User:Elboig

\novocalize \<al-salAm `alaykum>

S______ L____
MATH 1314

Lab 3

Page 440 (117)
${\begin{aligned}D&=5e^{-0.4h}\\(2)&=5e^{-0.4h}\\0.4&=e^{-0.4h}\\-0.4h&=ln0.4\\h&={\frac {ln0.4}{-0.4}}\approx 2.29{\mbox{ hours.}}\end{aligned}}$

Page 454 (11)
${\begin{aligned}lnx+ln(x+2)&=4\\ln((x+2)(x))&=4\\ln(x^{2}+2x)&=4\\x^{2}+2x&=e^{4}\\x^{2}+2x-e^{4}&=0\\\end{aligned}}$

${\begin{aligned}x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\\x&={\frac {-(2)\pm {\sqrt {(2)^{2}-4(1)(-e^{4})}}}{2(1)}}\\x&={\frac {-2\pm {\sqrt {4+4e^{4}}}}{2}}\\x&={\frac {-2\pm {\sqrt {4(1+e^{4})}}}{2}}\\x&={\frac {-2\pm 2{\sqrt {1+e^{4}}}}{2}}\\x&={\frac {2(-1\pm 1{\sqrt {1+e^{4}}})}{2}}\\x&=-1\pm {\sqrt {1+e^{4}}}\\\end{aligned}}$
${\begin{aligned}x&=-1+{\sqrt {1+e^{4}}}&x&=-1-{\sqrt {1+e^{4}}}\\x&\approx 6.456&x&\approx -8.456\\\end{aligned}}$

${\begin{aligned}\{6.456\}{\mbox{ Since substituting -8.456 for x causes the natural logs to be undefined.}}\\\end{aligned}}$

Lab 6-7

Page 463 (31)
${\begin{aligned}A&=P(1+{\frac {r}{n}})^{nt}\\2P&=P(1+{\frac {.08}{12}})^{12t}\\2&=(1+{\frac {.08}{12}})^{12t}\\2&=({\frac {12.08}{12}})^{12t}\\12t&=log_{\frac {12.08}{12}}2\\t&={\frac {log_{\frac {12.08}{12}}2}{12}}\approx 8.69{\mbox{ years}}\\\end{aligned}}$

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Page 463 (33)

(a)
${\begin{aligned}A&=P(1+{\frac {r}{n}})^{nt}\\150&=100(1+{\frac {.08}{12}})^{12t}\\1.5&=({\frac {12.08}{12}})^{12t}\\log1.5&=log({\frac {12.08}{12}})^{12t}\\log1.5&=12t\cdot log({\frac {12.08}{12}})\\t&={\frac {log{1.5}}{12log({\frac {12.08}{12}})}}&\approx 5.09{\mbox{ years}}\\\end{aligned}}$

(b)

${\begin{aligned}A&=Pe^{rt}\\150&=100e^{.08t}\\1.5&=e^{.08t}\\.08t&=ln{1.5}\\t&={\frac {ln{1.5}}{.08}}\approx 5.07{\mbox{ years}}\\\end{aligned}}$

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Page 472(1)

(a)
${\begin{aligned}P(t)&=500e^{0.02t}\\P(0)&=500e^{(0.02)(0)}=500{\mbox{ flies}}\\\end{aligned}}$

(b)
${\begin{aligned}&0.02{\mbox{ or }}2\%\\\end{aligned}}$

(c)
${\begin{aligned}P(10)&=500e^{(0.02)(10)}\approx 611{\mbox{ flies}}\\\end{aligned}}$

(d)
${\begin{aligned}P(t)&=500e^{0.02t}\\800&=500e^{0.02t}\\1.6&=e^{0.02t}\\0.02t&=ln{1.6}\\t&={\frac {ln{1.6}}{0.02}}\approx 23.5{\mbox{ days}}\\\end{aligned}}$

(e)
${\begin{aligned}P(t)&=500e^{0.02t}\\1000&=500e^{0.02t}\\2&=e^{0.02t}\\0.02t&=ln{2}\\t&={\frac {ln{2}}{0.02}}\approx 34.7{\mbox{ days}}\\\end{aligned}}$

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Page 472 (3)

(a)
${\begin{aligned}0.0244{\mbox{ or }}2.44\%\\\end{aligned}}$

(b)
${\begin{aligned}A(10)&=500e^{(-0.0244)(10)}\\A(10)&=500e^{-0.244}\approx 391.744{\mbox{ grams}}\\\end{aligned}}$

(c)
${\begin{aligned}400&=500e^{-0.0244t}\\.8&=e^{-0.0244t}\\-0.0244t&=ln{.8}\\t&={\frac {ln{.8}}{-0.0244}}\approx 9.145{\mbox{ grams}}\\\end{aligned}}$

(d)
${\begin{aligned}250&=500e^{-0.0244t}\\.5&=e^{.0.0244t}\\-0.0244t&=ln.5\\t&={\frac {ln.5}{-0.0244}}\approx 28.4{\mbox{ days}}\\\end{aligned}}$

Lab 8-9

Page 563 (77)
${\begin{aligned}30x+15y+3z&=78\\20x+2y+25z&=59\\2x+20y+32z&=75\\\end{aligned}}$

${\begin{bmatrix}30&15&3&78\\20&2&25&59\\2&20&32&75\\\end{bmatrix}}$

${\begin{bmatrix}1&0&0&1.5\\0&1&0&2\\0&0&1&1\\\end{bmatrix}}$

${\begin{aligned}x&=4000\\y&=4000\\z&=2000\\\end{aligned}}$

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Page 563 (79)
Let $x$ = amount in T bills.
Let $y$ = amount in T bonds.
Let $z$ = amount in c bonds.
${\begin{aligned}x&+y&+&z&&=10000\\.06x&+.07y&+&.08z&&=680\\x&&-&2z&&=0\\\end{aligned}}$

${\begin{bmatrix}1&1&1&10000\\.06&.07&.08&680\\1&0&-2&0\\\end{bmatrix}}$

${\begin{bmatrix}1&0&0&4000\\0&1&0&4000\\0&0&1&2000\\\end{bmatrix}}$

${\begin{aligned}x&=4000\\y&=4000\\z&=2000\\\end{aligned}}$

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Page 590 (53)

${\begin{aligned}A&={\begin{bmatrix}1&-1&1\\0&-2&1\\-2&-3&0\\\end{bmatrix}}X&={\begin{bmatrix}x\\y\\z\\\end{bmatrix}}B&={\begin{bmatrix}2\\2\\.5\\\end{bmatrix}}\end{aligned}}$
${\begin{aligned}A^{-1}B&={\begin{bmatrix}.5\\-.5\\1\end{bmatrix}}\end{aligned}}$
${\begin{aligned}x&=0.5\\y&=-0.5\\z&=1\\\end{aligned}}$

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Page 590 (57)

$A={\begin{bmatrix}1&1&1\\3&2&-1\\3&1&2\\\end{bmatrix}}B={\begin{bmatrix}2\\{\frac {7}{3}}\\{\frac {10}{3}}\\\end{bmatrix}}$
$a^{-1}B={\begin{bmatrix}{\frac {1}{3}}\\1\\{\frac {2}{3}}\\\end{bmatrix}}$
${\begin{aligned}x&={\frac {1}{3}}\\y&=1\\z&={\frac {2}{3}}\\\end{aligned}}$

Labs 4 & 10

Page 639 (13)

${\begin{aligned}a_{n}&=(-1)^{n+1}n^{2}\\a_{1}&=(-1)^{1+1}1^{2}=1\\a_{2}&=...=-4\\a_{3}&=...=9\\a_{4}&=...=-16\\a_{5}&=...=25\\\end{aligned}}$

Note: I made a simple program to help me on this one, adapting it for the next problem:

    Input x 

    ((-1)^(x+1))x^2

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Page 639 (15)

${\begin{aligned}a_{n}&={\frac {2^{n}}{3^{n}+1}}\\a_{1}&={\frac {2^{1}}{3^{1}+1}}={\frac {2}{4}}={\frac {1}{2}}\\a_{2}&=...={\frac {2}{5}}\\a_{3}&=...={\frac {2}{7}}\\a_{4}&=...={\frac {8}{41}}\\a_{5}&=...={\frac {8}{61}}\\\end{aligned}}$

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Page 646 (39)

${\begin{aligned}a&=2\\a_{n}&=70\\d&=2\\\end{aligned}}$

${\begin{aligned}n&=({\frac {a_{n}-a}{d}})+1\\&=({\frac {70-2}{2}})+1\\n&=35\\\end{aligned}}$

${\begin{aligned}S_{n}&={\frac {n}{2}}(a+a_{n})\\S_{35}&={\frac {35}{2}}(2+70)\\S_{35}&=1260\\\end{aligned}}$

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Page 646 (41)

${\begin{aligned}a&=5\\a_{n}&=49\\d&=4\\\end{aligned}}$

${\begin{aligned}n&=({\frac {a_{n}-a}{d}})+1\\n&=({\frac {49-5}{4}})+1\\n&=12\\\end{aligned}}$

${\begin{aligned}S_{n}&={\frac {n}{2}}(a+a_{n})\\S_{12}&={\frac {12}{2}}(5+49)\\S_{12}&=324\\\end{aligned}}$

Labs 11 & 12

Page 657 (59)

${\begin{aligned}a&=1\\r&={\frac {1}{3}}\\{\mbox{ }}\\S&={\frac {a}{1-r}}={\frac {1}{1-{\frac {1}{3}}}}={\frac {1}{\frac {2}{3}}}=1\cdot {\frac {3}{2}}={\frac {3}{2}}\\\end{aligned}}$

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Page 657 (61)

${\begin{aligned}a&=8\\r&={\frac {1}{2}}\\{\mbox{ }}\\S&={\frac {a}{1-r}}={\frac {8}{1-{\frac {1}{2}}}}={\frac {8}{\frac {1}{2}}}=8\cdot {\frac {2}{1}}=16\\\end{aligned}}$