heureka

12a^2-16a+48

$12a^2-16a+48 = 12 \cdot \left [~ \left( a-\frac23 \right) - \frac{ 2^{2.5} } {3} i ~\right] \left [~ \left( a-\frac23 \right) + \frac{ 2^{2.5} } {3} i ~\right]$

-6,-24,-96,...

$\begin{array}{rcl} a_1 &=& - 6 \qquad a_2 = -24 \qquad a_3 = -96\\ \end{array}\\ \begin{array}{rcl} \frac{a_2}{a_1} = \frac{-24}{- 6} &=& 4 \\ \frac{a_3}{a_2} = \frac{-96}{- 24} &=& 4 \\ \text{Geometric Sequence } r &=& 4 \\\\ \text{To find any term of a geometric sequence: } \boxed{~ \begin{array}{rcl} a_n &=& a_1\cdot r^{n-1} \end{array} ~}\\ a_1= -6\cdot 4^0 &=& -6\cdot 1 = -6 \\ a_2= -6\cdot 4^1 &=& -6\cdot 4 = -24 \\ a_3= -6\cdot 4^2 &=& -6\cdot 16 = -96 \\ a_4= -6\cdot 4^3 &=& -6\cdot 64 = -384 \\ a_5= -6\cdot 4^4 &=& -6\cdot 256 = -1536 \\ a_6= -6\cdot 4^5 &=& -6\cdot 1024 = -6144 \\ \cdots \\\\ \text{ or }a_1 &=& -6\\ \boxed{~ \begin{array}{rcl} a_n &=& a_{n-1}\cdot r \end{array} ~}\\ a_2 &=& -6 \cdot 4 = -24 \\ a_3 &=& -24 \cdot 4 = -96 \\ a_4 &=& -96 \cdot 4 = -384 \\ a_5 &=& -384 \cdot 4 = -1536 \\ a_6 &=& -1536 \cdot 4 = -6144 \\ \cdots \end{array}$

Hallo aditya.

Example: If n = 1 we have:

$3^{2n+2} - 8n -9 = 3^{2\cdot 1+2} - 8\cdot 1 -9 = 3^4-8-9 = 81-17 = 64 ={\color{red}2^6}$

$2^6 \text{ can be devide by } 2^m \text{, if }m=6.\\ So~the~mimimum~ m~ is ~ 6.\\ \text{The maximum m must be equal or greater than the minimum m }.\\ \text{So m is }6\text{ or greater}.$

m is not a constant.

A helicopter is ascending vertically with a speed of 2.50 m/s. At a height of 145 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

$\begin{array}{rcl} h &=& h_0+v_0\cdot t-\frac12\cdot g\cdot t^2 \qquad | \qquad h = 0\ m \\ 0 &=& h_0+v_0\cdot t-\frac12\cdot g\cdot t^2 \\ -\frac12\cdot g\cdot t^2 + v_0\cdot t +h_0 &=& 0\\\\ \boxed{~ \begin{array}{rrcl} ax^2+bx+c &=& 0 \\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\\\ \begin{array}{rcl} a=-\frac12\cdot g\qquad b=v_0 \qquad c = h_0 \\ \end{array}\\ t_{1,2} &=& {-v_0 \pm \sqrt{v_0^2-4(-\frac12\cdot g)\cdot h_0} \over 2(-\frac12\cdot g)} \\ t_{1,2} &=& {-v_0 \pm \sqrt{v_0^2+2g\cdot h_0} \over -g } \\ t_{1,2} &=& {v_0 \pm \sqrt{v_0^2+2g\cdot h_0} \over g } \\\\ \begin{array}{rcl} v_0= 2.50\ \frac{m}{s} \qquad h_0 = 145\ m \qquad g= 9.81\ \frac{m}{s^2}\\ \end{array}\\ t_{1,2} &=& {2.50 \pm \sqrt{2.50^2+2\cdot 9.81\cdot 145} \over 9.81 } \\\\ t_{1,2} &=& \frac{2.50 \pm 149.903802487} { 9.81 } \\ t_1 &=& \frac{2.50 + 53.3961609107} { 9.81 } \\ t_1 &=& 5.69787572994\ s\\\\ t_2 &=& \frac{2.50 - 53.3961609107} { 9.81 } \\ t_2 &=& -5.18819173402\ s \quad \text{ negative time, no solution} \end{array}$

t = 5,7 s

If a satellite with a mass of 500 kg is in an orbit 5 times the radius of earth,what is the centripetal force it experiences?Earths radius=6.37*10^6 satellites velocity=3,542.4 m/s

$\text{Centripetal force } = \text{ mass } \times \frac{ \text{ velocity}^2 } { \text{radius} }$

$\begin{array}{rcl} \text{Radius }\ r &=& 5\cdot 6370000\ m = 31850000\ m \\ \text{Mass }\ m &=& 500\ kg \\ \text{Velocity }\ v &=& 3542.4\ \frac{m}{s} \\ \text{Centripetal force } &=& m\cdot \frac{v^2}{r} \\ &=& 500\ kg \cdot \frac{(3542.4\ \frac{m}{s})^2}{31850000\ m} \\ &=& 500\cdot 0.39399051052\ kg \frac{m}{s^2} \\ &=& 196.995255259 \ kg \frac{m}{s^2} \\ &=& 196.995255259 \ N \\ \end{array}$

In combinatorial mathematics, a derangement is a permutation of the elements of a set, such that no element appears in its original position.

The number of derangements of a set of size n, usually written Dn, dn, or !n, is called the "derangement number" or "de Montmort number". (These numbers are generalized to rencontres numbers.) The subfactorial function (not to be confused with the factorial n!) maps n to !n.

No standard notation for subfactorials is agreed upon; n¡ is sometimes used instead of !n.

Simplify the expression. Write the answer in scientific notation. left parenthesis 9 times 10 Superscript negative 4 right parenthesis left parenthesis 6 times 10 Superscript negative 5 right parenthesis(9×10−4)(6×10−5)

$\begin{array}{rcl} (9\cdot 10^{-4}) \cdot (6\cdot 10^{-5}) &=& 9\cdot 10^{-4} \cdot 6\cdot 10^{-5}\\ &=& 9\cdot 6\cdot 10^{-4} \cdot 10^{-5} \\ &=& 9\cdot 6\cdot 10^{-4+(-5)} \\ &=& 9\cdot 6\cdot 10^{-9} \\ &=& 54\cdot 10^{-9} \\ &=& 5.4\cdot 10\cdot 10^{-9} \\ &=& 5.4\cdot 10^1\cdot 10^{-9} \\ &=& 5.4\cdot 10^{1-9} \\ &=& 5.4\cdot 10^{-8} \\ \end{array}$

On a certain college faculity, 4/7 of the professors are male, and the ratio of the professors older than 50 to the professors less than or equal to 50 years is 2:5. if 1/5 of the male professors are older than 50 years, what fraction of the female professors are less than or equal to 50 years ?

$\begin{array}{rcl} m_{>50} &=& \frac47\cdot \frac15 = \frac{4}{35} \\ m_{\le50} &=& \frac47\cdot \frac45 = \frac{16}{35} \\ m_{>50} + m_{\le50} &=& \frac{4}{35} + \frac{16}{35} = \frac{20}{35} \\\\ f_{>50} + f_{\le50} &=& 1- ( m_{>50} + m_{\le50} ) = 1- \frac{20}{35} = \frac{15}{35} \\ f_{>50} &=& \frac{15}{35} - f_{\le50} \\\\ \frac{m_{>50}+f_{>50}}{m_{\le50}+f_{\le50}} &=& \frac25 \\ 5\cdot ( m_{>50}+f_{>50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \qquad | \qquad f_{>50} = \frac{15}{35} - f_{\le50} \\ 5\cdot ( m_{>50}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( m_{\le50}+f_{\le50} ) \\ 5\cdot ( \frac{4}{35}+\frac{15}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ 5\cdot ( \frac{19}{35} - f_{\le50} ) &=& 2\cdot ( \frac{16}{35}+f_{\le50} ) \\ \frac{95}{35} - 5\cdot f_{\le50} &=& \frac{32}{35}+2\cdot f_{\le50} \\ 7\cdot f_{\le50} &=& \frac{95}{35} - \frac{32}{35} \\ 7\cdot f_{\le50} &=& \frac{63}{35} \\ f_{\le50} &=& \frac{63}{35\cdot 7 } \\ \text{female}_{\le50} &=& \frac{9}{35} \end{array}$

$\begin{array}{rcl} \hline \text{continuation}\\ \hline \\ \frac{ \text{female}_{\le50}}{f+m} &=& \frac{9}{35} \\ \text{female}_{\le50} &=& \frac{9}{35}\cdot (f+m) \quad | \quad :f \\ \frac{ \text{female}_{\le50}}{f} &=& \frac{9}{35}\cdot (1+\frac{m}{f}) \\\\ \boxed{~ \begin{array}{rcl} \frac{m}{m+f} &=& \frac{4}{7} \\ \frac{f}{m+f} &=& 1- \frac{4}{7} = \frac{3}{7}\\ \frac{ \frac{m}{m+f} } { \frac{f}{m+f} } &=& \frac{ \frac{4}{7} } { \frac{3}{7} }\\ \frac{ m } { f } &=& \frac43 \end{array} ~}\\\\ \frac{ \text{female}_{\le50}}{f} &=& \frac{9}{35}\cdot (1+ \frac43) \\ \frac{ \text{female}_{\le50}}{f} &=& \frac{9}{35}\cdot (\frac73) \\ \frac{ \text{female}_{\le50}}{f} &=& \frac{9\cdot 7}{35\cdot 3} \\ \frac{ \text{female}_{\le50}}{f} &=& \frac{3}{5} \\ \end{array}$

1. |x2-1|=|x-1|

$\begin{array}{rcll} |x^2-1| &=& |x-1| \qquad & (x^2-1) = (x-1)(x+1) \\ (|(x-1)(x+1)|)^2 &=& (|x-1|)^2 \qquad & (\text{square both sides}) \\ (~(x-1)(x+1)~)^2 &=& (x-1)^2 \qquad & (~(ab)^2 = a^2b^2~) \\ (x-1)^2(x+1)^2 &=& (x-1)^2 \\ (x-1)^2(x+1)^2-(x-1)^2 &=& 0\\ (x-1)^2\left[(x+1)^2-1 \right] &=& 0\\ (x-1)^2(x^2+2x+1-1) &=& 0\\ (x-1)^2(x^2+2x) &=& 0\\ (x-1)^2\cdot x \cdot (x+2) &=& 0\\ (x-1)\cdot (x-1)\cdot x \cdot (x+2) &=& 0 \end{array}\\ \begin{array}{rrcl} \\ 1. & x-1=0 && x=1 \\ 2. & x = 0 \\ 3. & x+2 = 0 && x = -2 \end{array}$

2. 3|x-2|=2|x-3|

$\begin{array}{rcll} 3|x-2| &=& 2|x-3| \\ (3|x-2|)^2 &=& (2|x-3|)^2 \qquad & (\text{square both sides}) \\ 3^2(x-2)^2 &=& 2^2(x-3)^2 \qquad & (~(ab)^2 = a^2b^2~) \\ 9(x-2)^2 &=& 4(x-3)^2 \\ 9(x^2-4x+4) &=& 4(x^2-6x+9) \\ 9x^2-36x+36 &=& 4x^2-24x+36 \\ 9x^2-36x &=& 4x^2-24x\\ 5x^2-12x &=& 0\\ x\cdot(5x-12) &=& 0\\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x = 0 \\ 2. & 5x-12=0 && x=\frac{12}{5} \end{array}$

3. |4-x|=5

$\begin{array}{rcll} |4-x| &=& 5 \\ (|4-x|)^2 &=& (5)^2 \qquad & (\text{square both sides}) \\ (4-x)^2 &=& 25 \qquad & (\pm\sqrt{}) \\ 4-x &=& \pm 5\\ x &=& 4 \pm 5 \\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x = 4+5 && x = 9 \\ 2. & x = 4-5 && x = -1 \end{array}$

4. |2x-7|=|x-5|

$\begin{array}{rcll} |2x-7| &=& |x-5| \\ (|2x-7|)^2 &=& (|x-5|)^2 \qquad & (\text{square both sides}) \\ (2x-7)^2 &=& (x-5)^2 \\ 4x^2-28x+49 &=& x^2-10x+25\\ 3x^2-18x+24 &=& 0\qquad & :3\\ x^2-6x+8 &=& 0\\ (x-4)(x-2) &=& 0 \end{array}\\ \begin{array}{rrcl} \\ 1. & x-4=0 && x = 4 \\ 2. & x-2=0 && x=2 \end{array}$

5. |x2-x|=|x2-1|

$\begin{array}{rcll} |x^2-x| &=& |x^2-1| \\\\ \qquad x^2-x = x(x-1) \\ \qquad (x^2-1) = (x-1)(x+1) \\\\ |x(x-1)| &=& |(x-1)(x+1)| \\ (~|x(x-1)|~)^2 &=& (~|(x-1)(x+1)|~)^2 & (\text{square both sides}) \\ (~x(x-1)~)^2 &=& (~(x-1)(x+1)~)^2 & (~(ab)^2 = a^2b^2~) \\ x^2(x-1)^2 &=& (x-1)^2(x+1)^2\\ x^2(x-1)^2 - (x-1)^2(x+1)^2 &=& 0\\ (x-1)^2 [ x^2 - (x+1)^2 ] &=& 0\\ (x-1)^2 ( x^2 - x^2-2x-1 ) &=& 0\\ (x-1)^2 ( -2x-1 ) &=& 0\\ (x-1)\cdot (x-1)\cdot ( -2x-1 ) &=& 0\\ \end{array}\\ \begin{array}{rrcl} \\ 1. & x-1=0 && x=1 \\ 2. & -2x-1 = 0 && x = -\frac{1}{2} \end{array}$

!5 = 44

$\begin{array}{rcll} !n = n!\sum \limits_{k=0}^{n} \frac{(-1)^k}{k!} = n!\cdot \left( ~ 1 - \frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n \frac{1}{n!} ~ \right) \end{array}$

Example:

$\begin{array}{rcll} !6 = 6!\cdot \left( ~ 1 - \frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} +\frac{1}{4!} - \frac{1}{5!} +\frac{1}{6!}~ \right) = 265 \end{array}$

$\begin{array}{rcll} !5 &=& 5!\cdot \left( ~ 1 - \frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} +\frac{1}{4!} - \frac{1}{5!} ~ \right) \\ &=& 120\cdot \left( ~ 1 - 1 +\frac{1}{2} - \frac{1}{6} +\frac{1}{24} - \frac{1}{120} ~ \right) \\ &=& 120\cdot \left( ~ \frac{1}{2} - \frac{1}{6} +\frac{1}{24} - \frac{1}{120} ~ \right) \\ &=& 120\cdot \left( ~ \frac{3\cdot 4\cdot 5 - 4\cdot 5 + 5 - 1 }{120} ~ \right) \\ &=& 3\cdot 4\cdot 5 - 4\cdot 5 + 5 - 1 \\ &=& 60 - 20 + 5 - 1 \\ \mathbf{!5} & \mathbf{=} & \mathbf{44 } \end{array}$