heureka

There are a total of 110 men and women in a hall. If 3/7 of the men leave the hall and another 40 women enter the hall,the ratio of the number of men to the number of women becomes 8:11. Find the number of women in the hall at first

$$\small{\text{ We set m = man, and w = women. In the hall are
}}\\$$

$$\small{\text{$
\begin{array}{rrclr}
& 110 &=& m + w \\
or &\mathbf{ m }& \mathbf{=} & \mathbf{110 - w} & (1)\\\\
\dfrac{3}{7} \text{ of the men leave the hall}: & m_{now} &=& m-\dfrac{3}{7}m\\
& m_{now} &=& \left( 1 - \dfrac{3}{7} \right)m \\
& m_{now} &=& \left(\dfrac{4}{7} \right)m \\\\
40 \text{ women enter the hall}: & w_{now} &=& w+40 \\\\
\text{ The Ratio of the number of men to the number of women becomes }: & \dfrac{8}{11} &=& \dfrac{m_{now}}{w_{now}} \\\\
& \dfrac{8}{11} &=& \dfrac{ \left(\dfrac{4}{7} \right)m }{ w+40 }
\qquad | \qquad m = 110-w \\\\
& \dfrac{8}{11} &=& \dfrac{ \left(\dfrac{4}{7} \right)(110-w) }{ w+40 } \\\\
& \dfrac{56}{44} &=& \dfrac{110-w}{ w+40 } \\\\
& 56\cdot(w+40) &=& 44\cdot (110-w) \\\\
& 56w+56\cdot 40 &=& 44\cdot 110 - 44w\\
& 100w &=& 44\cdot 110 - 56\cdot 40 \qquad | \qquad : 10\\\\
& 10w &=& 44\cdot 11 - 56\cdot 4\\
& 10w &=& 484- 224\\
& 10w &=& 260\\
& \mathbf{w} & \mathbf{=} & \mathbf{26}
\end{array}
$}}$$

The number of women in the hall at first is 26

what is x cubed in exponential form

$$\small{\text{$
\begin{array}{rcl}
x^3 &=& e^z \qquad | \qquad \ln{()}\\
\ln{(x^3)} &=& \ln{(e^z)}\\
3\ln{(x)} &=& z\ln{(e)}\qquad | \qquad \ln{(e)}=1\\
3\ln{(x)} &=& z\\\\
\mathbf{x^3} & \mathbf{=} & \mathbf{e^{3\ln{(x)}}}\\
\\
\hline
\\
\end{array}
$}}$$

2cos^2(x)-sin(x)-1=0

$$\small{\text{$
\begin{array}{rclr}
2\cos^2{(x)}-\sin{(x)}-1 &=& 0 \qquad | \qquad \cos^2{(x)}=1-\sin^2{(x)} \\
2 [ 1-\sin^2{(x)} ]-\sin{(x)}-1 &=& 0 \\
2 - 2 \sin^2{(x)} -\sin{(x)}-1 &=& 0 \\
- 2 \sin^2{(x)} -\sin{(x)}+1 &=& 0 \qquad | \qquad \cdot (-1)\\
2 \sin^2{(x)} +\sin{(x)}-1 &=& 0 \\
\left[ 2 \sin{(x)} -1 \right]\cdot \left[ \sin{(x)}+1 \right] &=& 0 \\\\
\underbrace{\left[
\textcolor[rgb]{1,0,0}{2 \sin{(x)} -1}\right]}_{=0}
\cdot
\underbrace{\left[
\sin{(x)}+1 \right]}_{=0}
&=& 0 \\\\
\sin{(x)}+1 &=& 0 \\
\sin{(x)} &=& -1\\
x &=& \arcsin(-1) \pm 2k\pi \\
\mathbf{x} & \mathbf{=} & \mathbf{-\dfrac{\pi}{2} \pm 2k\pi } \quad \mathbf{k}\in Z\\
\\
\hline
\\
\textcolor[rgb]{1,0,0}{2 \sin{(x)} -1} &\textcolor[rgb]{1,0,0}{=}&\textcolor[rgb]{1,0,0}{0} \\
2 \sin{(x)} &=&1 \\\\
\sin{(x)} &=&\dfrac{1}{2} \\
x &=& \arcsin\left(\dfrac{1}{2}\right) \pm 2k\pi \\
\mathbf{x} & \mathbf{=} & \mathbf{\dfrac{\pi} {6} \pm 2k\pi} \mathbf \quad {k}\in Z\\\\
\sin{(\pi-x)} &=&\dfrac{1}{2} \\
\pi-x &=& \arcsin\left(\dfrac{1}{2}\right) \pm 2k\pi \\
\pi-x &=& \dfrac{\pi} {6} \pm 2k\pi \\\\
\mathbf{x} & \mathbf{=} & \mathbf{\dfrac{5} {6}\pi \pm 2k\pi} \quad \mathbf{k}\in Z
\end{array}
$}}$$

$$\small{\text{
If $b$ is positive, what is the value of $b$ in the geometric sequence $9, a , 4, b$?}}\\
\small{\text{
Express your answer as a common fraction.
}}$$

$$\small{\text{We have $ u_1 = 9,~ u_2 = a, ~ u_3 = 4, and ~ u_4 = b
$ ~~\boxed{ \dfrac{u_n}{u_{n-1}} = constant. }
}}\\\\
\small{\text{ $
\begin{array}{rcl}
\dfrac{u_2}{u_1} &=& \dfrac{u_3}{u_2} \\\\
\dfrac{a}{9} &=& \dfrac{4}{a} \\\\
a^2 &=& 4\cdot 9 = 36 \\\\
\mathbf{a} & \mathbf{=} & \mathbf{6}\\\\\\
\dfrac{u_3}{u_2} &=& \dfrac{u_4}{u_3} \\\\
\dfrac{4}{a} &=& \dfrac{b}{4} \\\\
\dfrac{4}{6} &=& \dfrac{b}{4} \\\\
b &=& \dfrac{16}{6} \\\\
\mathbf{b} & \mathbf{=} & \mathbf{\dfrac{8}{3}}
\end{array}
$}}\\\\$$

$$\small{\text{
What is the next number in this sequence: $\dfrac12, 2, 5, 11, 23, 47, 95, \ldots$?
}}$$

$$\small{\text{$
t_1 = \dfrac{1}{2}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
t_n &=& 2^{n-1}t_1 +\sum \limits_{k=0}^{n-2}2^k \qquad | \qquad \sum \limits_{k=0}^{n-2}2^k = 2^{n-1}-1 \\\\
t_n &=& 2^{n-1}t_1 + 2^{n-1}-1 \\\\
t_n &=& 2^{n-1}(1+t_1)-1 \\\\
t_n &=& 2^{n-1}\left(1+\dfrac{1}{2}\right)-1 \\\\
t_n &=& 2^{n-1}\left(\dfrac{3}{2}\right)-1 \\\\
\mathbf{t_n} & \mathbf{=} & \mathbf{3\cdot 2^{n-2}-1} \\\\
t_8 &=& 3\cdot 2^6-1\\\\
t_8 &=& 3\cdot 64 - 1 \\\\
\mathbf{t_8} & \mathbf{=} & \mathbf{191}
\end{array}
$}}$$

Formel (2a+1)^4

wie sieht der Rechenweg für die obige Aufgabe aus. Muss ich hier die 1. binomische Formel erweitern?

NEIN!

$$\small{\text{$
\begin{array}{rcl}
(1+2a)^4 &=& [(1+2a)^2]^2 \qquad | \qquad (1+2a)^2 = 1+4a+4a^2 = 1+4a(1+a)\\\\
(1+2a)^4 &=& [1+4a(1+a) ]^2 = 1 + 2\cdot 4a(1+a) +[4a(1+a)]^2\\\\
(1+2a)^4 &=& 1 + 2\cdot 4a(1+a) +16a^2(1+a)^2\qquad | \qquad (1+a)^2 = 1+2a+a^2\\\\
(1+2a)^4 &=& 1 + 2\cdot 4a(1+a) +16a^2(1+2a+a^2)\\\\
(1+2a)^4 &=& 1 + 8a(1+a) +16a^2(1+2a+a^2)\\\\
(1+2a)^4 &=& 1 + 8a+8a^2 +16a^2 + 32a^3 + 16a^4\\\\
(1+2a)^4 &=& 1 + 8a+24a^2 + 32a^3 + 16a^4\\\\
\end{array}
$}}$$

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84. What is the 23rd term?

$$\small{\text{We have $t_x=t_5=9$ and $t_y=t_{32}=-84$ and we want $t_z=t_{23}=?$
}}\\\\
\small{\text{$\boxed{~~t_z = t_x\cdot \left( \dfrac{y-z}{y-x}\right)
+t_y\cdot \left(\dfrac{z-x}{y-x}\right) ~~}
$}}\\\\
\small{\text{$
\begin{array}{rcl}
t_{23} &=& t_5\cdot \left( \dfrac{32-23}{32-5}\right)
+t_{32}\cdot \left(\dfrac{23-5}{32-5}\right) \\\\
t_{23} &=& 9\cdot \left( \dfrac{9}{27}\right)
-84\cdot \left(\dfrac{18}{27}\right) \\\\
t_{23} &=& 9\cdot \left( \dfrac{1}{3}\right)
-84\cdot \left(\dfrac{2}{3}\right) \\\\
t_{23} &=& -\dfrac{159}{3}\\\\
\mathbf{t_{23}} & \mathbf{=} & \mathbf{-53}\\
\hline
\\
\end{array}
$}}\\\\$$

The 23rd term is -53

The sum of 40 consecutive integers is 100. What is the smallest of these 40 integers?

$$\small{\text{The sum of a arithmetic sequence is $
\boxed{~~S_n = t_1\cdot \binom{n}{1}+d\cdot \binom{n}{2}~~}
$}}\\\\
\small{\text{We have the sum of 40 consecutive integers is 100 so $d=1$ and $ n = 40 $ and $S_{40}=100$ we have :
}}\\
\small{\text{$
\begin{array}{rcl}
S_{40} = 100 &=& t_1\cdot \binom{40}{1} + \binom{40}{2}\\\\
t_1\cdot \binom{40}{1} &=& 100 - \binom{40}{2}\\\\
t_1 &=& \dfrac{100 - \binom{40}{2} }{\binom{40}{1}} \qquad | \qquad \binom{40}{2} = 39\cdot 20 = 780 \quad \binom{40}{1} = 40 \\\\
t_1 &=& \dfrac{100-780}{40} \\\\
t_1 &=& \dfrac{-680}{40}\\\\
t_1 &=& \dfrac{-68}{4}\\\\
t_1 &=& -\dfrac{68}{4}\\\\
\mathbf{t_1} &\mathbf{=}& \mathbf{-17}
\end{array}
$}}$$

The smallest of these 40 integers is -17

Question How do you solve for the intersection between two linear equations? How do you imput that formula in a scientific calculator using variables and constants?

For Example:

Input: solve(2x+3y=5, -4x+y=7)

you get : {x=-((8/7)), y=(17/7)}

Natalie went to the movies for $16.50 then spent half of her remaining money on jogging shoes. Next she bought lunch for $9.00 then spent half her remaining money on a skirt. After that she still had $21 left. How much money did she start with?

$$\small{\text{$
\dfrac{
\dfrac{x-\$16.50}{2} - \$9.00
}
{2}=\$21
$}} \\\\\\
\small{\text{$
\dfrac{x-\$16.50}{2} - \$9.00 = \$42
$}}\\\\
\small{\text{$
\dfrac{x-\$16.50}{2} = \$51
$}}\\\\
\small{\text{$
x-\$16.50 = \$102
$}}\\\\
\small{\text{$
x = \$118.50
$}}$$

She did start with $118.50