MATH 103 Solutions - Section 1.1 1 then x 1 1(a f(\u2212a = =\u2212 \u2212a a 1(b f(a\u22121 = \u22121 = a a \u221a 1 1(c f a = \u221a = 1\/2 = a\u22121\/2 a a C01S01.001 If f(x

MATH 103 Solutions - Section 1.1 1 then x 1 1(a f(u2212a =...

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Unformatted text preview: Section 1.1 1 , then: x 1 1 (a) f (−a) = =− ; −a a 1 (b) f (a−1 ) = −1 = a; a √ 1 1 (c) f ( a ) = √ = 1/2 = a−1/2 ; a a C01S01.001: If f (x) = (d) f (a2 ) = 1 = a−2 . a2 C01S01.002: If f (x) = x2 + 5, then: (a) f (−a) = (−a)2 + 5 = a2 + 5; (b) f (a−1 ) = (a−1 )2 + 5 = a−2 + 5 = 1 1 + 5a2 +5= ; 2 a a2 √ √ 2 (c) f ( a ) = ( a ) + 5 = a + 5; (d) f (a2 ) = (a2 )2 + 5 = a4 + 5. 1 , then: x2 + 5 1 1 (a) f (−a) = = 2 ; (−a)2 + 5 a +5 C01S01.003: If f (x) = (b) f (a−1 ) = 1 a2 1 1 · a2 = ; = = (a−1 )2 + 5 a−2 + 5 a−2 · a2 + 5 · a2 1 + 5a2 √ 1 1 (c) f ( a ) = √ 2 = ; a+5 ( a) + 5 (d) f (a2 ) = 1 (a2 )2 +5 = a4 1 . +5 √ C01S01.004: If f (x) = 1 + x2 + x4 , then: ! √ (a) f (−a) = 1 + (−a)2 + (−a)4 = 1 + a2 + a4 ; (b) f (a −1 )= ! 1+ (a−1 )2 + (a−1 )4 = √ 1+ a−2 + a−4 = √ √ a4 + a2 + 1 a4 + a2 + 1 a4 + a2 + 1 √ = = = ; a4 a2 a4 # √ √ 2 √ 4 √ (c) f ( a ) = 1 + ( a ) + ( a ) = 1 + a + a2 ; ! √ (d) f (a2 ) = 1 + (a2 )2 + (a4 )2 = 1 + a4 + a8 . " (a4 ) · (1 + a−2 + a−4 ) a4 " C01S01.005: If g(x) = 3x + 4 and g(a) = 5, then 3a + 4 = 5, so 3a = 1; therefore a = 13 . C01S01.006: If g(x) = 1 and g(a) = 5, then: 2x − 1 1 1 = 5; 2a − 1 1 = 5 · (2a − 1); 1 = 10a − 5; 10a = 6; a= C01S01.007: If g(x) = √ 3 . 5 x2 + 16 and g(a) = 5, then: ! a2 + 16 = 5; a2 + 16 = 25; a2 = 9; a = 3 or a = −3. C01S01.008: If g(x) = x3 − 3 and g(a) = 5, then a3 − 3 = 5, so a3 = 8. Hence a = 2. C01S01.009: If g(x) = √ 3 x + 25 = (x + 25)1/3 and g(a) = 5, then (a + 25)1/3 = 5; a + 25 = 53 = 125; a = 100. C01S01.010: If g(x) = 2x2 − x + 4 and g(a) = 5, then: 2a2 − a + 4 = 5; 2a2 − a − 1 = 0; (2a + 1)(a − 1) = 0; 2a + 1 = 0 or a − 1 = 0; a=− 1 or a = 1. 2 C01S01.011: If f (x) = 3x − 2, then f (a + h) − f (a) = [3(a + h) − 2] − [3a − 2] = 3a + 3h − 2 − 3a + 2 = 3h. C01S01.012: If f (x) = 1 − 2x, then 2 f (a + h) − f (a) = [1 − 2(a + h)] − [1 − 2a] = 1 − 2a − 2h − 1 + 2a = −2h. C01S01.013: If f (x) = x2 , then f (a + h) − f (a) = (a + h)2 − a2 = a2 + 2ah + h2 − a2 = 2ah + h2 = h · (2a + h). C01S01.014: If f (x) = x2 + 2x, then f (a + h) − f (a) = [(a + h)2 + 2(a + h)] − [a2 + 2a] = a2 + 2ah + h2 + 2a + 2h − a2 − 2a = 2ah + h2 + 2h = h · (2a + h + 2). C01S01.015: If f (x) = 1 , then x f (a + h) − f (a) = = C01S01.016: If f (x) = 1 1 a a+h − = − a+h a a(a + h) a(a + h) a − (a + h) −h = . a(a + h) a(a + h) 2 , then x+1 f (a + h) − f (a) = 2 2 2(a + 1) 2(a + h + 1) − = − a+h+1 a+1 (a + h + 1)(a + 1) (a + h + 1)(a + 1) = 2a + 2 2a + 2h + 2 (2a + 2) − (2a + 2h + 2) − = (a + h + 1)(a + 1) (a + h + 1)(a + 1) (a + h + 1)(a + 1) = 2a + 2 − 2a − 2h − 2 −2h = . (a + h + 1)(a + 1) (a + h + 1)(a + 1) C01S01.017: If x > 0 then f (x) = x x = = 1. |x| x If x < 0 then f (x) = x x = = −1. |x| −x We are given f (0) = 0, so the range of f is {−1, 0, 1}. That is, the range of f is the set consisting of the three real numbers −1, 0, and 1. C01S01.018: Given f (x) = [[3x]], we see that 3 f (x) = 0 if 0 ! x < 13 , f (x) = 1 if 1 3 ! x < 23 , f (2) = 2 if 2 3 ! x < 1; moreover, f (x) = −3 if − 1 ! x < − 23 , f (x) = −2 if − 2 3 ! x < − 13 , f (x) = −1 if − 1 3 ! x < 0. f (x) = 3m if m ! x < m + 13 , f (x) = 3m + 1 if m+ 1 3 ! x < m + 23 , f (x) = 3m + 2 if m+ 2 3 ! x < m + 1. In general, if m is any integer, then Because every integer is equal to 3m or to 3m + 1 or to 3m + 2 for some integer m, we see that the range of f includes the set Z of all integers. Because f can assume no values other than integers, we can conclude that the range of f is exactly Z. C01S01.019: Given f (x) = (−1)[[x]] , we first note that the values of the exponent [[x]] consist of all the integers and no other numbers. So all that matters about the exponent is whether it is an even integer or an odd integer, for if even then f (x) = 1 and if odd then f (x) = −1. No other values of f (x) are possible, so the range of f is the set consisting of the two numbers −1 and 1. C01S01.020: If 0 < x ! 1, then f (x) = 34. If 1 < x ! 2 then f (x) = 34 + 21 = 55. If 2 < x ! 3 then f (x) = 34 + 2 · 21 = 76. We continue in this way and conclude with the observation that if 11 < x < 12 then f (x) = 34 + 11 · 21 = 265. So the range of f is the set {34, 55, 76, 97, 118, 139, 160, 181, 202, 223, 244, 265}. C01S01.021: Given f (x) = 10 − x2 , note that for every real number x, x2 is defined, and for every such real number x2 , 10 − x2 is also defined. Therefore the domain of f is the set R of all real numbers. C01S01.022: Given f (x) = x3 + 5, we note that for each real number x, x3 is defined; moreover, for each such real number x3 , x3 + 5 is also defined. Thus the domain of f is the set R of all real numbers. √ C01S01.023: √ Given f (t) = t2 , we observe that for every real number t, t2 is defined and nonnegative, and hence that t2 is defined as well. Therefore the domain of f is the set R of all real numbers. √ $√ %2 $√ %2 C01S01.024: Given g(t) = t , we observe that t is defined exactly when t " 0. In this case, t is also defined, and hence the domain of g is the set [0, +∞) of all nonnegative real numbers. √ C01S01.025: Given f (x) = 3x − 5, we note that 3x − 5 is defined for all real numbers x, but that its square root will be defined when and only when 3x − 5 is nonnegative; that &is, when% 3x − 5 " 0, so that x " 53 . So the domain of f consists of all those real numbers x in the interval 53 , +∞ . 4 √ C01S01.026: Given g(t) = 3 t + 4 = (t + 4)1/3 , we note that t + 4 is defined for every real number t and the cube root of t + 4 is defined for every possible resulting value of t + 4. Therefore the domain of g is the set R of all real numbers. √ C01S01.027: Given f (t) = 1 − 2t, we observe that 1 − 2t is defined for every real number t, but that its square root is defined only when 1 − 2t is nonnegative. We solve the $inequality ' 1 − 2t " 0 to find that f (t) is defined exactly when t ! 12 . Hence the domain of f is the interval −∞, 12 . C01S01.028: Given g(x) = 1 , (x + 2)2 we see that (x + 2)2 is defined for every real number x, but that g(x), its reciprocal, will be defined only when (x + 2)2 $= 0; that is, when x + 2 $= 0. So the domain of g consists of those real numbers x $= −2. C01S01.029: Given f (x) = 2 , 3−x we see that 3 − x is defined for all real values of x, but that f (x), double its reciprocal, is defined only when 3 − x $= 0. So the domain of f consists of those real numbers x $= 3. C01S01.030: Given g(t) = " 2 , 3−t it is necessary that 3 − t be both nonzero (so that its reciprocal is defined) and nonnegative (so that the square root is defined). Thus 3 − t > 0, and therefore the domain of g consists of those real numbers t < 3. √ C01S01.031: Given f (x) = x2 + 9, observe that for each real number x, x2 + 9 is defined and, moreover, is positive. So its square root is defined for every real number x. Hence the domain of f is the set R of all real numbers. C01S01.032: Given h(z) = √ 1 , 4 − z2 we note that 4 − z 2 is defined for every real number z, but that its square root will be defined only if 4 − z 2 " 0. Moreover, the square root cannot be zero, else its reciprocal will be undefined, so we need to solve the inequality 4 − z 2 > 0; that is, z 2 < 4. The solution is −2 < z < 2, so the domain of h is the open interval (−2, 2). ! √ √ C01S01.033: Given f (x) = 4 − x , note first that we require x " 0 in order that x be defined. In √ addition, we require 4 − x " 0 so that its square root will be defined as well. So we solve [simultaneously] √ x " 0 and x ! 4 to find that 0 ! x ! 16. So the domain of f is the closed interval [0, 16]. C01S01.034: Given f (x) = " 5 x+1 , x−1 we require that x $= 1 so that the fraction is defined. In addition we require that the fraction be nonnegative so that its square root will be defined. These conditions imply that both numerator and denominator be positive or that both be negative; moreover, the numerator may also be zero. But if the denominator is positive then the [larger] numerator will be positive as well; if the numerator is nonpositive then the [smaller] denominator will be negative. So the domain of f consists of those real numbers for which either x − 1 > 0 or x + 1 ! 0; that is, either x > 1 or x ! −1. So the domain of f is the union of the two intervals (−∞, −1] and (1, +∞). Alternatively, it consists of those real numbers x not in the interval (−1, 1]. C01S01.035: Given: g(t) = t . |t| This fraction will be defined whenever its denominator is nonzero, thus for all real numbers t $= 0. So the domain of g consists of the nonzero real numbers; that is, the union of the two intervals (−∞, 0) and (0, +∞). C01S01.036: If a square has edge length x, then its area A is given by A = x2 and its perimeter P is given by P = 4x. To express A in terms of P : x = 14 P ; A = x2 = Thus to express A as a function of P , we write A(P ) = $ 1 %2 = 4P 2 1 16 P . 0 ! P < +∞. 2 1 16 P , (It will be convenient later in the course to allow the possibility that P , x, and A are zero. If this produces an answer that fails to meet real-world criteria for a solution, then that possibility can simply be eliminated when the answer to the problem is stated.) C01S01.037: If a circle has radius r, then its circumference C is given by C = 2πr and its area A by A = πr2 . To express C in terms of A, we first express r in terms of A, then substitute in the formula for C: A = πr ; 2 r= C = 2πr = 2π " " A ; π " A =2 π √ π2 A = 2 πA. π Therefore to express C as a function of A, we write √ C(A) = 2 πA, 0 ! A < +∞. √ It is also permissible simply to write C(A) = 2 πA without mentioning the ! domain, because the “default” domain is correct. In the first displayed equation we do not write r = ± A/π because we know that r is never negative. C01S01.38: If r denotes the radius of the sphere, then its volume is given by V = 43 πr3 and its surface area by S = 4πr2 . Hence 6 r= V = Answer: V (S) = 1 π 6 ( )3/2 S , π 1 2 " S ; π 4 3 4 1 πr = π · 3 3 8 ( )3/2 ( )3/2 1 S S = π . π 6 π 0 ! S < +∞. C01S01.039: To avoid decimals, we note that a change of 5◦ C is the same as a change of 9◦ F, so when the temperature is 10◦ C it is 32 + 18 = 50◦ F; when the temperature is 20◦ C then it is 32 + 2 · 18 = 68◦ F. 1 In general we get the Fahrenheit temperature F by adding 32 to the product of 10 C and 18, where C is the Celsius temperature. That is, 9 F = 32 + C, 5 and therefore C = 59 (F − 32). Answer: C(F ) = 5 (F − 32), 9 F > −459.67. C01S01.040: Suppose that a rectangle has base length x and perimeter 100. Let h denote the height of such a rectangle. Then 2x + 2h = 100, so that h = 50 − x. Because x " 0 and h " 0, we see that 0 ! x ! 50. The area A of the rectangle is xh, so that A(x) = x(50 − x), 0 ! x ! 50. C01S01.041: Let y denote the height of such a rectangle. The rectangle is inscribed in a circle of diameter 4, so the bottom side x √ and the left side y√ are the two legs of a right triangle with hypotenuse 4. Consequently x2 + y 2 = 16, so y = 16 − x2 (not − 16 − x2 because y " 0). Because x " 0 and y " 0, we see that 0 ! x ! 4. The rectangle has area A = xy, so ! A(x) = x 16 − x2 , 0 ! x ! 4. C01S042.042: We take the problem to mean that current production is 200 barrels per day per well, that if one new well is drilled then the 21 wells will produce 195 barrels per day per well; in general, that if x new wells are drilled then the 20 + x wells will produce 200 − 5x barrels per day per well. So total production would be p = (20 + x)(200 − 5x) barrels per day. But because 200 − 5x " 0, we see that x ! 40. Because x " 0 as well (you don’t “undrill” wells), here’s the answer: 0 ! x ! 40, p(x) = 4000 + 100x − 5x2 , x an integer. C01S01.043: The square base of the box measures x by x centimeters; let y denote its height (in centimeters). Because the volume of the box is 324 cm3 , we see that x2 y = 324. The base of the box costs 2x2 cents, each of its four sides costs xy cents, and its top costs x2 cents. So the total cost of the box is C = 2x2 + 4xy + x2 = 3x2 + 4xy. 7 (1) Because x > 0 and y > 0 (the box has positive volume), but because y can be arbitrarily close to zero (as well as x), we see also that 0 < x < +∞. We use the equation x2 y = 324 to eliminate y from Eq. (1) and thereby find that C(x) = 3x2 + 1296 , x 0 < x < +∞. C01S01.044: If the rectangle is rotated around its side S of length x to produce a cylinder, then x will also be the height of the cylinder. Let y denote the length of the two sides perpendicular to S; then y will be the radius of the cylinder; moreover, the perimeter of the original rectangle is 2x + 2y = 36. Hence y = 18 − x. Note also that x " 0 and that x ! 18 (because y " 0). The volume of the cylinder is V = πy 2 x, and so 0 ! x ! 18. V (x) = πx(18 − x)2 , C01S01.045: Let h denote the height of the cylinder. Its radius is r, so its volume is πr2 h = 1000. The total surface area of the cylinder is A = 2πr2 + 2πrh h= 1000 , πr2 (look inside the front cover of the book); so A = 2πr2 + 2πr · 1000 2000 = 2πr2 + . 2 πr r Now r cannot be negative; r cannot be zero, else πr2 h $= 1000. But r can be arbitrarily small positive as well as arbitrarily large positive (by making h sufficiently close to zero). Answer: A(r) = 2πr2 + 2000 , r 0 < r < +∞. C01S01.046: Let y denote the height of the box (in centimeters). Then 2x2 + 4xy = 600, so that y= 600 − 2x2 . 4x (1) The volume of the box is V = x2 y = (600 − 2x2 ) · x2 1 1 = (600x − 2x3 ) = (300x − x3 ) 4x 4 2 by Eq. (1). Also x >√0 by Eq. √ (1), but the maximum value of x is attained when Eq. (1) forces y to be zero, at which point x = 300 = 10 3. Answer: V (x) = 300x − x3 , 2 √ 0 < x ! 10 3. C01S01.047: The base of the box will be a square measuring 50 − 2x in. on each side, so the open-topped box will have that square as its base and four rectangular sides each measuring 50 − 2x by x (the height of the box). Clearly 0 ! x and 2x ! 50. So the volume of the box will be V (x) = x(50 − 2x)2 , 8 0 ! x ! 25. C01S01.048: Recall that A(x) = x(50 − x), 0 ! x ! 50. Here is a table of a few values of the function A at some special numbers in its domain: x 0 A 0 5 225 10 400 15 525 20 600 25 625 30 600 35 525 40 400 45 225 50 0 It appears that when x = 25 (so the rectangle is a square), the rectangle has maximum area 625. C01S01.049: Recall that the total daily production of the oil field is p(x) = (20 + x)(200 − 5x) if x new wells are drilled (where x is an integer satisfying 0 ! x ! 40). Here is a table of all of the values of the production function p: x 0 p 4000 1 4095 2 4180 3 4 5 6 7 4255 4320 4375 4420 4455 x 8 p 4480 9 4495 10 4500 11 12 13 14 15 4495 4480 4455 4420 4375 x 16 p 4320 17 4255 18 4180 19 20 21 22 23 4095 4000 3895 3780 3655 x 24 p 3520 25 3375 26 3220 27 28 29 30 31 3055 2880 2695 2500 2295 x 32 p 2080 33 1855 34 1620 35 36 1375 1120 37 855 38 580 39 295 and, finally, p(40) = 0. Answer: Drill ten new wells. C01S01.050: The surface area A of the box of Example 8 was A(x) = 2x2 + 500 , x The restrictions x " 1 and y " 1 imply that 1 ! x ! three places, are given in the following table. x 1 A 502 2 258 3 185 4 157 0 < x < ∞. √ 125. A small number of values of A, rounded to 5 6 7 8 9 10 11 150 155 169 191 218 250 287 It appears that A is minimized when x = y = 5. C01S01.051: If x is an integer, then Ceiling(x) = x and −Floor(−x) = −(−x) = x. If x is not an integer, then choose the integer n so that n < x < n + 1. Then Ceiling(x) = n + 1, −(n + 1) < −x < −n, and −Floor(−x) = −[−(n + 1)] = n + 1. In both cases we see that Ceiling(x) = −Floor(−x). C01S01.052: The range of Round(x) is the set Z of all integers. If k is a nonzero constant, then as x varies through all real number values, so does kx. Hence the range of Round(kx) is Z if k $= 0. If k = 0 then the range of Round(kx) consists of the single number zero. 9 C01S01.053: By the result of Problem 52, the range of Round(10x) is the set of all integers, so the range 1 1 of g(x) = 10 Round(10x) is the set of all integral multiple of 10 . 1 C01S01.054: What works for π will work for every real number; let Round2(x) = 100 Round(100x). To be certain that this is correct (we will verify it only for positive numbers), write the [positive] real number x in the form x = k+ t h m + + + r, 10 100 1000 where k is a nonnnegative integer, t (the “tenths” digit) is a nonnegative integer between 0 and 9, h (the “hundredths” digit) is a nonnegative integer between 0 and 9, as is m, and 0 ! r < 0.001. Then Round2(x) = 1 100 Floor(100x + 0.5) = 1 100 Floor(100k + 10t + h + 1 10 (m + 5) + 100r). If 0 ! m ! 4, the last expression becomes 1 t h (100k + 10t + h) = k + + , 100 10 100 which is the correct two-digit rounding of x. If 5 ! m ! 9, it becomes 1 t h+1 (100k + 10t + h + 1) = k + + , 100 10 100 also the correct two-digit rounding of x in this case. 1 C01S01.055: Let Round4(x) = 10000 Round(10000x). To verify that Round4 has the desired property for [say] positive values of x, write such a number x in the form x=k+ d1 d2 d3 d4 d5 + + + + + r, 10 100 1000 10000 100,000 where k is a nonnegative integer, each di is an integer between 0 and 9, and 0 ! r < 0.00001. Application of Round4 to x then produces 1 Floor(10000k + 1000d1 + 100d2 + 10d3 + d4 + 10000 1 10 (d5 + 5) + 10000r). Then consideration of the two cases 0 ! d5 ! 4 and 5 ! d5 ! 9 will show that Round4 produces the correct four-place rounding of x in both cases. C01S01.056: Let Chop4(x) = 1 10000 Floor(10000x). x=k+ Suppose that x > 0. Write x in the form d1 d2 d3 d4 + + + + r, 10 100 1000 10000 where k is a nonnegative integer, each of the di is an integer between 0 and 9, and 0 ! r < 0.0001. Then Chop4(x) produces 1 10000 Floor(10000k = + 1000d1 + 100d2 + 10d3 + d4 + 10000r) 1 10000 (10000k + 1000d1 + 100d2 + 10d3 + d4 ) 10 because 10000r < 1. It follows that Chop4 has the desired effect. C01S01.057: x y 0.0 1.0 0.2 0.44 0.4 −0.04 0.6 −0.44 The sign change occurs between x = 0.2 and x = 0.4. x y 0.20 0.44 0.25 0.3125 0.30 0.19 0.35 0.0725 The sign change occurs between x = 0.35 and x = 0.40. x y 0.35 0.0725 0.36 0.0496 0.37 0.0269 0.38 0.0044 From this point on, the data for y will be rounded. x y 0.380 0.0044 0.382 −0.0001 0.384 −0.0045 0.386 −0.0090 0.8 −0.76 1.0 −1.0 0.40 −0.04 0.39 −0.0179 0.40 −0.04 0.388 −0.0135 0.390 −0.0179 Answer (rounded to two places): 0.38. The quadratic formula yields the two roots of these is approximately 0.381966011250105151795. 1 2 $ 3± √ % 5 ; the smaller Problems 58 through 66 are worked in the same way as Problem 57. C01S01.058: The sign change intervals are [2, 3], [2.6, 2.8], [2.60, 2.64], and [2.616, 2.624]. Answer: √ % $ 1 3 + 5 ≈ 2.62. 2 C01S01.059: √The sign change intervals are [1, 2], [1.2, 1.4], [1.20, 1.24], [1.232, 1.240], and [1.2352, 1.2368]. Answer: −1 + 5 ≈ 1.24. C01S01.060: The sign change intervals are [−4, −3], [−3.4, −3.2], [−3.24, −3.20], [−3.240, −3.232], and √ [−3.2368, −3.2352]. Answer: −1 − 5 ≈ −3.24. C01S01.061: The √ sign $ % change intervals are [0, 1], [0.6, 0.8], [0.68, 0.72], [0.712, 0.720], and [0.7184, 0.7200]. 1 Answer: 4 7 − 17 ≈ 0.72. C01S01.062: The √ sign $ % change intervals are [2, 3], [2.6, 2.8], [2.76, 2.80], [2.776, 2.784], and [2.7792, 2.7808]. Answer: 14 7 + 17 ≈ 2.78. C01S01.063: The √ sign $ % change intervals are [3, 4], [3.2, 3.4], [3.20, 3.24], [3.208, 3.216], and [3.2080, 3.2096]. 1 Answer: 2 11 − 21 ≈ 3.21. C01S01.064: The √ sign $ % change intervals are [7, 8], [7.6, 7.8], [7.76, 7.80], [7.784, 7.792], and [7.7904, 7.7920]. Answer: 12 11 + 21 ≈ 7.79. C01S01.065: The sign change intervals √are [1,% 2], [1.6, 1.8], [1.60, 1.64], [1.608, 1.616], [1.6144, 1.6160], $ and [1.61568, 1.61600]. Answer: 16 −23 + 1069 ≈ 1.62. C01S01.066: The sign change$ intervals −9], [−9.4, −9.2], [−9.32, −9.28], [−9.288, −9.280], and √ are [−10, % [−9.2832, −9.2816]. Answer: 16 −23 − 1069 ≈ −9.28. 11 Section 1.2 C01S02.001: The slope of L is m = (3 − 0)/(2 − 0) = 32 , so L has equation y−0= 3 (x − 0); 2 that is, 2y = 3x. C01S02.002: Because L is vertical and (7, 0) lies on L, every point on L has Cartesian coordinates (7, y) for some number y (and every such point lies on L). Hence an equ...
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