heureka

Find the first term, the common difference, and tn for a series having Sn= 4n−3n^2.

$\begin{array}{lcll} \boxed{~ \begin{array}{lcll} S_n &=& 4n-3n^2 \end{array} ~}\\\\ \end{array}\\ \begin{array}{lcll} t_n &=& S_n - S_{n-1} \\ t_n &=&4n-3n^2 - [ 4(n-1)-3(n-1)^2 ] \\ t_n &=&4n-3n^2 - 4(n-1) + 3(n-1)^2 \\ t_n &=&4n-3n^2 - 4n + 4 + 3(n^2-2n+1) \\ t_n &=&-3n^2 + 4 + 3(n^2-2n+1) \\ t_n &=&-3n^2 + 4 + 3n^2-6n+3 \\ t_n &=& 4 -6n+3 \\ \mathbf{t_n} & \mathbf{=} & \mathbf{7 -6n} \\\\ t_1 &=& 7 - 6\cdot 1 \\ t_1 &=& 7-6\\ \mathbf{t_1} &\mathbf{=}& \mathbf{1}\\\\ d &=& t_n - t_{n-1} \\ d &=& 7-6n -[ 7-6(n-1)]\\ d &=& 7-6n - 7 + 6(n-1)\\ d &=&-6n + 6(n-1)\\ d &=&-6n + 6n -6\\ \mathbf{d} &\mathbf{=}&\mathbf{-6}\\ \end{array}$

Sorry:

$\sin{( 2x )} = 2\cdot \sin{( x )}\cdot \cos{( x )}\\ \frac12 \cdot \sin{( 2x )}= \frac12 \cdot2\cdot \sin{( x )}\cdot \cos{( x )}\\ \frac12 \cdot \sin{( 2x )}=\sin{( x )}\cdot \cos{( x )}\\$

45,300 at 21% compounded monthly for 8 3/4 years

compounded monthly

$\begin{array}{lcll} K_{n} &=& K_0 \cdot (1 + \frac{i}{12} )^n \qquad n = \text{Number of the months} \\\\ K_{n} &=& K_0 \cdot (1 + \frac{i}{12} )^n \qquad K_0 = $45300 \quad i=21\% \quad n = 8\cdot 12 + \frac34 \cdot 12 = 105 \\ K_{105} &=& 45300 \cdot (1 + \frac{\frac{21}{100}}{12} )^{105} \\ K_{105} &=& 45300 \cdot (1 + \frac{21}{100\cdot 12} )^{105} \\ K_{105} &=& 45300 \cdot (1 + 0.01750000000 )^{105} \\ K_{105} &=& 45300 \cdot (1.01750000000 )^{105} \\ K_{105} &=& 45300 \cdot 6.18178475531 \\ \mathbf{K_{105}} & \mathbf{=} & \mathbf{$280034.849416} \\ \end{array}$

The angle of elevation of the top of a tree is found to be 23° at one point and 76° at a point 49 feet nearer the tree. How high is the tree if both observation points and the base of the tree are on the same horizontal plane?

$\begin{array}{rcll} \tan{(23^\circ)} &=& \frac{h}{x} \\ x &=& \frac{h}{ \tan{(23^\circ)}} \\\\ \tan{(76^\circ)} &=& \frac{h}{x-49} \\ \tan{(76^\circ)} &=& \frac{h}{\frac{h}{ \tan{(23^\circ)}}-49} \\ \left( \frac{h}{ \tan{(23^\circ)}}-49 \right)\cdot \tan{(76^\circ)} &=& h\\ h\cdot \frac{\tan{(76^\circ)}} { \tan{(23^\circ)}}-49\cdot \tan{(76^\circ)}&=& h\\ h\cdot \left( \frac{\tan{(76^\circ)}} { \tan{(23^\circ)}} - 1 \right) &=& 49\cdot \tan{(76^\circ)}\\ h\cdot \left( \frac{\tan{(76^\circ)}-\tan{(23^\circ)} } { \tan{(23^\circ)}} \right) &=& 49\cdot \tan{(76^\circ)}\\ h &=& 49\cdot \left( \frac{ \tan{(76^\circ)} \cdot \tan{(23^\circ)} } { \tan{(76^\circ)}-\tan{(23^\circ)} } \right) \\ h &=& 49\cdot \left( \frac{4.01078093354 \cdot 0.42447481621 } { 4.01078093354-0.42447481621 } \right) \\ h &=& 49\cdot \left( \frac{1.70247549962 } { 3.58630611733 } \right) \\ h &=& 49\cdot 0.47471561097 \\ h &=& 23.2610649376\ \text{feet} \\ \end{array}$

The tree is 23.2610649376 feet high

A rectangle is 152 by 383 feet. Determine the length of the diagonal and the angle between the shorter side and the diagonal.

$\begin{array}{rcll} \text{diagonal} &=& \sqrt{ 383^2+152^2} \\ &=& \sqrt{ 169793 } \\ &=& 412.059461729\ \text{feet} \\\\ \tan{(\varphi)} &=& \frac{383}{152} \\ &=& 2.51973684211 \\ \varphi &=& \arctan{(2.51973684211)}\\ \varphi &=& 68.3535133795^\circ \end{array}$

how do i isolate x in y=b∗x^a?

$\begin{array}{rcll} y &=& b\cdot x^a \qquad & | \qquad :b\\ \frac{y}{b} &=& x^a \qquad & | \qquad 1^{\frac{1}{a}}\\ \left( \frac{y}{b} \right)^{\frac{1}{a}} &=& x^{ a\cdot \frac{1}{a} } \\ \left( \frac{y}{b} \right)^{\frac{1}{a}} &=& x^{\frac{a}{a} } \\ \left( \frac{y}{b} \right)^{\frac{1}{a}} &=& x^{1 } \\ \left( \frac{y}{b} \right)^{\frac{1}{a}} &=& x \\ \end{array}$

A broken clock is set corectly at 12:00 noon. However, it registers only 20 minutes for each hour. In how many hours will it again register the correct time  ?

A)12

B)18  analog clock(12h)

C)24 

D)30

E)36  digital clock (12h)

1. Analog Clock:

 $\begin{array}{rcll} \text{angular velocity real time: } \omega_1 &=& \frac{2\pi}{12\ h} \\ \text{angular velocity display time: } \omega_2 &=& \frac{2\pi}{36\ h} \\ w_2\cdot t - w_1\cdot t &=& n\cdot 2\pi \\ \frac{2\pi}{12\ h}\cdot t - \frac{2\pi}{36\ h}\cdot t &=& n\cdot 2\pi \qquad &| \qquad :2\pi \\ \frac{1}{12\ h}\cdot t - \frac{1}{36\ h}\cdot t &=& n \\ t\cdot (\frac{1}{12\ h} - \frac{1}{36\ h} ) &=& n \\ t\cdot ( \frac{36-12}{12\cdot 36} ) &=& n \\ t\cdot ( \frac{24}{12\cdot 36} ) &=& n \\ t\cdot ( \frac{2}{36} ) &=& n \\ t\cdot ( \frac{1}{18} ) &=& n \\ t &=& 18\cdot n \\ \end{array}$

next correct time, if n = 1, then t = 18 hours.

2. Digital Clock(24h):

$\begin{array}{rcll} \text{angular velocity real time: } \omega_1 &=& \frac{2\pi}{24\ h} \\ \text{angular velocity display time: } \omega_2 &=& \frac{2\pi}{72\ h} \\ w_2\cdot t - w_1\cdot t &=& n\cdot 2\pi \\ \frac{2\pi}{24 h}\cdot t - \frac{2\pi}{72\ h}\cdot t &=& n\cdot 2\pi \qquad &| \qquad :2\pi \\ \frac{1}{24\ h}\cdot t - \frac{1}{72\ h}\cdot t &=& n \\ t\cdot (\frac{1}{24\ h} - \frac{1}{72\ h} ) &=& n \\ t\cdot ( \frac{72-24}{24\cdot 72} ) &=& n \\ t\cdot ( \frac{48}{24\cdot 72} ) &=& n \\ t\cdot ( \frac{2}{72} ) &=& n \\ t\cdot ( \frac{1}{36} ) &=& n \\ t &=& 36\cdot n \\ \end{array}$

next correct time, if n = 1, then t = 36 hours.

can someone please tell me how can i find the roots in these  9x^4+16x^3-76x^2-8x+32=0 numerically? since i was told it can't be done algebraically.

cos 2x / cos^2(x) * sin^2(x) (integrals) I want to know how to solve it

$\small{ \begin{array}{rcll} \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \frac{ \cos{( 2x )} \cdot [ 1-\cos^2{(x)} ] } { \cos^2{(x)} } \qquad &| \qquad \sin^2{(x)} = 1-\cos^2{(x)} \\ &=& \cos{( 2x )} \cdot \left[ \frac{1-\cos^2{(x)}} {\cos^2{(x)}} \right] \\ &=& \cos{( 2x )} \cdot \left[ \frac{1}{\cos^2{(x)}} -1 \right] \\ &=& \frac{\cos{( 2x )} }{\cos^2{(x)}} -\cos{( 2x )} \qquad &| \qquad \cos{( 2x )} = \cos^2{(x)}- \sin^2{(x)}\\ &=& \frac{ \cos^2{(x)}- \sin^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \frac{\sin^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \frac{ 1-\cos^2{(x)} }{\cos^2{(x)}} -\cos{( 2x )} \\ &=& 1- \left[ \frac{ 1 } {\cos^2{(x)} }-1 \right] -\cos{( 2x )} \\ &=& 2- \frac{ 1 } {\cos^2{(x)} } -\cos{( 2x )} \\ \end{array} }$

$\begin{array}{rcll} \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \int 2\ dx -\int \frac{ 1 } {\cos^2{(x)} }\ dx -\int \cos{( 2x )}\ dx \\ \end{array}$

$\begin{array}{rcll} \text{1.} \qquad \int 2\ dx &=& 2\int dx \\ \int 2\ dx &=& 2x \end{array}$

$\begin{array}{rcll} \text{2.} \qquad \int \frac{ 1 } {\cos^2{(x)} }\ dx &=& \int \frac{ \sin^2{(x)}+\cos^2{(x)} }{\cos^2{(x)}} \ dx \\ &=& \int (\tan^2{(x)} + 1 )\ dx\\ && \boxed{~ \begin{array}{rcll} \text{we need: } y &=& \tan{(x)} \\ y &=& \frac{\sin{(x)}} {\cos{(x)}} \\ y' &=& \frac{\sin{(x)}} {\cos{(x)}} \left[ \frac{\cos{(x)}} {\sin{(x)}} - \frac{-\sin{(x)}}{\cos{(x)}} \right] \\ y' &=& \tan{(x)} \left[ \cot{(x)} + \tan{(x)} \right] \\ y' &=& 1+ \tan^2{(x)} \\ \end{array} ~}\\ &=& \int (\tan^2{(x)} + 1 )\ dx \\ \text{we substitute:} ~ u &=& \tan{(x)}\\ du &=&\left( 1+\tan^2{(x)} \right)\ dx\\ &=& \int (\tan^2{(x)} + 1 ) \frac{du}{1+\tan^2{(x)}} \\ &=& \int du\\ &=& u\\ \int \frac{ 1 } {\cos^2{(x)} }\ dx &=& \tan{(x)}\\ \end{array}$

$\begin{array}{rcll} \text{3.} \qquad \int \cos{( 2x )}\ dx \\ \text{we substitute:} ~ u &=& 2x\\ du &=&2\ dx\\ \int \cos{( 2x )}\ dx &=& \int \cos{( u )} \frac{du}{2} \\ &=& \frac12 \cdot \int \cos{( u )} \ du \\ &=& \frac12 \cdot \sin{( u )} \\ &=& \frac12 \cdot \sin{( 2x )} \qquad &| \qquad \sin{( 2x )} = 2\cdot \sin{( 2x )}\cdot \cos{( x )} \\ &=& \frac12 \cdot 2\cdot \sin{( 2x )}\cdot \cos{( x )} \\ \int \cos{( 2x )}\ dx &=& \sin{( x )}\cdot \cos{( x )} \\ \end{array}$

$\begin{array}{rcll} \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& \int 2\ dx -\int \frac{ 1 } {\cos^2{(x)} }\ dx -\int \cos{( 2x )}\ dx \\\\ \int \frac{ \cos{( 2x )} \cdot \sin^2{(x)} } { \cos^2{(x)} } &=& 2x - \tan{(x)}-\sin{( x )}\cdot \cos{( x )} + c \end{array}$

Bestimmen Sie alle Werte z, für die folgende Gleichungen gelten:

tan(z)=2

$\begin{array}{rcll} \tan{(z)} &=& 2 \qquad & | \qquad \arctan{()} \\ z &=& \arctan{(2)} \pm k\cdot 180^\circ \\ z &=& 63.4349488229^\circ \pm k\cdot 180^\circ \qquad k = 0,1,2,3, \cdots \\ z &=& \cdots , -116.565051177^\circ, 63.4349488229^\circ, 243.434948823^\circ, \cdots \end{array}$

Der tangens ist periodisch mit einer Periodenlänge von $180^\circ$