heureka

Could you explain that a bit more simply please?

Changing the order of factors does not change the product,
i.e. if a and b are two real numbers, then $a \cdot b = b\cdot a$ .

$\begin{array}{rcll} \dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } &=& \dfrac{12\cdot 5 \cdot 4} {20\cdot 6} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12\cdot 5 \cdot 4} {6\cdot 20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12}{6}\cdot \dfrac{5 \cdot 4}{20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12}{6}\cdot \dfrac{20}{20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \quad | \quad \dfrac{12}{6} = 2 \quad \dfrac{20}{20} = 1\\\\ &=& 2\cdot 1 \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& 2\cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& 2\cdot \dfrac{m^7 }{m^5} \cdot \dfrac{ n^8 }{n^4 } \cdot p^4 \end{array}$

$\boxed{~ \begin{array}{c} \text{Formula Quotient rule with same base:}\\ ~\frac{a^n} {a^m} = a^{n-m} \\ ~\text{Example:}\\ ~\frac{2^5}{ 2^3 } = 2^{5-3} = 2^2= 2\cdot 2 = 4 \end{array} ~}$

$\begin{array}{rcll} \dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } &=& 2\cdot (m^7\cdot m^{-5}) \cdot ( n^8 \cdot n^{-4})\cdot p^4 \\\\ &=& 2\cdot m^{7-5} \cdot n^{8-4}\cdot p^4 \\\\ \mathbf{\dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } }&\mathbf{=}& \mathbf{2\cdot m^{2} \cdot n^{4}\cdot p^4 } \\\\ \end{array}$

$\boxed{~ \begin{array}{c} \text{Formula Product rule with same exponent:}\\ ~a^n\cdot b^n = (a\cdot b)^{n} \\ ~\text{Example:}\\ ~3^2\cdot 4^2 =(3\cdot 4)^2 = 12^2 = 12\cdot 12 = 144 \end{array} ~}$

$\begin{array}{rcll} \mathbf{\dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } }&\mathbf{=}& \mathbf{2\cdot m^{2} \cdot (n\cdot p)^4 } \\\\ \end{array}$

86 * 9 + 68 = 896    ist eine falsche Aussage !

Wie heißt eine richtige Aussage, wenn alle  8  Ziffern und die  Zeichen   *   +  und  =  wieder verwendet werden müssen  (  alles aber nur  einmal ! )

$\begin{array}{rcll} 68 \cdot 9 + 86 &=& 698 \\ 96 \cdot 8 + 98 &=& 866 \\ 98 \cdot 6 + 98 &=& 686 \end{array}$

12m⁷ x 5n⁸ x 4p⁴               

_____________          

20m⁵ x 6n⁴

How do i simplify it? I don't see how you can.

$\begin{array}{rcll} \dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } &=& \dfrac{12\cdot 5 \cdot 4} {20\cdot 6} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12\cdot 5 \cdot 4} {6\cdot 20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12}{6}\cdot \dfrac{5 \cdot 4}{20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& \dfrac{12}{6}\cdot \dfrac{20}{20} \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& 2\cdot 1 \cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& 2\cdot \dfrac{ m^7 \cdot n^8 \cdot p^4 }{ m^5 \cdot n^4 } \\\\ &=& 2\cdot \dfrac{m^7 }{m^5} \cdot \dfrac{ n^8 }{n^4 } \cdot p^4 \\\\ &=& 2\cdot (m^7\cdot m^{-5}) \cdot ( n^8 \cdot n^{-4})\cdot p^4 \\\\ &=& 2\cdot m^{7-5} \cdot n^{8-4}\cdot p^4 \\\\ \mathbf{\dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } }&\mathbf{=}& \mathbf{2\cdot m^{2} \cdot n^{4}\cdot p^4 } \\\\ \mathbf{\dfrac{ 12m^7 \cdot 5n^8 \cdot 4p^4 }{ 20m^5 \cdot 6n^4 } }&\mathbf{=}& \mathbf{2\cdot m^{2} \cdot (n\cdot p)^4 } \\\\ \end{array}$

(-200000*e^2.5)*(-0.5) ?

$\begin{array}{rcll} ( -200000\cdot e^{2.5} )\cdot(-0.5) &=& ( 200000\cdot e^{2.5} )\cdot(0.5) \quad & | \qquad (-)\cdot(-) = (+) \\ &=& ( 200000\cdot e^{2.5} )\cdot(\frac12) \\ &=& \frac{200000}{2} \cdot e^{2.5} \quad & | \qquad e^{2.5} = 12.182493960703471\dots\\ &=& 100000 \cdot e^{2.5} \\ &=& 100000 \cdot 12.182493960703471\dots \\ \mathbf{( -200000\cdot e^{2.5} )\cdot(-0.5) }&\mathbf{=}& \mathbf{1218249.3960703471\dots} \end{array}$

x^2+3x+4=0 what does the x equal? please solve

$\boxed{~ \begin{array}{rcll} ax^2+bx+c &=& 0\\ x &=& \dfrac{-b \pm \sqrt{b^2-4ac}} {2a} \\ \end{array} ~}$

$\begin{array}{rcll} x^2+3x+4 &=& 0 \qquad & a=1 \quad b=3 \quad c = 4 \\\\ x &=& \frac{-3 \pm \sqrt{3^2-4\cdot 1 \cdot 4} }{2\cdot 1} \\\\ x &=& \frac{-3 \pm \sqrt{9-16} } {2} \\\\ x &=& \frac{-3 \pm \sqrt{-7} } {2} \qquad &| \quad \sqrt{-7}=\sqrt{-1\cdot 7}=\sqrt{-1}\sqrt{7}\\\\ x &=& \frac{-3 \pm \sqrt{-1}\sqrt{7} } {2} \qquad &| \quad \sqrt{-1} = i \\\\ x &=& \frac{-3 \pm i\sqrt{7} } {2} \\\\ \mathbf{x_1} &\mathbf{=}& \mathbf{-\frac32 + \frac{\sqrt{7}}{2}\cdot i } \\\\ \mathbf{x_2} &\mathbf{=}& \mathbf{-\frac32 - \frac{\sqrt{7}}{2}\cdot i } \end{array}$

Algebra 1.

For f = { (-5,2), (-4,-1), (3,-8), (-2,0), (4,-9) } find the indicated value.

f(-4) =-1 _______(-4,-1)___________________

f(3) = -8________(3,-8)___________________

f(4) = -9________(4,-9)___________________

f(-5) =2 ________(-5,2)___________________

Find the four consecutive even whole numbers in 372

$\begin{array}{rcll} \text{Number }~1 ~(n_1) &=& 2\cdot i \\ \text{Number } ~2~(n_2) &=& 2\cdot i +2\\ \text{Number } ~3~(n_3) &=& 2\cdot i +4\\ \text{Number } ~4~(n_4)&=& 2\cdot i +6\\ \end{array}$

$\begin{array}{rcll} n_1+n_2+n_3+n_4 &=& 372 \\ (2\cdot i) +(2\cdot i +2)+(2\cdot i +4)+(2\cdot i +6) &=& 372 \\ 4\cdot 2\cdot i +2 +4+6 &=& 372 \\ 4\cdot 2\cdot i +12 &=& 372 \quad | \quad -12\\ 4\cdot 2\cdot i &=& 372-12\\ 4\cdot 2\cdot i &=& 360\quad | \quad :4\\ 2\cdot i &=& \frac{360}{4}\\ 2\cdot i &=& 90 \end{array}$

$\begin{array}{rcll} n_1 &=& 2\cdot i \\ n_1 &=& 90 \\\\ n_2 &=& 2\cdot i +2\\ n_2 &=& 90+2 \\ n_2 &=& 92 \\\\ n_3 &=& 2\cdot i +4\\ n_3 &=& 90 +4\\ n_3 &=& 94\\\\ n_4 &=& 2\cdot i +6\\ n_4 &=& 90 +6\\ n_4 &=& 96\\\\ n_1+n_2+n_3+n_4 &=& 372 \\ 90+92+94+96 &=& 372 \end{array}$

5.5 lbs.= -?-oz.

A man spends one fifth of the money in his wallet . He then spends one fifth of what remains in the wallet. He spends $36.00 in all. How much money did he have to begin with?

x = Money to begin

$\begin{array}{c|l|r} \hline & \text{wallet} &\text{giving dollar} \\ \hline & x & 0 \\ &(x)-\frac15\cdot x & \frac15\cdot x \\ &(x-\frac15\cdot x) -\frac15(x-\frac15\cdot x) & \frac15(x-\frac15\cdot x)\\ \hline \text{sum} && $36\\ \hline \end{array}$

$\begin{array}{rcl} \frac15\cdot x + \frac15(x-\frac15\cdot x) &=& $36 \quad & | \qquad \cdot 5 \\ x + x-\frac15\cdot x &=& $180 \\ 2x-\frac15\cdot x &=& $180 \quad & | \qquad \cdot 5 \\ 10x-x &=& $900 \\ 9x &=& $900 \quad & | \qquad :9 \\ \mathbf{ x } & \mathbf{=} & \mathbf{$100} \\ \end{array}$

To begin he has $100.

$\begin{array}{c|l|r} \hline & \text{wallet} &\text{giving dollar} \\ \hline & \mathbf{$100} & 0 \\ & $100-$20=$80 & \frac{$100}{5}=$20 \\ & $80-$16=$64 & \frac{$80}{5} =$16\\ \hline \text{sum} && $36\\ \hline \end{array}$

b. Show, using index laws, that 3¹⁰=9⁵

$\begin{array}{lrcll} & 3^{{\color{red}10}} &=& 9^5 \quad & | \qquad {\color{red}10} = 2\cdot 5\\ & 3^{2\cdot 5} &=& 9^5 \\\\ \hline \\ \text{formula: } & \boxed{~ \begin{array}{rcll} (a^b)^c = a^{b\cdot c} = a^{c\cdot b} = (a^c)^b \end{array} ~} \\ \\ \hline \\ & (3^2)^5 &=& 9^5 \quad & | \qquad 3^{2\cdot 5} = (3^2)^5\\ & ({\color{green}3^2})^5 &=& 9^5 \quad & | \qquad {\color{green}3^2} = 3\cdot 3 = 9\\ & 9^5 &=& 9^5 \\ \end{array}$