ISM: Introductory Mathematical Analysis
34. w = f(d) = ad + b, f(0) = 21, 6.3 a = slope = = 0.63. Thus 10 w = f(d) = 0.63d + b. Since f(0) = 21, we have 20 = 0.63(0) + b, so b = 21. Hence w = 0.63d + 21. When d = 55, then w = 0.63(55) + 21 = 34.65 + 21 = 55.65 kg.
Section 2.8
0.6 0.5 0.4 0.3 0.2 0.1 −2
−1
0
−0.1
1 2 3 4 5 Number of years after 2006
6
−0.2
t −t 80 − 68 12 1 m= 2 1 = = = . c2 − c1 172 − 124 48 4
35. a.
1 1 t − 68 = (c − 124) , t − 68 = c − 31 , or 4 4 1 t = c + 37 . 4
b. Since c is the number of chirps per minute, 1 1 then c is the number of chirps in 4 4 minute or 15 seconds. Thus from part (a), to estimate temperature add 37 to the number of chirps in 15 seconds.
h(t ) = −16t 2 + 32t + 28, a = –16, b = 32, and c = 8. Since a < 0, the parabola opens downward. The x -coordinate of the vertex is b 32 − =− = 1 . The 2a 2(−16) y-coordinate of the vertex is
( )
h(1) = −16 12 + 32(1) + 8 = 24 . Thus, the vertex
is (1, 24). Since c = 8, the y-intercept is (0, 8). To find the x-intercepts we set y = h(t) = 0. 0 = −16t 2 + 32t + 8 t=
Apply It 2.8 27. P ( x) is a quadratic function with a = −0.083, b = 0.39 and c = 0.09. Since a < 0, the graph is a parabola that opens downward.
The vertex is at x = −
28. In the quadratic function
b 0.39 =− = 2.35, 2a 2(−0.083)
and the y-value is P (2.35) = −0.083(2.35)2 + 0.39(2.35) + 0.09 ≈ 0.55.
=
−32 ± 322 − 4(−16)(8) −b ± b 2 − 4ac = 2a 2(−16)
−32 ± 1536 −32 ± 16 6 6 = = 1± −32 −32 2
⎛ 6 ⎞ , 0 ⎟ and Thus, the x-intercepts are ⎜1 + ⎜ ⎟ 2 ⎝ ⎠ ⎛ 6 ⎞ , 0⎟ . ⎜⎜1 − ⎟ 2 ⎝ ⎠ 30
The y-intercept is (0, 0.09). The x-intercepts are where P ( x) = 0 . Using the quadratic formula, we find
–5
−0.39 ± (0.39)2 − 4(−0.083)(0.09) 2(−0.083) ≈ −0.22, 4.92
x=
So the x-intercepts are (−0.22, 0) and (4.92, 0).
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