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When the fraction 1/288 is expressed in base 12,

is it terminating or repeating? Explain.

$\begin{array}{|rcll|} \hline \frac{1}{288} &=& a\cdot 12^{-1} + b\cdot 12^{-2}+ c\cdot 12^{-3}+ d\cdot 12^{-4}+ e\cdot 12^{-5} \dots \quad & | \quad \cdot 12 \\ \frac{12}{288}=\frac{1}{24} &=& a + \underbrace{ b\cdot 12^{-1}+ c\cdot 12^{-2}+ d\cdot 12^{-3}+ e\cdot 12^{-4} \dots }_{\text{decimal places}=\frac{1}{24}}\\ integer(\frac{1}{24}) &=& a \quad \Rightarrow \quad a = 0\\\\ \frac{1}{24} -a &=& b\cdot 12^{-1}+ c\cdot 12^{-2}+ d\cdot 12^{-3}+ e\cdot 12^{-4} \dots \\ \frac{1}{24} -0 &=& b\cdot 12^{-1}+ c\cdot 12^{-2}+ d\cdot 12^{-3}+ e\cdot 12^{-4} \dots \\ \frac{1}{24} &=& b\cdot 12^{-1}+ c\cdot 12^{-2}+ d\cdot 12^{-3}+ e\cdot 12^{-4} \dots \quad & | \quad \cdot 12 \\ \frac{12}{24}=\frac{1}{2} &=& b + \underbrace{ c\cdot 12^{-1}+ d\cdot 12^{-2}+ e\cdot 12^{-3} \dots }_{\text{decimal places}=\frac{1}{2}}\\ integer(\frac{1}{2}) &=& b \quad \Rightarrow \quad b = 0\\\\ \frac{1}{2}-b &=& c\cdot 12^{-1}+ d\cdot 12^{-2}+ e\cdot 12^{-3} \dots \\ \frac{1}{2}-0 &=& c\cdot 12^{-1}+ d\cdot 12^{-2}+ e\cdot 12^{-3} \dots \\ \frac{1}{2} &=& c\cdot 12^{-1}+ d\cdot 12^{-2}+ e\cdot 12^{-3} \dots \quad & | \quad \cdot 12 \\ \frac{12}{2}= 6.0000000\dots &=& c + \underbrace{ d\cdot 12^{-1}+ e\cdot 12^{-2} \dots }_{\text{decimal places}=0}\\ integer(6.0000000\dots) &=& c \quad \Rightarrow \quad c = 6 \\\\ \frac{1}{288} = 0.006_{12}\\ \hline \end{array}$

The product of two integers is -144.
The sum of the integers is -7.
What are the two integers?

$\begin{array}{|rcll|} \hline n_1 + n_2 &=& -7 \\ n_2 &=& -7 - n_1 \\\\ n_1\cdot n_2 &=& -144 \\ n_1\cdot (-7 - n_1) &=& -144 \\ -7\cdot n_1 - n_1^2 &=& -144 \quad & | \quad \cdot(-1)\\ 7\cdot n_1 + n_1^2 &=& 144 \quad & | \quad -144\\ 7\cdot n_1 + n_1^2 -144 &=& 0\\ n_1^2 + 7 n_1 -144 &=& 0\\ n_{1,2} &=& \frac{-7\pm\sqrt{(-7)^2-4\cdot(-144)} }{2} \\ n_{1,2} &=& \frac{-7\pm\sqrt{49+4\cdot 144} }{2} \\ n_{1,2} &=& \frac{-7\pm\sqrt{625} }{2} \quad & | \quad \sqrt{625} = 25\\ n_{1,2} &=& \frac{-7\pm 25 }{2}\\ n_1 = \frac{-7+25}{2} &\text{or}& n_1 = \frac{-7-25}{2} \\ n_1 = 9 &\text{or}& n_1 = -16 \\\\ n_2 = -7 - n_1 &\text{or}& n_2 = -7 - n_1 \\ n_2 = -7 - 9 &\text{or}& n_2 = -7 - (-16) \\ n_2 = -16 &\text{or}& n_2 = -7 +16 \\ & & n_2 =9 \\ \hline \end{array}$

The two integers are 9 and -16

(2n+3)! / (2n+4)!

$\begin{array}{|rcll|} \hline (2n+4)! &=& (2n+3)!\cdot (2n+3+1) \\ &=& (2n+3)!\cdot (2n+4) \\\\ && \frac{(2n+3)!} { (2n+4)!}\\ &=& \frac{(2n+3)!} { (2n+3)!\cdot (2n+4) }\\ &=& \frac{1} { (2n+4) }\\ \hline \end{array}$

f(t)=4cost+2sin2t, [0,pi/2]

What is ABSOLUTE MAX?

$\begin{array}{|rcll|} \hline f(t) &=& 4\cdot \cos(t) + 2\cdot \sin(2t) \qquad 0\le t\le \frac{\pi }{2}\\ f'(t) &=& 4\cdot[-\sin(t)] + 2\cdot \cos(2t) \cdot 2 \\ f'(t) &=& -4\cdot \sin(t) + 4\cdot \cos(2t) & | \quad \text{find min/max } f'(t) = 0 \\ 0 &=& -4\cdot \sin(t) + 4\cdot \cos(2t) & | \quad : 4 \\ 0 &=& - \sin(t) + \cos(2t) \\ \cos(2t) - \sin(t) &=& 0 & | \quad \cos(2t) = \cos^2(t)-\sin^2(t)\\ && & | \quad = 1-\sin^2(t)-\sin^2(t) \\ && & | \quad = 1-2\sin^2(t)\\ \cos(2t) - \sin(t) &=& 0 & | \quad \cos(2t) = 1-2\sin^2(t)\\ 1-2\sin^2(t) - \sin(t) &=& 0 & | \quad \cdot (-1) \\ -1+2\sin^2(t) + \sin(t) &=& 0 \\ 2\sin^2(t) + \sin(t) -1 &=& 0 & | \quad \text{substitute: } u = \sin(t)\\ 2u^2 + u -1 &=& 0 \\ u_{1,2} = \frac{ -1\pm \sqrt{(-1)^2-4\cdot2\cdot (-1)} } { 2\cdot 2 }\\ u_{1,2} = \frac{ -1\pm \sqrt{1+8} } {4 }\\ u_{1,2} = \frac{ -1\pm \sqrt{9} } {4 }\\ u_{1,2} = \frac{ -1\pm 3 } {4 }\\ u_1 &=& \frac{-1+3}{4} \\ u_1 & =& \frac12 \\ u_2 &=& \frac{-1-3}{4} \\ u_2 & =& -1 \\ t_1 &=& \arcsin(u_1) \\ t_1 &=& \arcsin(\frac12) \\ \mathbf{t_1 }& \mathbf{=} & \mathbf{30^{\circ} \text{ or } t_1 = \frac{\pi}{6} }\\ t_2 &=& \arcsin(u_2) \\ t_2 &=& \arcsin(-1) \\ t_2 &=&-90^{\circ} \text{ or } t_2 = -\frac{\pi}{2} & | \quad \text{no solution because } 0\le t\le \frac{\pi }{2}\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline f(t) &=& 4\cdot \cos(t) + 2\cdot \sin(2t) \\ f(30^{\circ}) &=& 4\cdot \cos(30^{\circ}) + 2\cdot \sin(2\cdot 30^{\circ}) \\ f(30^{\circ}) &=& 4\cdot \cos(30^{\circ}) + 2\cdot \sin(60^{\circ}) \\ f(30^{\circ}) &=& 4\cdot \frac{ \sqrt{3} } {2} + 2\cdot \frac{ \sqrt{3} } {2} \\ f(30^{\circ}) &=& 2\cdot \sqrt{3} + \sqrt{3} \\ f(30^{\circ}) &=& 3\cdot \sqrt{3} \\ f(30^{\circ}) &=& 5.19615242271 \\ \hline \end{array}$

Max or min at $\mathbf{t } \mathbf{=} \mathbf{30^{\circ} ( = \frac{\pi}{6} } ) \qquad f(30^{\circ}) = 5.196\dots \quad ?$
$\begin{array}{|rcll|} \hline y"(t) &=& -4\cdot \cos(t)-4\cdot\sin(2t)\cdot 2 \\ y"(t) &=& -4\cdot \cos(t)-8\cdot\sin(2t)\\ y"(30^{\circ}) &=& -4\cdot \cos(30^{\circ})-8\cdot\sin(2\cdot 30^{\circ})\\ y"(30^{\circ}) &=& -4\cdot \cos(30^{\circ})-8\cdot\sin(60^{\circ})\\ y"(30^{\circ}) &=& -4\cdot \frac{ \sqrt{3} } {2}-8\cdot\frac{ \sqrt{3} } {2}\\ y"(30^{\circ}) &=& -12\cdot \frac{ \sqrt{3} } {2} \\ y"(30^{\circ}) &=& -6\cdot \sqrt{3} & | \quad \text{maximum because } y''(30^{\circ}) < 0\\ \hline \end{array}$

Max at $\mathbf{t } \mathbf{=} \mathbf{30^{\circ} ( = \frac{\pi}{6} } ) \qquad f(30^{\circ}) = 5.196\dots \quad$

A tall tree; which is situated close to the point A on a site; must be retained. You have been asked to calculate the height of a tree you position a theodolite 45m from its base. The observed angle of elevation to the top of the tree is 21 degrees, 37', 12". Given that the instrument is 1.5m off the ground, calculate the height of the tree.

$\begin{array}{|rcll|} \hline \text{The height of the tree } &=& 1.5\ m + 45\ m \cdot \tan(21^{\circ}\ 37'\ 12 ") \\ &=& 1.5\ m + 45\ m \cdot \tan(21+\frac{37+\frac{12}{60}}{60 }) \\ &=& 1.5\ m + 45\ m \cdot \tan(21.62^{\circ} ) \\ &=& 1.5\ m + 45\ m \cdot 0.39633184969 \\ &=& 1.5\ m +17.8349332359\ m \\ &=& 19.3349332359\ m \\ \hline \end{array}$

The height of the tree is 19.335 m

3sqrt((cospi/2 + isinpi/2)).

can someone help me with this to get into polar form

$\begin{array}{|rcll|} \hline && 3\cdot \sqrt{ \cos(\frac{\pi}{2}) + i\cdot \sin(\frac{\pi}{2}) } \\ &=& 3\cdot [ \cos(\frac{\pi}{2}) + i\cdot \sin(\frac{\pi}{2}) ]^{\frac12}\\\\ && \text{Euler's formula: } & e^{ix} = \cos(x)+i\cdot \sin(x) \\ && & e^{i\frac{\pi}{2}} = \cos(\frac{\pi}{2})+i\cdot \sin(\frac{\pi}{2}) \\ &=& 3\cdot [ \cos(\frac{\pi}{2}) + i\cdot \sin(\frac{\pi}{2}) ]^{\frac12}\\ &=& 3\cdot ( e^{i\frac{\pi}{2}} )^{\frac12}\\ &=& 3\cdot e^{i\frac{\pi}{2}\cdot \frac12} \\ &=& 3\cdot e^{ i\frac{\pi}{4} } \\\\ && \text{Euler's formula: } & e^{ix} = \cos(x)+i\cdot \sin(x) \\ && & e^{i\frac{\pi}{4}} = \cos(\frac{\pi}{4})+i\cdot \sin(\frac{\pi}{4}) \\ &=& 3\cdot [ \cos(\frac{\pi}{4})+i\cdot \sin(\frac{\pi}{4}) ] \\ \hline \end{array}$

what is the next 2 terms -9 -8 -4 5 21 46

$\begin{array}{|rcll|} \hline a_1 &=& -9 \\ a_2 &=& -9 + 1^2 = -8 \\ a_3 &=& -9 + 1^2 + 2^2 = -4 \\ a_4 &=& -9 + 1^2 + 2^2 + 3^2= 5 \\ a_5 &=& -9 + 1^2 + 2^2 + 3^2 + 4^2 = 21 \\ a_6 &=& -9 + 1^2 + 2^2 + 3^2 + 4^2 +5^2 = 46 \\ a_7 &=& -9 + 1^2 + 2^2 + 3^2 + 4^2 +5^2 +6^2 = 82 \\ a_8 &=& -9 + 1^2 + 2^2 + 3^2 + 4^2 +5^2 +6^2 +7^2= 131 \\ \cdots \\ a_n &=& -9 +\underbrace{ (1^2 + 2^2 + 3^2 + 4^2 +5^2 +6^2 +7^2\cdots )}_{\text{sum of the square numbers}} \\ a_n &=& -9 + \underbrace{\frac{(n-1)\cdot n \cdot (n-\frac12 )}{3}}_{\text{sum of the square numbers}} \quad n = 1,2,3, \cdots \\ \hline \end{array}$

Bei einem Test hatten 4 Schüler ein "Sehr gut" ; das sind 16% der Klasse.

Wie viele schüler sind insgesamt in der Klasse?

$\begin{array}{|rcll|} \hline x \cdot 16\ \% &=& 4 \\ x \cdot \frac{16}{100} &=& 4 \qquad | \qquad \cdot \frac{100}{16} \\ x &=& 4 \cdot \frac{100}{16} \\ x &=& \frac{100}{4} \\ x &=& 25 \\ \hline \end{array}$

In der Klasse sind insgesamt 25 Schüler.

How many 4-digit numbers have no repeating digits?

1.  $\begin{array}{|lcll|} \hline \{0,1,2,3,4,5,6,7,8,9\} \\ \text{4-digit numbers}: \\ \binom{10}{4} * 4! = \frac{10!}{6!} = 7*8*9*10 = 5040 \\ \hline \end{array}$

2.  $\begin{array}{|lcll|} \hline \{1,2,3,4,5,6,7,8,9\} \\ \text{3-digit numbers}: \\ \binom{9}{3}* 3! = \frac{9!}{6!} = 7*8*9 = 504 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 5040 \qquad - \underbrace{504}_{\text{zero cannot be the first digit}} = 4536\\ \hline \end{array}$

(a + a^2 + a^3)^4=a^12+4 a^11+10 a^10+16 a^9+19 a^8+16 a^7+10 a^6+4 a^5+a^4
In the expanded part, the term "19a^8", how is the coefficient 19 arrived at? Thanks for any help.

Formula trinomial:

$\begin{array}{|rcll|} \hline (a+b+c)^n &=& \sum \limits_{i=0}^{n} \sum \limits_{j=0}^{n-i} \frac{n!}{i!~j!~(n-i-j)!} \cdot a^ib^jc^{n-i-j} \\ \hline \end{array}$

We have:

$\begin{array}{|rcll|} \hline (a + a^2 + a^3)^4 &=& \sum \limits_{i=0}^{4} \sum \limits_{j=0}^{4-i} \frac{4!}{i!~j!~(4-i-j)!} \cdot a^i(a^2)^j(a^3)^{4-i-j} \\ &=& \sum \limits_{i=0}^{4} \sum \limits_{j=0}^{4-i} \frac{4!}{i!~j!~(4-i-j)!} \cdot a^{i+2j+3\cdot(4-i-j)} \\ &=& \sum \limits_{i=0}^{4} \sum \limits_{j=0}^{4-i} \frac{4!}{i!~j!~(4-i-j)!} \cdot a^{12-2i-j} \\ \hline \end{array}$

We need the coefficient $c_8$ of $a^8$:

So

$\begin{array}{|rcll|} \hline a^{12-2i-j} &=& a^8 \\ 12-2i-j &=& 8 \\ \mathbf{j} &\mathbf{=}& \mathbf{4-2i}\\ \hline \end{array}$

$\begin{array}{|rcll|} \hline c_8\cdot a^8 &=& \sum \limits_{i=0}^{4} \sum \limits_{j=0}^{4-i} \frac{4!}{i!~j!~(4-i-j)!} \cdot a^{8} \quad | \quad j=4-2i\\ c_8\cdot a^8 &=& a^{8} \cdot \sum \limits_{i=0}^{4} \frac{4!}{i!~(4-2i)!~[4-i-(4-2i)]!} \\ c_8\cdot a^8 &=& a^{8} \cdot \sum \limits_{i=0}^{4} \frac{4!}{i!~(4-2i)!~i!} \\ && 4-2i \ge 0 \\ && 4 \ge 2i \\ && 2 \ge i \\ && \mathbf{i \le 2}\\ c_8\cdot a^8 &=& a^{8} \cdot \sum \limits_{i=0}^{2} \frac{4!}{i!~(4-2i)!~i!} \\ c_8\cdot a^8 &=& a^{8} \cdot \left[ \frac{4!}{0!~(4-2\cdot 0)!~0!} +\frac{4!}{1!~(4-2\cdot 1)!~1!} +\frac{4!}{2!~(4-2\cdot(2))!~2!} \right] \\ c_8\cdot a^8 &=& a^{8} \cdot ( \frac{4!}{4!} +\frac{4!}{2!} +\frac{4!}{2!~2!} ) \\ c_8\cdot a^8 &=& a^{8} \cdot ( 1 +3\cdot 4 +\frac{3\cdot 4}{2} ) \\ c_8\cdot a^8 &=& a^{8} \cdot ( 1 +12 + 6 ) \\ c_8\cdot a^8 &=& a^{8} \cdot 19 \\ c_8\cdot a^8 &=& \mathbf{19}\cdot a^{8} \\ \hline \end{array}$