heureka

Could you help me with this question?

5(sqrt5) can be written in the form 5^k

What is the value of k

$\begin{array}{rcll} && 5\cdot \sqrt{5} \qquad &| \qquad \sqrt{5} = 5^{\frac12} \\ &=& 5\cdot 5^{\frac12} \qquad &| \qquad 5 = 5^1 \\ &=& 5^1 \cdot 5^{\frac12} \qquad &| \qquad a^b\cdot a^c = a^{b+c} \\ &=& 5^{1+\frac12} \qquad &| \qquad 1 = \frac22\\ &=& 5^{\frac22+\frac12}\qquad &| \qquad \frac22+\frac12 = \frac{2+1}{2} \\ &=& 5^{\frac{2+1}{2}} \\ &=& 5^{\frac{3}{2}} \qquad &| \qquad \frac{3}{2} = 1,5\\ &=& 5^{1,5} \\\\ \mathbf{k}& \mathbf{=}& \mathbf{1,5} \end{array}$

whats 70x{40x120:[70-800:(90-136:13)]}   I need help fast please

1x2x3x4x5x6x7……x1990x1991'a answer have a lot of zero at the  end. What is the first number that is not 0 counting from the right?

see: LEGENDRE factorial in prime factors disassemble

1. factor 2

$\begin{array}{rclr} \frac{1991}{2} &=& 995.5 & 995\\ \frac{1991}{2^2} &=& 497.75 & 497\\ \frac{1991}{2^3} &=& 248.875 & 248\\ \frac{1991}{2^4} &=& 124.437 & 124\\ \frac{1991}{2^5} &=& 62.21875 & 62\\ \frac{1991}{2^6} &=& 31.109375 & 31\\ \frac{1991}{2^7} &=& 15.5546875 & 15\\ \frac{1991}{2^8} &=& 7.77734375 & 7\\ \frac{1991}{2^9} &=& 3.888671875 & 3 \\ \frac{1991}{2^{10}} &=& 1.94433593750 & 1\\ \frac{1991}{2^{11}} &=& 0.97216796875 & 0\\ && \text{sum} & = 1983 \end{array}$

2. facor 5

$\begin{array}{rclr} \frac{1991}{5} &=& 398.2 & 398\\ \frac{1991}{5^2} &=& 79.64 & 79\\ \frac{1991}{5^3} &=& 15.928 & 15\\ \frac{1991}{5^4} &=& 3.1856 & 3\\ \frac{1991}{5^5} &=& 0.63712 & 0\\ && \text{sum} & = 495 \end{array}$

...

$1991! = 2^{1983} \times 3^{991} \times 5^{ \color{red}459} \times 7^{329} \times \dots \times 1987^1$

because the exponent of 2 is not less  than the exponent of 5 - 1983 is greater than 459 - we have 459 zeros at the  end of 1991!

Complete the equation of the graphed linear function. Write the slope in decimal form.            

y=? x+?

$\text{line } y = m\cdot x +b$

1. slope ?

$m = \frac{ \Delta y } {\Delta x} \\ m = \frac{4-0}{0-(-5)}\\ m=\frac{4}{5}$

2. y - intersection ?

b = 4

$y = m\cdot x +b \\ \boxed{~ y = \frac45 x+ 4 ~}$

Proof:

if x = 0   =>   y = 4   okay

if y = 0   =>  $\begin{array}{rcll} 0 &=& \frac45 x+ 4 \\ \frac45 x &=& -4 \\ x &=& -\frac54 \cdot 4\\ x &=& -5 \qquad \text{ okay } \end{array}$

Wie viele Verfahren gibt es, mit denen man die Variablen durch 2 Gleichungen herausfinden kann? Bis jetzt kenne ich: Additionsverfahren Gleichsetzungsverfahren Grafikverfahren Einsetzungverfahren.

Es gibt viele Verfahren, die mit Matrizen arbeiten, zum Beispiel nach Gauß-Jordan.

In diesem Falle, wenn es sich nur um 2 Gleichungen mit 2 Unbekannten handelt kann man auch einfach die Lösung nach dem Kramer-Verfahren aufschreiben. In diesem Falle haben wir eine Zählerdeterminate und eine Nennerdeterminante für x und y.

Kramer-Verfahren:

$\begin{array}{rcll} a\cdot x + b\cdot y &=& c \\ d\cdot x + e\cdot y &=& f \\ \hline x &=& \frac{ \begin{vmatrix} c & b \\ f & e \end{vmatrix} } { \begin{vmatrix} a & b \\ d & e \end{vmatrix} } \\ x &=& \frac{c\cdot e-f\cdot b}{a\cdot e-d\cdot b} \\\\ y &=& \frac{ \begin{vmatrix} a & c \\ d & f \end{vmatrix} } { \begin{vmatrix} a & b \\ d & e \end{vmatrix} } \\ y &=& \frac{a\cdot f-d\cdot c}{a\cdot e-d\cdot b} \\\\ \end{array}$

Find the value(s) of y such that the triangle with the given vertices has an area of 2 square units. (Enter your answers as a comma-separated list.)

(−5, 1), (0, 5), (−4, y)

y= ?

$\begin{array}{lcll} \vec{A} = \binom{a_x}{a_y} = \binom{-5}{1} \qquad \vec{B} = \binom{b_x}{b_y} = \binom{0}{5} \qquad \vec{C} = \binom{c_x}{c_y} = \binom{-4}{y} \\\\ \vec{B}-\vec{A} = \binom{b_x}{b_y}-\binom{a_x}{a_y} = \binom{b_x-a_x}{b_y-a_y} = \binom{0-(-5)}{5-1} = \binom{5}{4} \\ \vec{C}-\vec{A} = \binom{c_x}{c_y}-\binom{a_x}{a_y} = \binom{c_x-a_x}{c_y-a_y} = \binom{-4-(-5)}{y-1} = \binom{1}{y-1} \\ \\ \end{array}\\ \begin{array}{rcll} 2\cdot A_{\text{triangle}} &=& | (\vec{B}-\vec{A}) \times (\vec{C}-\vec{A}) | \\ &=& | \binom{5}{4} \times \binom{1}{y-1} | \\ &=& | 5\cdot(y-1) - 4\cdot 1 | \\ &=& | 5\cdot y -5 - 4 | \\ &=& | 5\cdot y -9 | \\ \\ \hline \\ 2\cdot A_{\text{triangle}} &=& + ( 5\cdot y_2 -9 ) \\ 5\cdot y_2 &=& 9+ 2\cdot A_{\text{triangle}} \\ y_2 &=& \frac{9+ 2\cdot A_{\text{triangle}}}{5} \qquad &| \qquad A_{\text{triangle}} = 2 \\ y_2 &=& \frac{9+ 2\cdot 2}{5}\\ y_2 &=& \frac{13}{5}\\ \mathbf{y_2} & \mathbf{=} & \mathbf{2.6} \\ \\ \hline \\ 2\cdot A_{\text{triangle}} &=& - ( 5\cdot y_1 -9 ) \\ 2\cdot A_{\text{triangle}} &=& - 5\cdot y_1 +9 \\ 5\cdot y_1 &=& 9 - 2\cdot A_{\text{triangle}} \\ y_1 &=& \frac{9 - 2\cdot A_{\text{triangle}}} {5} \qquad &| \qquad A_{\text{triangle}} = 2 \\ y_1 &=& \frac{9 - 2\cdot2} {5} \\ y_1 &=& \frac{5} {5} \\ \mathbf{y_1} & \mathbf{=} & \mathbf{1} \\ \end{array}$

Thanks for that answer, heureka.........

could you explain how you derived your final formula?

...so far:

$\begin{array}{rcll} a_{n}= \left\{ \begin{array}{ll} 2\cdot n-2 = (2\cdot n - 2) + 0\qquad & (\text{n odd})\\ 2\cdot n-3 = (2\cdot n - 2) -1 \qquad & (\text{n even}) \end{array} \right. \end{array}$

...composite:

$\begin{array}{rcll} a_n = (2\cdot n - 2) -1 \cdot s \end{array}$

s is a switch. We need a function for a switch.

if n is even, s is on or equal 1,

and if n is odd, s is off or equal 0.

As the next we see we can switch with $(-1)^n$

This function does switch beween -1 and 1 if n is odd or if n is even.

$\begin{array}{lcr} (-1)^{\text{odd number}} &=& -1 \\ (-1)^{\text{even number}} &=& 1 \end{array}$

We must modify this "switch" function to get our function.

But it is easy to do this.

First step we add 1.

$\begin{array}{lcr} 1 + (-1)^{\text{odd number}} &=& -1+1 = 0\\ 1 + (-1)^{\text{even number}}&=& 1+1 = 2\\ \end{array}$

Second step, the number 2 must be number 1, so we divide by 2
$\begin{array}{lcr} \dfrac{1 + (-1)^{\text{odd number}} } {2} &=& \frac{0}{2} = 0\\\\ \dfrac{1 + (-1)^{\text{even number}} } {2}&=& \frac{2}{2} = 1\\ \end{array}$

So our switch s is ready:

$\begin{array}{lrcr} & s &=& \dfrac{1 + (-1)^n } {2} \\ & s &=& \frac12 [~ 1+(-1)^n ~]\\ n & \text{ is odd} & s = 0 \\ n & \text{ is even} & s = 1 \end{array}$

what are the next two numbers? 0,1,4,5,8,?,?

without mistakes:

$\begin{array}{r|rcll} n &\\ \hline 1& a_1 &=& 0\\ & && & +1 \\ 2& a_2 &=& 1\\ & && & +3 \\ 3& a_3 &=& 4 \\ & && & +1 \\ 4& a_4 &=& 5 \\ & && & +3 \\ 5& a_5 &=& 8 \\ & && & +1 \\ 6& a_6 &=& 9 \\ & && & +3 \\ 7& a_7 &=& 12 \\ & && & +1 \\ 8& a_8 &=& 13 \\ & && & +3 \\ 9& a_9 &=& 16 \\ & && & +1 \\ 10& a_{10} &=& 17 \\ \cdots \end{array}$

$\begin{array}{rcll} a_{n}= \left\{ \begin{array}{rl} 2\cdot n-2 \qquad & (\text{n odd})\\ 2\cdot n-3 \qquad & (\text{n even}) \end{array} \right. \end{array}$

$\color{red} \boxed{~ \color{black} \begin{array}{rcll} a_n &=& (2\cdot n - 2) -\frac12 [~ 1+(-1)^n ~] \end{array} ~}$

what are the next two numbers? 0,1,4,5,8,?,?

$\begin{array}{rcll} a_1 &=& 0\\ && & +1 \\ a_2 &=& 1\\ && & +3 \\ a_3 &=& 4 \\ && & +1 \\ a_4 &=& 5 \\ && & +3 \\ a_5 &=& 8 \\ && & +1 \\ a_6 &=& 9 \\ && & +3 \\ a_7 &=& 12 \\ && & +1 \\ a_8 &=& 13 \\ && & +3 \\ a_9 &=& 16 \\ && & +1 \\ a_{10} &=& 17 \\ \cdots \end{array}$

$\begin{array}{rcll} a_{n+1}= \left\{ \begin{array}{rl} 2\cdot a_n-2 \qquad & (\text{n odd})\\ 2\cdot a_n-3 \qquad & (\text{n even}) \end{array} \right. \end{array}$

$\color{red} \boxed{~ \color{black} \begin{array}{rcll} a_n &=& (2\cdot n - 2) -\frac12 [~ 1+(-1)^n ~] \end{array} ~}$

The swedish population is at approximately 9 851 852 atm and the worlds population was 7 423 223 700 last time i checked. So how many precentage is the swedish population in relation to the world? 0,13%? Because i wonder, if we pretend the world is a smaller amount, how many people would there have to be to only have one swede?

$\begin{array}{rcll} \dfrac{ 9851852 }{ 7423223700} &=& 0.00132716626 \\ &=& 0.133\ \% \\\\ \dfrac{ 7423223700}{ 9851852 } &=& 753.485101075\\ &\approx& 753 \text{ people per one swede } \end{array}$