asinus

 \(Thank \ you, \ Melody!\)   !

ln i =??

\(i = \sqrt{-1}\)

 i ist keine reelle Zahl.

Der ln von i ist eine komplexe Zahl.

ln i = 1.57079633 i

Simplify 64 1/6

\(64\frac{1}{6}=\frac{64 \times 6+1}{6}=\frac{385}{6}=64.166\overline6 \)      !

6+ 4/5b= 9/10b

1.

\(6+\frac{4}{5b}=\frac{9}{10b}\)

\(\frac{4}{5b}-\frac{9}{10b} =-6 \)

\(\frac{8-9}{10b} =-6 \)

\(-1=-60b\)

\(b=\frac{1}{60} \)

2.

\(6+\frac{4}{5}b=\frac{9}{10}b\)

\(\frac{4}{5}b-\frac{9}{10}b=-6 \)

\(\frac{8b-9b}{10}=-6 \)

\(-b=-60\)

\(b=60\)

Find out the right result, but take the head and not the computer.

  !

1+2+3+4+5+6+...= -1/12 ????????

\(Die \ Summe \ einer \ Folge \ aus \ positiven \ Zahlen \ ist \ nie \ negativ.\)

  !

-27x ..... =-3

1.  -27x  \(bedeutet "-27x" \)

\(-27x=-3 \)     [   / -27 beidseits,danach kürzen

\(x=\frac{1}{9}\) 

2. \(\times \ bedeutet "mal"\)

\(27\times ...=-3\)    [  /27 beidseits

\(...=\frac{-3}{27}\)            [ kürzen

\(...=-\frac{1}{9}\)

\(27\times -\frac{1}{9}=-3\)

  !

Of course you're right, CPhill.

(1+ (6 (sqrt (2)) / 5 = (1+√72)/5

I do not correct, we have your right solution.
Thanks for the correction.   !

Here's the graph 

x^2+y^2 = 36

y = (1+ (6(sqrt(2))/5)x-6

\(x^2+y^2=36\)

\(y=(1+6\times \frac{\sqrt{2}}{5})x-6=2.69705627484x-6\)

\(y^2=7.27411254965x^2-32.3646752981x+36\)

\(x^2+7.27411254965x^2-32.3646752981x+36=36\)

\(8.27411254965x^2-32.3646752981x=0\)

\(x\times(8.27411254965x-32.3646752981)=0\)

\(x_1=0\)

\(y_1=\pm\sqrt{36}=\pm6\)  (+6 entfällt)

\(x_2= \frac{32.3646752981}{8.27411254965}\)

\(x_2=3.91155850297\)

\(y_2=\pm\sqrt{36-3.91155850297^2}\)

\(y_2=\pm4.54969340481\)   (-4.54969... entfällt)

  !