heureka

log(75÷16)-2log(5÷9)+log(32÷243)=log(2)

$\begin{array}{rcll} \log{( \frac{75}{16} )}-2\cdot \log{(\frac{5}{9} )}+ \log{( \frac{32}{243} )} &\overset{?}{=}& \log{(2)} \\ \log{( \frac{75}{16} )}- \log{(\frac{5^2}{9^2} )}+ \log{( \frac{32}{243} )} &\overset{?}{=}& \log{(2)} \\ \log{( \frac{75}{16}\cdot \frac{1}{\frac{5^2}{9^2} } \cdot \frac{32}{243} )} &\overset{?}{=}& \log{(2)} \\ \log{( \frac{75}{16}\cdot \frac{9^2}{ 5^2 } \cdot \frac{32}{243} )} &\overset{?}{=}& \log{(2)} \\ \log{( \frac{ 75\cdot 9^2 \cdot 32 } {16\cdot 5^2\cdot 243} ) } &\overset{?}{=}& \log{(2)} \\ \log{( \frac{ 75\cdot 81 \cdot 32 } {16\cdot 25\cdot 243} ) } &\overset{?}{=}& \log{(2)} \\ \log{( \frac{75}{25} \cdot \frac{32}{16}\cdot \frac{81}{243} ) } &\overset{?}{=}& \log{(2)} \\ \log{( \frac{3}{1} \cdot \frac{2}{1}\cdot \frac{1}{3} ) } &\overset{?}{=}& \log{(2)} \\ \log{( 2 ) } &=& \log{(2)} \\ \end{array}$

5*4*3 1\2

$\begin{array}{rcll} 5\cdot 4 \cdot 3~ \frac12 &=& 60\ \frac12 \\ &=& 60+\frac12 \\ &=& 60.5 \\ \end{array}$

what is the formula for the volume of a sphere with diameter d?

r = Radius

d = diameter

$\begin{array}{rcll} V &=& \dfrac43 \cdot \pi \cdot r^3 \quad & | \quad r = \dfrac{d}{2}\\\\ V &=& \dfrac43 \cdot \pi \cdot \left(\dfrac{d}{2} \right)^3 \\\\ V &=& \dfrac43 \cdot \pi \cdot \dfrac{d^3}{2^3} \\\\ V &=& \dfrac43 \cdot \pi \cdot \dfrac{d^3}{8} \\\\ V &=& \dfrac13 \cdot \pi \cdot \dfrac{d^3}{2} \\\\ V &=& \dfrac16 \cdot \pi \cdot d^3 \\\\ && \boxed{~ \begin{array}{rcll} V &=& \dfrac{\pi}{6} \cdot d^3 \end{array} ~} \end{array}$

solve for m? y=mx+b??

$\begin{array}{rcll} y &=& mx+b \quad & | \quad -b \\ y-b &=& mx \quad & | \quad :x \\ \dfrac{y-b}{x} &=& m \\ m &=& \dfrac{y-b}{x} \end{array}$

Formel V= Pi x d³ : 6

d = ?

$\begin{array}{rcll} V &=& \dfrac{ \pi \cdot d^3 }{6} \quad & | \quad \cdot 6 \\\\ 6V &=& \pi \cdot d^3 \quad & | \quad :\pi \\\\ \dfrac{6V}{\pi} &=& d^3\\\\ d^3 &=& \dfrac{6V}{\pi} \quad & | \quad \sqrt[3]{} \\\\ d &=& \sqrt[3]{ \dfrac{6V}{\pi} } \end{array}$

10a⁵ : 8a⁴

weis es jemand?

$\begin{array}{rcll} \frac{ 10\cdot a^5} { 8\cdot a^4 } &=& \frac{10}{8} \cdot \frac{a^5}{a^4} \\\\ &=& \frac{2\cdot 5}{2\cdot 4} \cdot \frac{a^4\cdot a}{a^4} \\\\ &=& \frac22 \cdot \frac54 \cdot \frac{a^4}{a^4} \cdot a \\\\ &=& 1 \cdot \frac54 \cdot 1 \cdot a \\\\ &=& \frac54 \cdot a \\\\ \end{array}$

simplify: [ ((e^10x - e^-10x)/2) - ((e^10x + e^-10x)/2) ]¹⁰

need help!!!

$\begin{array}{rcll} \left( \frac{e^{10x} - e^{-10x} } {2} - \frac{e^{10x} + e^{-10x} } {2} \right)^{10} &=& \left[ \frac{e^{10x} - e^{-10x} -( e^{10x} + e^{-10x} )} {2} \right]^{10} \\\\ &=& \left( \frac{e^{10x} - e^{-10x} - e^{10x} - e^{-10x} } {2} \right)^{10} \\\\ &=& \left( \frac{e^{10x}- e^{10x} - e^{-10x} - e^{-10x} } {2} \right)^{10} \\\\ &=& \left( \frac{ 0 - e^{-10x} - e^{-10x} } {2} \right)^{10} \\\\ &=& \left( \frac{ - e^{-10x} - e^{-10x} } {2} \right)^{10} \\\\ &=& \left( \frac{- 2\cdot e^{-10x} } {2} \right)^{10} \\\\ &=& \left( -e^{-10x} \right)^{10} \\\\ &=& \left[ (-1)\cdot e^{-10x} \right]^{10} \\\\ &=& (-1)^{10}\cdot e^{-10\cdot 10 x} \\\\ &=& 1\cdot e^{-10\cdot 10 x} \\\\ &=& 1\cdot e^{-100 x} \\\\ \mathbf{ \left( \frac{e^{10x} - e^{-10x} } {2} - \frac{e^{10x} + e^{-10x} } {2} \right)^{10} } &\mathbf{=}& \mathbf{ e^{-100 x} } \end{array}$

Saul works two part-time jobs. Last week, he worked 15 hours at a shoe store, and 21 hours as a telemarketer. His total pay was $289.50. The week before, he made $356.00 for 20 hours at the shoe store and 24 hours telemarketing. Find his hourly wage for each job.

hourly wage in shoe store $=w_s$

hourly wage in telemarket $=w_t$

$\small{ \begin{array}{lrcll} (1) & 20\ h \cdot w_s + 24\ h \cdot w_t &=& $\ 356.00 \quad & | \quad : 4\\ & 5 \ h \cdot w_s + 6 \ h \cdot w_t &=& $\ 89.00 \\ (2) & 15\ h \cdot w_s + 21\ h \cdot w_t &=& $\ 289.50 \quad & | \quad :3 \\ & 5\ h \cdot w_s + 7\ h \cdot w_t &=& $\ 96.50 \\ \hline \\ (2)-(1): & 5\ h \cdot w_s + 7\ h \cdot w_t - 5 \ h \cdot w_s - 6 \ h \cdot w_t&=& $\ 96.50 - $\ 89.00\\ & 5\ h \cdot w_s- 5 \ h \cdot w_s + 7\ h \cdot w_t - 6 \ h \cdot w_t&=& $\ 96.50 - $\ 89.00\\ & 7\ h \cdot w_t - 6 \ h \cdot w_t&=& $\ 96.50 - $\ 89.00\\ & 1\ h \cdot w_t &=& $\ 7.50\\ & \mathbf{w_t} &\mathbf{=}& \mathbf{ \frac{$\ 7.50} { 1\ h } }\\ \hline \\ & 5 \ h \cdot w_s + 6 \ h \cdot w_t &=& $\ 89.00 \quad & | \quad w_t = \frac{$\ 7.50} { 1\ h }\\ & 5 \ h \cdot w_s + 6 \ h \cdot ( \frac{$\ 7.50} { 1\ h }) &=& $\ 89.00 \\ & 5 \ h \cdot w_s + $\ 45.00 &=& $\ 89.00 \quad & | \quad -$\ 45.00\\ & 5 \ h \cdot w_s &=& $\ 89.00 -$\ 45.00\\ & 5 \ h \cdot w_s &=& $\ 44.00 \quad & | \quad : 5\ h\\ & w_s &=& \frac{ $\ 44.00 } { 5\ h } \\ & \mathbf{w_s} &\mathbf{=}& \mathbf{\frac{$\ 8.80} { 1\ h }} \\ \end{array} }$

why is there no pair of linear functions that can multiply together to result f(x)= x^2+1

why is there no pair of linear functions that can multiply together to result f(x)= x^2+1

There is a pair of linear functions: 

$\begin{array}{rcll} x^2+1 &=& (x-i)\cdot (x+i) \qquad & | \qquad i \text{ is the imaginary number} \end{array}$

$\begin{array}{rcll} x^2+1 &=& (x-i)\cdot (x+i) \\ &=& x^2 - i^2 \qquad & | \qquad i^2 = -1 \\ &=& x^2 - (-1) \\ &=& x^2 +1 \end{array}$