tolong dong kak
masih bocil saya soal nya
masih bocil saya soal nya
[tex]3x - 10° + 2x + 5° = 180° \\ 5x - 5° = 180° \\ 5x = 180° + 5° \\ 5x = 185° \\ x = \frac{185°}{5°} = 37° \\ \\ pelurus \: < boc = < aoc \\ 3x - 10° \\ 3.37° - 10° \\ 111° - 10° \\ 101°[/tex]
[answer.2.content]![Matematika (+50) KuMat - Kuis Matematika<br />
Materi: Logaritma<br /><br />
Diketahui:<br />
[tex]\large\text{$\begin{aligned}&{}^{\sf2}\log\left({}^{\sf8}\log x\right)={}^{\sf8}\log\left({}^{\sf2}\log x\right)\end{aligned}$}[/tex]<br /><br />
Pertanyaan:<br />
[tex]\Large\text{$\begin{aligned}&\left({}^{2}\log{x}\right)^2=.....?\end{aligned}$}[/tex]](https://1.bp.blogspot.com/-FxxoZkxtk_U/WKAvV2Gd6iI/AAAAAAAACd0/UTrLC-1x23k1E0oqRLErM2XliW2XDiBlQCK4B/s1600/dif.jpg "Matematika (+50) KuMat - Kuis Matematika<br />
Materi: Logaritma<br /><br />
Diketahui:<br />
[tex]\large\text{$\begin{aligned}&{}^{\sf2}\log\left({}^{\sf8}\log x\right)={}^{\sf8}\log\left({}^{\sf2}\log x\right)\end{aligned}$}[/tex]<br /><br />
Pertanyaan:<br />
[tex]\Large\text{$\begin{aligned}&\left({}^{2}\log{x}\right)^2=.....?\end{aligned}$}[/tex]")
