation(block: false, alt: "11")[$11$] million; about #math.equation(block: false, alt: "8.3 %")[$8.3 upright(%)$] ; about #math.equation(block: false, alt: "8.3 %")[$8.3 upright(%)$]
+ About #math.equation(block: false, alt: "8.3 %")[$8.3 upright(%)$]


        
      
      
        A population is in a period of #strong[supergrowth] if its rate of growth, #math.equation(block: false, alt: "R")[$R$], at any time is proportional to #math.equation(block: false, alt: "P to the power k")[$P^(k)$], where #math.equation(block: false, alt: "P")[$P$] is the population at that time and #math.equation(block: false, alt: "k")[$k$] is a constant greater than #math.equation(block: false, alt: "1")[$1$]. Suppose #math.equation(block: false, alt: "R")[$R$] is given by


        #math.equation(block: true, alt: "R equals 0.015 P to the power 1.2")[$R = 0.015 P^(1.2)$]
        where #math.equation(block: false, alt: "P")[$P$] is measured in thousands and #math.equation(block: false, alt: "R")[$R$] is measured in thousands per year.


        + Find #math.equation(block: false, alt: "R")[$R$] when #math.equation(block: false, alt: "P equals 20")[$P = 20$], when #math.equation(block: false, alt: "P equals 40")[$P = 40$], and when #math.equation(block: false, alt: "P equals 60")[$P = 60$].
+ What will the population be when its rate of growth is #math.equation(block: false, alt: "5000")[$5000$] per year?
+ Graph #math.equation(block: false, alt: "R")[$R$] and use your graph to verify your answers to parts (a) and (b).


      
      For Problems 41–50, solve


      
        #math.equation(block: true, alt: "6 t to the power minus 3 equals the fraction 3 over 500")[$6 t^(− 3) = display(frac(3, 500))$]


        
          #math.equation(block: true, alt: "t equals 10")[$t = 10$]


        
      
      
        #math.equation(block: true, alt: "3.5 minus 2.4 p to the power minus 2 equals minus 6.1")[$3.5 − 2.4 p^(− 2) = − 6.1$]


      
      
        #math.equation(block: true, alt: "the cube root of x plus 1 equals 2")[$root(3, x + 1) = 2$]


        
          #math.equation(block: true, alt: "x equals 7")[$x = 7$]


        
      
      
        #math.equation(block: true, alt: "x to the power 2 / 3 plus 2 equals 6")[$x^(2 / 3) + 2 = 6$]


      
      
        #math.equation(block: true, alt: "open parenthesis x minus 1 close parenthesis to the power minus 3 / 2 equals the fraction 1 over 8")[$( x − 1 )^(− 3 / 2) = display(frac(1, 8))$]


        
          #math.equation(block: true, alt: "x equals 5")[$x = 5$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 2 x plus 1 close parenthesis to the power minus 1 / 2 equals the fraction 1 over 3")[$( 2 x + 1 )^(− 1 / 2) = display(frac(1, 3))$]


      
      
        #math.equation(block: true, alt: "8 the 4th root of x plus 6 equals 24")[$8 root(4, x + 6) = 24$]


        
          #math.equation(block: true, alt: "x equals 75")[$x = 75$]


        
      
      
        #math.equation(block: true, alt: "9.8 equals 7 the cube root of z minus 4")[$9.8 = 7 root(3, z − 4)$]


      
      
        #math.equation(block: true, alt: "the fraction 2 over 3 open parenthesis 2 y plus 1 close parenthesis to the power 0.2 equals 6")[$display(frac(2, 3)) ( 2 y + 1 )^(0.2) = 6$]


        
          #math.equation(block: true, alt: "y equals 29 , 524")[$y = 29 , 524$]


        
      
      
        #math.equation(block: true, alt: "1.3 w to the power 0.3 plus 4.7 equals 5.2")[$1.3 w^(0.3) + 4.7 = 5.2$]


      
      For Problems 51–54, solve the formula for the indicated variable.


      
        #math.equation(block: false, alt: "t equals the square root of the fraction 2 v over g")[$t = sqrt(display(frac(2 v, g))) " " " " " "$], for #math.equation(block: false, alt: "g")[$g$]


        
          #math.equation(block: true, alt: "g equals the fraction 2 v over t squared")[$g = display(frac(2 v, t^(2)))$]


        
      
      
        #math.equation(block: false, alt: "q minus 1 equals 2 the square root of the fraction r squared minus 1 over 3")[$q − 1 = 2 sqrt(display(frac(r^(2) − 1, 3))) " " " " " "$], for #math.equation(block: false, alt: "r")[$r$]


      
      
        #math.equation(block: false, alt: "R equals the fraction 1 plus the square root of p squared plus 1 over 2")[$R = display(frac(1 + sqrt(p^(2) + 1), 2)) " " " " " "$], for #math.equation(block: false, alt: "p")[$p$]


        
          #math.equation(block: true, alt: "p equals plus or minus 2 the square root of R squared minus R")[$p = ± 2 sqrt(R^(2) − R)$]


        
      
      
        #math.equation(block: false, alt: "q equals the cube root of the fraction 1 plus r squared over 2")[$q = root(3, display(frac(1 + r^(2), 2))) " " " " " "$], for #math.equation(block: false, alt: "r")[$r$]


      
      For Problems 55–60, simplify by applying the laws of exponents.


      
        #math.equation(block: true, alt: "open parenthesis 7 t close parenthesis cubed open parenthesis 7 t close parenthesis to the power minus 1")[$( 7 t )^(3) ( 7 t )^(− 1)$]


        
          #math.equation(block: true, alt: "49 t squared")[$49 t^(2)$]


        
      
      
        #math.equation(block: true, alt: "the fraction 36 r to the power minus 2 s over 9 r to the power minus 3 s to the power 4")[$display(frac(36 r^(− 2) s, 9 r^(− 3) s^(4)))$]


      
      
        #math.equation(block: true, alt: "the fraction open parenthesis 2 k to the power minus 1 close parenthesis to the power minus 4 over 4 k to the power minus 3")[$display(frac(( 2 k^(− 1) )^(− 4), 4 k^(− 3)))$]


        
          #math.equation(block: true, alt: "the fraction k to the power 7 over 64")[$display(frac(k^(7), 64))$]


        
      
      
        #math.equation(block: true, alt: "open parenthesis 2 w to the power minus 3 close parenthesis open parenthesis 2 w to the power minus 3 close parenthesis to the power 5 open parenthesis minus 5 w squared close parenthesis")[$( 2 w^(− 3) ) ( 2 w^(− 3) )^(5) ( − 5 w^(2) )$]


      
      
        #math.equation(block: true, alt: "the fraction 8 a to the power minus 3 / 4 over a to the power minus 11 / 4")[$display(frac(8 a^(− 3 / 4), a^(− 11 / 4)))$]


        
          #math.equation(block: true, alt: "8 a squared")[$8 a^(2)$]


        
      
      
        #math.equation(block: true, alt: "b to the power 2 / 3 open parenthesis 4 b to the power minus 2 / 3 minus b to the power 1 / 3 close parenthesis")[$b^(2 / 3) ( 4 b^(− 2 / 3) − b^(1 / 3) )$]


      
      
        When the Concorde landed at Heathrow Airport in London, the width, #math.equation(block: false, alt: "w")[$w$], of the sonic boom felt on the ground is given in kilometers by the following formula:


        #math.equation(block: true, alt: "w equals 4 open parenthesis the fraction T h over m close parenthesis to the power 1 / 2")[$w = 4 attach(( frac(T h, m) ), t: 1 / 2)$]
        where #math.equation(block: false, alt: "T")[$T$] stands for the temperature on the ground in kelvins, #math.equation(block: false, alt: "h")[$h$] is the altitude of the Concorde when it breaks the sound barrier, and #math.equation(block: false, alt: "m")[$m$] is the drop in temperature for each gain in altitude of one kilometer.


        + Find the width of the sonic boom if the ground temperature was #math.equation(block: false, alt: "293")[$293$] K, the altitude of the Concorde was #math.equation(block: false, alt: "15")[$15$] kilometers, and the temperature drop was #math.equation(block: false, alt: "4")[$4$] K per kilometer of altitude.
+ Graph #math.equation(block: false, alt: "w")[$w$] as a function of #math.equation(block: false, alt: "h")[$h$] if #math.equation(block: false, alt: "T equals 293")[$T = 293$] and #math.equation(block: false, alt: "m equals 4")[$m = 4$].


        
          + #math.equation(block: false, alt: "132.6")[$132.6$] km
+ #figure(figph[square root function], alt: "square root function", caption: none)


        
      
      
        The manager of an office supply store must decide how many of each item in stock she should order. The Wilson lot size formula gives the most cost-efficient quantity, #math.equation(block: false, alt: "Q")[$Q$], as a function of the cost, #math.equation(block: false, alt: "C")[$C$], of placing an order, the number of items, #math.equation(block: false, alt: "N")[$N$], sold per week, and the weekly inventory cost, #math.equation(block: false, alt: "I")[$I$], per item (cost of storage, maintenance, and so on).


        #math.equation(block: true, alt: "Q equals open parenthesis the fraction 2 C N over I close parenthesis to the power 1 / 2")[$Q = attach(( frac(2 C N, I) ), t: 1 / 2)$]
        + How many reams of computer paper should she order if she sells on average #math.equation(block: false, alt: "80")[$80$] reams per week, the weekly inventory cost for a ream is \$#math.equation(block: false, alt: "0.20")[$0.20$], and the cost of ordering, including delivery charges, is \$#math.equation(block: false, alt: "25")[$25$]?
+ Graph #math.equation(block: false, alt: "Q")[$Q$] as a function of #math.equation(block: false, alt: "N")[$N$] if #math.equation(block: false, alt: "C equals 25")[$C = 25$] and #math.equation(block: false, alt: "I equals 0.2")[$I = 0.2$].


      
      
        Two businesswomen start a small company to produce saddle bags for bicycles. The number of saddle bags, #math.equation(block: false, alt: "q")[$q$], they can produce depends on the amount of money, #math.equation(block: false, alt: "m")[$m$], they invest and the number of hours of labor, #math.equation(block: false, alt: "w")[$w$], they employ, according to the Cobb-Douglas formula


        #math.equation(block: true, alt: "q equals 0.6 m to the power 1 / 4 w to the power 3 / 4")[$q = 0.6 m^(1 / 4) w^(3 / 4)$]
        where #math.equation(block: false, alt: "m")[$m$] is measured in thousands of dollars.


        + If the businesswomen invest \$#math.equation(block: false, alt: "100 , 000")[$100 , 000$] and employ #math.equation(block: false, alt: "1600")[$1600$] hours of labor in their first month of production, how many saddle bags can they expect to produce?
+ With the same initial investment, how many hours of labor would they need in order to produce #math.equation(block: false, alt: "200")[$200$] saddle bags?


        
          + #math.equation(block: false, alt: "480")[$480$]
+ #math.equation(block: false, alt: "498")[$498$]


        
      
      
        A child who weighs #math.equation(block: false, alt: "w")[$w$] pounds and is #math.equation(block: false, alt: "h")[$h$] inches tall has a surface area (in square inches) given approximately by


        #math.equation(block: true, alt: "S equals 8.5 h to the power 0.35 w to the power 0.55")[$S = 8.5 h^(0.35) w^(0.55)$]
        + What is the surface area of a child who weighs #math.equation(block: false, alt: "60")[$60$] pounds and is #math.equation(block: false, alt: "40")[$40$] inches tall?
+ What is the weight of a child who is #math.equation(block: false, alt: "50")[$50$] inches tall and whose surface area is #math.equation(block: false, alt: "397")[$397$] square inches?


      
      
        The cost, #math.equation(block: false, alt: "C")[$C$], of insulating the ceiling in a building depends on the thickness of the insulation and the area of the ceiling. The table shows values of #math.equation(block: false, alt: "C equals f open parenthesis t , A close parenthesis")[$C = f ( t , A )$], where #math.equation(block: false, alt: "t")[$t$] is the thickness of the insulation and #math.equation(block: false, alt: "A")[$A$] is the area of the ceiling.


        #figure(table(
  columns: 7, align: left, inset: 6pt,
  [Cost of Insulation (dollars)], [], [], [], [], [], [],
  [], [Area (sq m)], [], [], [], [], [],
  [Thickness #linebreak() (cm) #linebreak()], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "400")[$400$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "600")[$600$]],
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "72")[$72$]], [#math.equation(block: false, alt: "144")[$144$]], [#math.equation(block: false, alt: "216")[$216$]], [#math.equation(block: false, alt: "288")[$288$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "432")[$432$]],
  [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "180")[$180$]], [#math.equation(block: false, alt: "270")[$270$]], [#math.equation(block: false, alt: "360")[$360$]], [#math.equation(block: false, alt: "450")[$450$]], [#math.equation(block: false, alt: "540")[$540$]],
  [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "108")[$108$]], [#math.equation(block: false, alt: "216")[$216$]], [#math.equation(block: false, alt: "324")[$324$]], [#math.equation(block: false, alt: "432")[$432$]], [#math.equation(block: false, alt: "540")[$540$]], [#math.equation(block: false, alt: "648")[$648$]],
  [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "126")[$126$]], [#math.equation(block: false, alt: "252")[$252$]], [#math.equation(block: false, alt: "378")[$378$]], [#math.equation(block: false, alt: "504")[$504$]], [#math.equation(block: false, alt: "630")[$630$]], [#math.equation(block: false, alt: "756")[$756$]],
  [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "144")[$144$]], [#math.equation(block: false, alt: "288")[$288$]], [#math.equation(block: false, alt: "432")[$432$]], [#math.equation(block: false, alt: "576")[$576$]], [#math.equation(block: false, alt: "720")[$720$]], [#math.equation(block: false, alt: "864")[$864$]],
  [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "162")[$162$]], [#math.equation(block: false, alt: "324")[$324$]], [#math.equation(block: false, alt: "486")[$486$]], [#math.equation(block: false, alt: "648")[$648$]], [#math.equation(block: false, alt: "810")[$810$]], [#math.equation(block: false, alt: "972")[$972$]],
))


        + What does it cost to insulate a ceiling with an area of #math.equation(block: false, alt: "500")[$500$] square meters with #math.equation(block: false, alt: "5")[$5$] cm of insulation? Write your answer in function notation.
+ Solve the equation #math.equation(block: false, alt: "864 equals f open parenthesis t , 600 close parenthesis")[$864 = f ( t , 600 )$] and interpret your answer.
+ Consider the row corresponding to a thickness of #math.equation(block: false, alt: "4")[$4$] cm. How does the cost of insulating the ceiling depend on the area of the ceiling?
+ Consider the column corresponding to an area of #math.equation(block: false, alt: "100")[$100$] square meters. How does the cost depend on the thickness of the insulation?
+ Given that the cost varies jointly with the thickness of the insulation and the area of the ceiling, write an equation for cost as a function of area and thickness of insulation.
+ Use your formula from part (e) to determine the cost of insulating a building with #math.equation(block: false, alt: "10")[$10$] centimeters of insulation if the area of the ceiling is #math.equation(block: false, alt: "800")[$800$] square meters.


        
          + \$#math.equation(block: false, alt: "450")[$450$]
+ #math.equation(block: false, alt: "t equals 8")[$t = 8$]: It costs \$#math.equation(block: false, alt: "864")[$864$] to insulate a ceiling with #math.equation(block: false, alt: "8")[$8$] cm of insulation over an area of #math.equation(block: false, alt: "600")[$600$] square meters.
+ #math.equation(block: false, alt: "C equals 0.72 A")[$C = 0.72 A$]
+ #math.equation(block: false, alt: "C equals 18 T")[$C = 18 T$]
+ #math.equation(block: false, alt: "C equals 0.18 A T")[$C = 0.18 A T$]
+ \$#math.equation(block: false, alt: "1440")[$1440$]


        
      
      
        The volume, #math.equation(block: false, alt: "V")[$V$], of a quantity of helium depends on both the temperature and the pressure of the gas. The table shows values of #math.equation(block: false, alt: "V equals f open parenthesis P , T close parenthesis")[$V = f ( P , T )$] for temperature in kelvins and pressure in atmospheres.


        #figure(table(
  columns: 7, align: left, inset: 6pt,
  [Volume (cubic meters)], [], [], [], [], [], [],
  [], [Temperature (K)], [], [], [], [], [],
  [Pressure #linebreak() (atmospheres) #linebreak()], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "150")[$150$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "250")[$250$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "350")[$350$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "27")[$27$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "45")[$45$]], [#math.equation(block: false, alt: "54")[$54$]], [#math.equation(block: false, alt: "63")[$63$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "13.5")[$13.5$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "22.5")[$22.5$]], [#math.equation(block: false, alt: "27")[$27$]], [#math.equation(block: false, alt: "31.5")[$31.5$]],
  [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "21")[$21$]],
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "4.5")[$4.5$]], [#math.equation(block: false, alt: "6.75")[$6.75$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "11.25")[$11.25$]], [#math.equation(block: false, alt: "13.5")[$13.5$]], [#math.equation(block: false, alt: "15.75")[$15.75$]],
))


        + What is the volume of helium when the pressure is #math.equation(block: false, alt: "4")[$4$] atmospheres and the temperature is #math.equation(block: false, alt: "350")[$350$] K? Write your answer in function notation.
+ Solve the equation #math.equation(block: false, alt: "15 equals f open parenthesis 3 , T close parenthesis")[$15 = f ( 3 , T )$] and interpret your answer.
+ Consider the row corresponding to #math.equation(block: false, alt: "2")[$2$] atmospheres. How is the volume related to the absolute temperature?
+ Consider the column corresponding to #math.equation(block: false, alt: "300")[$300$] K. How is the volume related to the pressure?
+ Given that the volume of the gas varies directly with temperature and inversely with pressure, write an equation for volume as a function of temperature and pressure.
+ Use your formula from part (e) to determine the volume of the helium at #math.equation(block: false, alt: "50")[$50$] K and pressure of #math.equation(block: false, alt: "0.4")[$0.4$] atmospheres.


      
      
        In his hiking guidebook, #emph[Afoot and Afield in Los Angeles County], Jerry Schad notes that the number of people on a wilderness trail is inversely proportional to "the square of the distance and the cube of the elevation gain from the nearest road."


        + Choose variables and write a formula for this relationship.
+ On a sunny Saturday afternoon, you count #math.equation(block: false, alt: "42")[$42$] people enjoying the Rock Pool at Malibu Creek State Park. The Rock Pool is #math.equation(block: false, alt: "1.5")[$1.5$] miles from the main parking lot, and the trail includes an elevation gain of #math.equation(block: false, alt: "250")[$250$] feet. Calculate the constant of variation in your formula from part (a).  #linebreak() Hint: Convert the elevation gain to miles.
+ Lookout Trail leads #math.equation(block: false, alt: "1.9")[$1.9$] miles from the parking lot and involves an elevation gain of #math.equation(block: false, alt: "500")[$500$] feet. How many people would you expect to encounter at the end of the trail?


        
          + #math.equation(block: false, alt: "N equals the fraction k over d squared E cubed")[$N = display(frac(k, d^(2) E^(3)))$], where #math.equation(block: false, alt: "N")[$N$] is number of people, #math.equation(block: false, alt: "d")[$d$] is distance in miles from the road, #math.equation(block: false, alt: "E")[$E$] is the elevation gain, and #math.equation(block: false, alt: "k")[$k$] is the constant of variation.
+ #math.equation(block: false, alt: "k approximately equals 0.01")[$k ≈ 0.01$]
+ #math.equation(block: false, alt: "3")[$3$]


        
      
      
        A company's monthly production, #math.equation(block: false, alt: "P")[$P$], depends on the capital, #math.equation(block: false, alt: "C")[$C$], the company has invested and the amount of labor, #math.equation(block: false, alt: "L")[$L$], available each month. The Cobb-Douglas model for production assumes that #math.equation(block: false, alt: "P")[$P$] varies jointly with #math.equation(block: false, alt: "C to the power a")[$C^(a)$] and #math.equation(block: false, alt: "L to the power b")[$L^(b)$], where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are positive constants less than #math.equation(block: false, alt: "1")[$1$]. The Aztech Chip Company invested #math.equation(block: false, alt: "625")[$625$] units of capital and hired #math.equation(block: false, alt: "256")[$256$] workers and produces #math.equation(block: false, alt: "8000")[$8000$] computer chips each month.


        + Suppose that #math.equation(block: false, alt: "a equals 0.25")[$a = 0.25$], #math.equation(block: false, alt: "b equals 0.75")[$b = 0.75$]. Find the constant of variation and a formula giving #math.equation(block: false, alt: "P")[$P$] in terms of #math.equation(block: false, alt: "C")[$C$] and #math.equation(block: false, alt: "L")[$L$].
+ If Aztech increases its labor force to #math.equation(block: false, alt: "300")[$300$] workers, what production level can they expect?
+ If Aztech maintains its labor force at #math.equation(block: false, alt: "256")[$256$] workers, what amount of capital outlay would be required for monthly production to reach #math.equation(block: false, alt: "16 , 000")[$16 , 000$] computer chips?