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Soal-Soal Matematika/Eksponen dan logaritma
0
7607
108328
100063
2025-07-07T09:56:14Z
Akuindo
8654
108328
wikitext
text/x-wiki
== Bentuk dan sifat eksponen ==
bentuk: a<sup>b</sup> = c
sifat:
#a<sup>m</sup>xa<sup>n</sup> = a<sup>m+n</sup>
#a<sup>m</sup>:a<sup>n</sup> = a<sup>m-n</sup>
#(a<sup>m</sup>)<sup>n</sup> = a<sup>mxn</sup>
#<math>\sqrt[n]{a^m}</math> = a<sup>m/n</sup>
#a<sup>m</sup>xb<sup>m</sup> = (axb)<sup>m</sup>
#a<sup>m</sup>:b<sup>m</sup> = (a:b)<sup>m</sup>
#a<sup>-m</sup> = <math>\frac{1}{a^m}</math>
#a<sup>m<sup>n</sup></sup> = a<sup>(m<sup>n</sup>)</sup>
#(a<sup>m</sup>)<sup>n</sup> ≠ a<sup>m<sup>n</sup></sup>
#0<sup>a</sup> = 0 dengan a > 0
#a<sup>0</sup> = 1 dengan a ≠ 0
#0:a = 0 dengan a ≠ 0
#a<sup>1</sup> = a
#a<sup>-1</sup> = <math>\frac{1}{a}</math> dengan a ≠ 0
#a<sup>a</sup> = b<sup>b</sup> dimana a = b
; Tambahan
: x<sup>x<sup>……</sup><sup>n</sup></sup> = n maka <math>x = \sqrt[n]{n}</math> dengan x ≥ 1
: x<sup>x<sup>n</sup></sup> = n maka x = n dengan x ≥ 1
: Hasil x dari kedua persamaan yaitu x<sup>2</sup> = 4 dan x = <math>\sqrt{4}</math> adalah berbeda. Untuk x<sup>2</sup> = 4 maka hasilnya x = <math>\pm{2}</math> sedangkan x = <math>\sqrt{4}</math> hasilnya x = 2.
== Bentuk dan sifat logaritma ==
bentuk: <sup>a</sup>log c = b dengan {a,c} > 0 dan a ≠ 1
sifat:
#log a+log b = log (axb)
#log a-log b = log (a:b)
#log a:log b = <sup>b</sup>log a
#<sup>b</sup>log a x <sup>a</sup>log c = <sup>b</sup>log c
#<sup>b</sup>log a = log a:log b
#<sup>b</sup>log a = 1:<sup>a</sup>log b
#<math>b^{^b log a}</math> = a
#<sup>b</sup>log a<sup>m</sup> = m x <sup>b</sup>log a
#<math>^{b^n}log a</math> = 1/n <sup>b</sup>log a
#<math>^{b^n}log a^m</math> = m/n <sup>b</sup>log a
#log a<sup>log b</sup> = log b<sup>log a</sup> = log b x log a
#a<sup>log b</sup> = b<sup>log a</sup>
#<sup>a</sup>log 1 = 0
#<sup>a</sup>log a = 1
contoh soal
Berapa nilai x dari
# x<sup>2x<sup>6</sup></sup> = 36
# x<sup>x<sup>2</sup></sup> = 16
# 6<sup>x</sup> + x = 219
# 2<sup>x</sup> + 2x = 8
# 3<sup>x</sup> + 5x = 101
;jawaban
# x<sup>2x<sup>6</sup></sup> = 36
* x<sup>2x<sup>6</sup></sup> = 6<sup>2</sup>
* (x<sup>2x<sup>6</sup></sup>)<sup>3</sup> = (6<sup>2</sup>)<sup>3</sup>
* (x<sup>6</sup>)<sup>x<sup>6</sup></sup> = 6<sup>6</sup>
* x<sup>6</sup> = 6
* <math>x = \sqrt[6]{6}</math>
# x<sup>x<sup>2</sup></sup> = 16
* x<sup>x<sup>2</sup></sup> = 4<sup>2</sup>
* (x<sup>x<sup>2</sup></sup>)<sup>2</sup> = (4<sup>2</sup>)<sup>2</sup>
* (x<sup>2</sup>)<sup>x<sup>2</sup></sup> = 4<sup>4</sup>
* x<sup>2</sup> = 4
* <math>x = \sqrt{4}</math>
* <math>x = 2</math>
#
; cara biasa
* 6<sup>x</sup> + x = 219
* 6<sup>x</sup> = 219 - x
* 6<sup>219</sup> x 6<sup>x</sup> = (219 - x) 6<sup>219</sup>
* 6<sup>219</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 6<sup>3</sup> x 6<sup>216</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 216 x 6<sup>216</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 216 = 219 - x
* x = 3
; fungsi lambert (W(a x e<sup>a</sup>) = a; e = bilangan euler)
* 6<sup>x</sup> + x = 219
* 6<sup>x</sup> = 219 - x
* 1 = (219 - x) x 6<sup>-x</sup>
* 6<sup>219</sup> = (219 - x) x 6<sup>-x</sup> x 6<sup>219</sup>
* 6<sup>219</sup> = (219 - x) x 6<sup>(219-x)</sup>
* 6<sup>219</sup> = (219 - x) x e<sup>ln 6(219-x)</sup>
* 6<sup>219</sup> = (219 - x) x e<sup>(219-x)ln 6</sup>
* 6<sup>219</sup> x ln 6 = (219 - x) x ln 6 x e<sup>(219-x)ln 6</sup>
* W(6<sup>219</sup> x ln 6) = W((219 - x) x ln 6 x e<sup>(219-x)ln 6</sup>)
* W(6<sup>219</sup> x ln 6) = (219 - x) x ln 6
* W(6<sup>3</sup> x 6<sup>216</sup> x ln 6) = (219 - x) x ln 6
* W(216 x ln 6 x 6<sup>216</sup>) = (219 - x) x ln 6
* W(216 x ln 6 x e<sup>(ln 6 x 216)</sup>) = (219 - x) x ln 6
* W(216 x ln 6 x e<sup>(216 x ln 6)</sup>) = (219 - x) x ln 6
* 216 x ln 6 = (219 - x) x ln 6
* 216 = 219 - x
* x = 3
#
; cara biasa
* 2<sup>x</sup> + 2x = 8
* 2<sup>x</sup> = 8 - 2x
* 2<sup>x</sup> = 2(4 - x)
* 2<sup>4</sup> x 2<sup>x</sup> = 2(4 - x) 2<sup>4</sup>
* <math>\frac{2^4}{2}</math> = (4 - x) 2<sup>(4 - x)</sup>
* 2<sup>3</sup> = (4 - x) 2<sup>(4 - x)</sup>
* 2 x 2<sup>2</sup> = (4 - x) 2<sup>(4 - x)</sup>
* 2 = 4 - x
* x = 2
; fungsi lambert
* 2<sup>x</sup> + 2x = 8
* 2<sup>x</sup> = 8 - 2x
* 1 = (8 - 2x) x 2<sup>-x</sup>
* 2<sup>4</sup> = 2(4 - x) x 2<sup>-x</sup> x 2<sup>4</sup>
* 2<sup>3</sup> = (4 - x) x 2<sup>(4-x)</sup>
* 2<sup>3</sup> = (4 - x) x e<sup>ln 2(4-x)</sup>
* 2<sup>3</sup> = (4 - x) x e<sup>(4-x)ln 2</sup>
* 2<sup>3</sup> x ln 2 = (4 - x) x ln 2 x e<sup>(4-x)ln 2</sup>
* W(2<sup>3</sup> x ln 2) = W((4 - x) x ln 2 x e<sup>(4-x)ln 2</sup>)
* W(2<sup>3</sup> x ln 2) = (4 - x) x ln 2
* W(2 x 2<sup>2</sup> x ln 2) = (4 - x) x ln 2
* W(2 x ln 2 x 2<sup>2</sup>) = (4 - x) x ln 2
* W(2 x ln 2 x e<sup>(ln 2 x 2)</sup>) = (4 - x) x ln 2
* W(2 x ln 2 x e<sup>(2 x ln 2)</sup>) = (4 - x) x ln 2
* 2 x ln 2 = (4 - x) x ln 2
* 2 = 4 - x
* x = 2
#
; cara biasa
* 3<sup>x</sup> + 5x = 101
* 3<sup>x</sup> = 101 - 5x
* 3<sup>x</sup> = 5(<math>\frac{101}{5}</math> - x)
* 3<sup><math>\frac{101}{5}</math></sup> x 3<sup>x</sup> = 5(<math>\frac{101}{5}</math> - x) 3<sup><math>\frac{101}{5}</math></sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{1}{5}</math> 3<sup><math>\frac{20}{5}</math></sup> 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{1}{5}</math> 3<sup>4</sup> 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{81}{5}</math> x 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{81}{5}</math> = <math>\frac{101}{5}</math> - x
* x = <math>\frac{101}{5} - \frac{81}{5}</math>
* x = <math>\frac{20}{5}</math>
* x = 4
; fungsi lambert
* 3<sup>x</sup> + 5x = 101
* 3<sup>x</sup> = 101 - 5x
* 1 = (101 - 5x) x 3<sup>-x</sup>
* 3<sup><math>\frac{101}{5}</math></sup> = 5(<math>\frac{101}{5}</math> - x) x 3<sup>-x</sup> x 3<sup><math>\frac{101}{5}</math></sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x 3<sup>(<math>\frac{101}{5}</math>-x)</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x e<sup>ln 3(<math>\frac{101}{5}</math>-x)</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3 = (<math>\frac{101}{5}</math> - x) x ln 3 x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>
* W(<math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3) = W((<math>\frac{101}{5}</math> - x) x ln 3 x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>)
* W(<math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{\frac{20}{5}} \cdot 3^{\frac{81}{5}}}{5}</math> x ln 3) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{\frac{20}{5}}}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{4}}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x e<sup>(ln 3 x <math>\frac{81}{5}</math>)</sup>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x e<sup>(<math>\frac{81}{5}</math> x ln 3)</sup>) = (<math>\frac{101}{5}</math> - x) x ln 3
* <math>\frac{81}{5}</math> x ln 3 = (<math>\frac{101}{5}</math> - x) x ln 3
* <math>\frac{81}{5}</math> = <math>\frac{101}{5}</math> - x
* x = <math>\frac{101}{5}</math> - <math>\frac{81}{5}</math>
* x = <math>\frac{20}{5}</math>
* x = 4
; contoh soal
# Tentukan nilai dari <math>\sqrt[3]{5+2\sqrt{3}}+\sqrt[3]{5-2\sqrt{3}}</math>!
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\sqrt[3]{5+2\sqrt{3}}+\sqrt[3]{5-2\sqrt{3}}
\end{align}
</math>
</div></div>
[[Kategori:Soal-Soal Matematika]]
jl8wk6al2qmjt8twtlpken0mo5nq95m
108330
108328
2025-07-07T10:17:43Z
Akuindo
8654
108330
wikitext
text/x-wiki
== Bentuk dan sifat eksponen ==
bentuk: a<sup>b</sup> = c
sifat:
#a<sup>m</sup>xa<sup>n</sup> = a<sup>m+n</sup>
#a<sup>m</sup>:a<sup>n</sup> = a<sup>m-n</sup>
#(a<sup>m</sup>)<sup>n</sup> = a<sup>mxn</sup>
#<math>\sqrt[n]{a^m}</math> = a<sup>m/n</sup>
#a<sup>m</sup>xb<sup>m</sup> = (axb)<sup>m</sup>
#a<sup>m</sup>:b<sup>m</sup> = (a:b)<sup>m</sup>
#a<sup>-m</sup> = <math>\frac{1}{a^m}</math>
#a<sup>m<sup>n</sup></sup> = a<sup>(m<sup>n</sup>)</sup>
#(a<sup>m</sup>)<sup>n</sup> ≠ a<sup>m<sup>n</sup></sup>
#0<sup>a</sup> = 0 dengan a > 0
#a<sup>0</sup> = 1 dengan a ≠ 0
#0:a = 0 dengan a ≠ 0
#a<sup>1</sup> = a
#a<sup>-1</sup> = <math>\frac{1}{a}</math> dengan a ≠ 0
#a<sup>a</sup> = b<sup>b</sup> dimana a = b
; Tambahan
: x<sup>x<sup>……</sup><sup>n</sup></sup> = n maka <math>x = \sqrt[n]{n}</math> dengan x ≥ 1
: x<sup>x<sup>n</sup></sup> = n maka x = n dengan x ≥ 1
: Hasil x dari kedua persamaan yaitu x<sup>2</sup> = 4 dan x = <math>\sqrt{4}</math> adalah berbeda. Untuk x<sup>2</sup> = 4 maka hasilnya x = <math>\pm{2}</math> sedangkan x = <math>\sqrt{4}</math> hasilnya x = 2.
== Bentuk dan sifat logaritma ==
bentuk: <sup>a</sup>log c = b dengan {a,c} > 0 dan a ≠ 1
sifat:
#log a+log b = log (axb)
#log a-log b = log (a:b)
#log a:log b = <sup>b</sup>log a
#<sup>b</sup>log a x <sup>a</sup>log c = <sup>b</sup>log c
#<sup>b</sup>log a = log a:log b
#<sup>b</sup>log a = 1:<sup>a</sup>log b
#<math>b^{^b log a}</math> = a
#<sup>b</sup>log a<sup>m</sup> = m x <sup>b</sup>log a
#<math>^{b^n}log a</math> = 1/n <sup>b</sup>log a
#<math>^{b^n}log a^m</math> = m/n <sup>b</sup>log a
#log a<sup>log b</sup> = log b<sup>log a</sup> = log b x log a
#a<sup>log b</sup> = b<sup>log a</sup>
#<sup>a</sup>log 1 = 0
#<sup>a</sup>log a = 1
contoh soal
Berapa nilai x dari
# x<sup>2x<sup>6</sup></sup> = 36
# x<sup>x<sup>2</sup></sup> = 16
# 6<sup>x</sup> + x = 219
# 2<sup>x</sup> + 2x = 8
# 3<sup>x</sup> + 5x = 101
;jawaban
# x<sup>2x<sup>6</sup></sup> = 36
* x<sup>2x<sup>6</sup></sup> = 6<sup>2</sup>
* (x<sup>2x<sup>6</sup></sup>)<sup>3</sup> = (6<sup>2</sup>)<sup>3</sup>
* (x<sup>6</sup>)<sup>x<sup>6</sup></sup> = 6<sup>6</sup>
* x<sup>6</sup> = 6
* <math>x = \sqrt[6]{6}</math>
# x<sup>x<sup>2</sup></sup> = 16
* x<sup>x<sup>2</sup></sup> = 4<sup>2</sup>
* (x<sup>x<sup>2</sup></sup>)<sup>2</sup> = (4<sup>2</sup>)<sup>2</sup>
* (x<sup>2</sup>)<sup>x<sup>2</sup></sup> = 4<sup>4</sup>
* x<sup>2</sup> = 4
* <math>x = \sqrt{4}</math>
* <math>x = 2</math>
#
; cara biasa
* 6<sup>x</sup> + x = 219
* 6<sup>x</sup> = 219 - x
* 6<sup>219</sup> x 6<sup>x</sup> = (219 - x) 6<sup>219</sup>
* 6<sup>219</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 6<sup>3</sup> x 6<sup>216</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 216 x 6<sup>216</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 216 = 219 - x
* x = 3
; fungsi lambert (W(a x e<sup>a</sup>) = a; e = bilangan euler)
* 6<sup>x</sup> + x = 219
* 6<sup>x</sup> = 219 - x
* 1 = (219 - x) x 6<sup>-x</sup>
* 6<sup>219</sup> = (219 - x) x 6<sup>-x</sup> x 6<sup>219</sup>
* 6<sup>219</sup> = (219 - x) x 6<sup>(219-x)</sup>
* 6<sup>219</sup> = (219 - x) x e<sup>ln 6(219-x)</sup>
* 6<sup>219</sup> = (219 - x) x e<sup>(219-x)ln 6</sup>
* 6<sup>219</sup> x ln 6 = (219 - x) x ln 6 x e<sup>(219-x)ln 6</sup>
* W(6<sup>219</sup> x ln 6) = W((219 - x) x ln 6 x e<sup>(219-x)ln 6</sup>)
* W(6<sup>219</sup> x ln 6) = (219 - x) x ln 6
* W(6<sup>3</sup> x 6<sup>216</sup> x ln 6) = (219 - x) x ln 6
* W(216 x ln 6 x 6<sup>216</sup>) = (219 - x) x ln 6
* W(216 x ln 6 x e<sup>(ln 6 x 216)</sup>) = (219 - x) x ln 6
* W(216 x ln 6 x e<sup>(216 x ln 6)</sup>) = (219 - x) x ln 6
* 216 x ln 6 = (219 - x) x ln 6
* 216 = 219 - x
* x = 3
#
; cara biasa
* 2<sup>x</sup> + 2x = 8
* 2<sup>x</sup> = 8 - 2x
* 2<sup>x</sup> = 2(4 - x)
* 2<sup>4</sup> x 2<sup>x</sup> = 2(4 - x) 2<sup>4</sup>
* <math>\frac{2^4}{2}</math> = (4 - x) 2<sup>(4 - x)</sup>
* 2<sup>3</sup> = (4 - x) 2<sup>(4 - x)</sup>
* 2 x 2<sup>2</sup> = (4 - x) 2<sup>(4 - x)</sup>
* 2 = 4 - x
* x = 2
; fungsi lambert
* 2<sup>x</sup> + 2x = 8
* 2<sup>x</sup> = 8 - 2x
* 1 = (8 - 2x) x 2<sup>-x</sup>
* 2<sup>4</sup> = 2(4 - x) x 2<sup>-x</sup> x 2<sup>4</sup>
* 2<sup>3</sup> = (4 - x) x 2<sup>(4-x)</sup>
* 2<sup>3</sup> = (4 - x) x e<sup>ln 2(4-x)</sup>
* 2<sup>3</sup> = (4 - x) x e<sup>(4-x)ln 2</sup>
* 2<sup>3</sup> x ln 2 = (4 - x) x ln 2 x e<sup>(4-x)ln 2</sup>
* W(2<sup>3</sup> x ln 2) = W((4 - x) x ln 2 x e<sup>(4-x)ln 2</sup>)
* W(2<sup>3</sup> x ln 2) = (4 - x) x ln 2
* W(2 x 2<sup>2</sup> x ln 2) = (4 - x) x ln 2
* W(2 x ln 2 x 2<sup>2</sup>) = (4 - x) x ln 2
* W(2 x ln 2 x e<sup>(ln 2 x 2)</sup>) = (4 - x) x ln 2
* W(2 x ln 2 x e<sup>(2 x ln 2)</sup>) = (4 - x) x ln 2
* 2 x ln 2 = (4 - x) x ln 2
* 2 = 4 - x
* x = 2
#
; cara biasa
* 3<sup>x</sup> + 5x = 101
* 3<sup>x</sup> = 101 - 5x
* 3<sup>x</sup> = 5(<math>\frac{101}{5}</math> - x)
* 3<sup><math>\frac{101}{5}</math></sup> x 3<sup>x</sup> = 5(<math>\frac{101}{5}</math> - x) 3<sup><math>\frac{101}{5}</math></sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{1}{5}</math> 3<sup><math>\frac{20}{5}</math></sup> 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{1}{5}</math> 3<sup>4</sup> 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{81}{5}</math> x 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{81}{5}</math> = <math>\frac{101}{5}</math> - x
* x = <math>\frac{101}{5} - \frac{81}{5}</math>
* x = <math>\frac{20}{5}</math>
* x = 4
; fungsi lambert
* 3<sup>x</sup> + 5x = 101
* 3<sup>x</sup> = 101 - 5x
* 1 = (101 - 5x) x 3<sup>-x</sup>
* 3<sup><math>\frac{101}{5}</math></sup> = 5(<math>\frac{101}{5}</math> - x) x 3<sup>-x</sup> x 3<sup><math>\frac{101}{5}</math></sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x 3<sup>(<math>\frac{101}{5}</math>-x)</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x e<sup>ln 3(<math>\frac{101}{5}</math>-x)</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3 = (<math>\frac{101}{5}</math> - x) x ln 3 x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>
* W(<math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3) = W((<math>\frac{101}{5}</math> - x) x ln 3 x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>)
* W(<math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{\frac{20}{5}} \cdot 3^{\frac{81}{5}}}{5}</math> x ln 3) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{\frac{20}{5}}}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{4}}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x e<sup>(ln 3 x <math>\frac{81}{5}</math>)</sup>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x e<sup>(<math>\frac{81}{5}</math> x ln 3)</sup>) = (<math>\frac{101}{5}</math> - x) x ln 3
* <math>\frac{81}{5}</math> x ln 3 = (<math>\frac{101}{5}</math> - x) x ln 3
* <math>\frac{81}{5}</math> = <math>\frac{101}{5}</math> - x
* x = <math>\frac{101}{5}</math> - <math>\frac{81}{5}</math>
* x = <math>\frac{20}{5}</math>
* x = 4
; contoh soal
# Tentukan nilai dari <math>\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}</math>!
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} \\
\text{misal } \sqrt[3]{5+2\sqrt{13}}=a \text{ dan } \sqrt[3]{5-2\sqrt{13}}=b \text{ maka menjadi } a+b=p \\
a+b &= p \\
(a+b)^3 &= p^3 \\
a^3+b^3+3ab(a+b) &= p^3 \\
a^3+b^3 &= (\sqrt[3]{5+2\sqrt{13}})^3+(\sqrt[3]{5-2\sqrt{13}})^3 \\
&= 5+2\sqrt{13}+5-2\sqrt{13} \\
&= 10 \\
ab &= (\sqrt[3]{5+2\sqrt{13}})\cdot(\sqrt[3]{5-2\sqrt{13}}) \\
&= \sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13}} \\
&= \sqrt[3]{25-52} \\
&= \sqrt[3]{-27} \\
&= -3 \\
a^3+b^3+3ab(a+b) &= p^3 \\
10+3(-3)p &= p^3 \\
10-9p &= p^3 \\
p^3+9p-10 &= 0 \\
(p-1)(p^2+p+10) &= 0 \\
p=1 &\text{ atau (definit positif) } \\
\end{align}
</math>
</div></div>
[[Kategori:Soal-Soal Matematika]]
7ziltd4f23jasz9jh5jgf22hzr2oz1n
108331
108330
2025-07-07T10:19:12Z
Akuindo
8654
108331
wikitext
text/x-wiki
== Bentuk dan sifat eksponen ==
bentuk: a<sup>b</sup> = c
sifat:
#a<sup>m</sup>xa<sup>n</sup> = a<sup>m+n</sup>
#a<sup>m</sup>:a<sup>n</sup> = a<sup>m-n</sup>
#(a<sup>m</sup>)<sup>n</sup> = a<sup>mxn</sup>
#<math>\sqrt[n]{a^m}</math> = a<sup>m/n</sup>
#a<sup>m</sup>xb<sup>m</sup> = (axb)<sup>m</sup>
#a<sup>m</sup>:b<sup>m</sup> = (a:b)<sup>m</sup>
#a<sup>-m</sup> = <math>\frac{1}{a^m}</math>
#a<sup>m<sup>n</sup></sup> = a<sup>(m<sup>n</sup>)</sup>
#(a<sup>m</sup>)<sup>n</sup> ≠ a<sup>m<sup>n</sup></sup>
#0<sup>a</sup> = 0 dengan a > 0
#a<sup>0</sup> = 1 dengan a ≠ 0
#0:a = 0 dengan a ≠ 0
#a<sup>1</sup> = a
#a<sup>-1</sup> = <math>\frac{1}{a}</math> dengan a ≠ 0
#a<sup>a</sup> = b<sup>b</sup> dimana a = b
; Tambahan
: x<sup>x<sup>……</sup><sup>n</sup></sup> = n maka <math>x = \sqrt[n]{n}</math> dengan x ≥ 1
: x<sup>x<sup>n</sup></sup> = n maka x = n dengan x ≥ 1
: Hasil x dari kedua persamaan yaitu x<sup>2</sup> = 4 dan x = <math>\sqrt{4}</math> adalah berbeda. Untuk x<sup>2</sup> = 4 maka hasilnya x = <math>\pm{2}</math> sedangkan x = <math>\sqrt{4}</math> hasilnya x = 2.
== Bentuk dan sifat logaritma ==
bentuk: <sup>a</sup>log c = b dengan {a,c} > 0 dan a ≠ 1
sifat:
#log a+log b = log (axb)
#log a-log b = log (a:b)
#log a:log b = <sup>b</sup>log a
#<sup>b</sup>log a x <sup>a</sup>log c = <sup>b</sup>log c
#<sup>b</sup>log a = log a:log b
#<sup>b</sup>log a = 1:<sup>a</sup>log b
#<math>b^{^b log a}</math> = a
#<sup>b</sup>log a<sup>m</sup> = m x <sup>b</sup>log a
#<math>^{b^n}log a</math> = 1/n <sup>b</sup>log a
#<math>^{b^n}log a^m</math> = m/n <sup>b</sup>log a
#log a<sup>log b</sup> = log b<sup>log a</sup> = log b x log a
#a<sup>log b</sup> = b<sup>log a</sup>
#<sup>a</sup>log 1 = 0
#<sup>a</sup>log a = 1
contoh soal
Berapa nilai x dari
# x<sup>2x<sup>6</sup></sup> = 36
# x<sup>x<sup>2</sup></sup> = 16
# 6<sup>x</sup> + x = 219
# 2<sup>x</sup> + 2x = 8
# 3<sup>x</sup> + 5x = 101
;jawaban
# x<sup>2x<sup>6</sup></sup> = 36
* x<sup>2x<sup>6</sup></sup> = 6<sup>2</sup>
* (x<sup>2x<sup>6</sup></sup>)<sup>3</sup> = (6<sup>2</sup>)<sup>3</sup>
* (x<sup>6</sup>)<sup>x<sup>6</sup></sup> = 6<sup>6</sup>
* x<sup>6</sup> = 6
* <math>x = \sqrt[6]{6}</math>
# x<sup>x<sup>2</sup></sup> = 16
* x<sup>x<sup>2</sup></sup> = 4<sup>2</sup>
* (x<sup>x<sup>2</sup></sup>)<sup>2</sup> = (4<sup>2</sup>)<sup>2</sup>
* (x<sup>2</sup>)<sup>x<sup>2</sup></sup> = 4<sup>4</sup>
* x<sup>2</sup> = 4
* <math>x = \sqrt{4}</math>
* <math>x = 2</math>
#
; cara biasa
* 6<sup>x</sup> + x = 219
* 6<sup>x</sup> = 219 - x
* 6<sup>219</sup> x 6<sup>x</sup> = (219 - x) 6<sup>219</sup>
* 6<sup>219</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 6<sup>3</sup> x 6<sup>216</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 216 x 6<sup>216</sup> = (219 - x) 6<sup>(219 - x)</sup>
* 216 = 219 - x
* x = 3
; fungsi lambert (W(a x e<sup>a</sup>) = a; e = bilangan euler)
* 6<sup>x</sup> + x = 219
* 6<sup>x</sup> = 219 - x
* 1 = (219 - x) x 6<sup>-x</sup>
* 6<sup>219</sup> = (219 - x) x 6<sup>-x</sup> x 6<sup>219</sup>
* 6<sup>219</sup> = (219 - x) x 6<sup>(219-x)</sup>
* 6<sup>219</sup> = (219 - x) x e<sup>ln 6(219-x)</sup>
* 6<sup>219</sup> = (219 - x) x e<sup>(219-x)ln 6</sup>
* 6<sup>219</sup> x ln 6 = (219 - x) x ln 6 x e<sup>(219-x)ln 6</sup>
* W(6<sup>219</sup> x ln 6) = W((219 - x) x ln 6 x e<sup>(219-x)ln 6</sup>)
* W(6<sup>219</sup> x ln 6) = (219 - x) x ln 6
* W(6<sup>3</sup> x 6<sup>216</sup> x ln 6) = (219 - x) x ln 6
* W(216 x ln 6 x 6<sup>216</sup>) = (219 - x) x ln 6
* W(216 x ln 6 x e<sup>(ln 6 x 216)</sup>) = (219 - x) x ln 6
* W(216 x ln 6 x e<sup>(216 x ln 6)</sup>) = (219 - x) x ln 6
* 216 x ln 6 = (219 - x) x ln 6
* 216 = 219 - x
* x = 3
#
; cara biasa
* 2<sup>x</sup> + 2x = 8
* 2<sup>x</sup> = 8 - 2x
* 2<sup>x</sup> = 2(4 - x)
* 2<sup>4</sup> x 2<sup>x</sup> = 2(4 - x) 2<sup>4</sup>
* <math>\frac{2^4}{2}</math> = (4 - x) 2<sup>(4 - x)</sup>
* 2<sup>3</sup> = (4 - x) 2<sup>(4 - x)</sup>
* 2 x 2<sup>2</sup> = (4 - x) 2<sup>(4 - x)</sup>
* 2 = 4 - x
* x = 2
; fungsi lambert
* 2<sup>x</sup> + 2x = 8
* 2<sup>x</sup> = 8 - 2x
* 1 = (8 - 2x) x 2<sup>-x</sup>
* 2<sup>4</sup> = 2(4 - x) x 2<sup>-x</sup> x 2<sup>4</sup>
* 2<sup>3</sup> = (4 - x) x 2<sup>(4-x)</sup>
* 2<sup>3</sup> = (4 - x) x e<sup>ln 2(4-x)</sup>
* 2<sup>3</sup> = (4 - x) x e<sup>(4-x)ln 2</sup>
* 2<sup>3</sup> x ln 2 = (4 - x) x ln 2 x e<sup>(4-x)ln 2</sup>
* W(2<sup>3</sup> x ln 2) = W((4 - x) x ln 2 x e<sup>(4-x)ln 2</sup>)
* W(2<sup>3</sup> x ln 2) = (4 - x) x ln 2
* W(2 x 2<sup>2</sup> x ln 2) = (4 - x) x ln 2
* W(2 x ln 2 x 2<sup>2</sup>) = (4 - x) x ln 2
* W(2 x ln 2 x e<sup>(ln 2 x 2)</sup>) = (4 - x) x ln 2
* W(2 x ln 2 x e<sup>(2 x ln 2)</sup>) = (4 - x) x ln 2
* 2 x ln 2 = (4 - x) x ln 2
* 2 = 4 - x
* x = 2
#
; cara biasa
* 3<sup>x</sup> + 5x = 101
* 3<sup>x</sup> = 101 - 5x
* 3<sup>x</sup> = 5(<math>\frac{101}{5}</math> - x)
* 3<sup><math>\frac{101}{5}</math></sup> x 3<sup>x</sup> = 5(<math>\frac{101}{5}</math> - x) 3<sup><math>\frac{101}{5}</math></sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{1}{5}</math> 3<sup><math>\frac{20}{5}</math></sup> 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{1}{5}</math> 3<sup>4</sup> 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{81}{5}</math> x 3<sup><math>\frac{81}{5}</math></sup> = (<math>\frac{101}{5}</math> - x) 3<sup>(<math>\frac{101}{5}</math> - x)</sup>
* <math>\frac{81}{5}</math> = <math>\frac{101}{5}</math> - x
* x = <math>\frac{101}{5} - \frac{81}{5}</math>
* x = <math>\frac{20}{5}</math>
* x = 4
; fungsi lambert
* 3<sup>x</sup> + 5x = 101
* 3<sup>x</sup> = 101 - 5x
* 1 = (101 - 5x) x 3<sup>-x</sup>
* 3<sup><math>\frac{101}{5}</math></sup> = 5(<math>\frac{101}{5}</math> - x) x 3<sup>-x</sup> x 3<sup><math>\frac{101}{5}</math></sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x 3<sup>(<math>\frac{101}{5}</math>-x)</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x e<sup>ln 3(<math>\frac{101}{5}</math>-x)</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> = (<math>\frac{101}{5}</math> - x) x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>
* <math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3 = (<math>\frac{101}{5}</math> - x) x ln 3 x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>
* W(<math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3) = W((<math>\frac{101}{5}</math> - x) x ln 3 x e<sup>(<math>\frac{101}{5}</math>-x)ln 3</sup>)
* W(<math>\frac{3^{\frac{101}{5}}}{5}</math> x ln 3) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{\frac{20}{5}} \cdot 3^{\frac{81}{5}}}{5}</math> x ln 3) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{\frac{20}{5}}}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{3^{4}}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x <math>3^{\frac{81}{5}}</math>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x e<sup>(ln 3 x <math>\frac{81}{5}</math>)</sup>) = (<math>\frac{101}{5}</math> - x) x ln 3
* W(<math>\frac{81}{5}</math> x ln 3 x e<sup>(<math>\frac{81}{5}</math> x ln 3)</sup>) = (<math>\frac{101}{5}</math> - x) x ln 3
* <math>\frac{81}{5}</math> x ln 3 = (<math>\frac{101}{5}</math> - x) x ln 3
* <math>\frac{81}{5}</math> = <math>\frac{101}{5}</math> - x
* x = <math>\frac{101}{5}</math> - <math>\frac{81}{5}</math>
* x = <math>\frac{20}{5}</math>
* x = 4
; contoh soal
# Tentukan nilai dari <math>\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}</math>!
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}} \\
\text{misal } \sqrt[3]{5+2\sqrt{13}}=a \text{ dan } \sqrt[3]{5-2\sqrt{13}}=b \text{ maka menjadi } a+b=p \\
a+b &= p \\
(a+b)^3 &= p^3 \\
a^3+b^3+3ab(a+b) &= p^3 \\
a^3+b^3 &= (\sqrt[3]{5+2\sqrt{13}})^3+(\sqrt[3]{5-2\sqrt{13}})^3 \\
&= 5+2\sqrt{13}+5-2\sqrt{13} \\
&= 10 \\
ab &= (\sqrt[3]{5+2\sqrt{13}})\cdot(\sqrt[3]{5-2\sqrt{13}}) \\
&= \sqrt[3]{(5+2\sqrt{13})(5-2\sqrt{13})} \\
&= \sqrt[3]{25-52} \\
&= \sqrt[3]{-27} \\
&= -3 \\
a^3+b^3+3ab(a+b) &= p^3 \\
10+3(-3)p &= p^3 \\
10-9p &= p^3 \\
p^3+9p-10 &= 0 \\
(p-1)(p^2+p+10) &= 0 \\
p=1 &\text{ atau (definit positif) } \\
\end{align}
</math>
</div></div>
[[Kategori:Soal-Soal Matematika]]
ne99m632hxu6opw4o20mwc9trfw1g88
Soal-Soal Matematika/Bangun datar
0
23127
108313
108312
2025-07-06T12:05:20Z
Akuindo
8654
108313
wikitext
text/x-wiki
Bangun datar disebut juga dimensi dua yang memiliki sebidang serta beberapa rusuknya. bangun ini memiliki simetri lipat, simetri putar dan sumbu simetri
{| class="wikitable"
|+
|-
! bangun datar !! simetri lipat !! simetri putar !! sumbu simetri
|-
| persegi || 4 || 4 || 4
|-
| segi-n || || ||
|-
| persegi panjang || 2 || 2 || 2
|-
| segitiga sama sisi || 3 || 3 || 3
|-
| segitiga sama kaki || 1 || 0 || 1
|-
| segitiga siku-siku || 1 || 0 || 1
|-
| lingkaran || takhingga || takhingga || takhingga
|-
| jajar genjang || 0 || 2 || 0
|-
| belah ketupat || 2 || 2 || 2
|-
| trapesium sama kaki || 1 || 0 || 1
|-
| trapesium siku-siku || 0 || 0 || 0
|-
| layang-layang || 1 || 0 || 1
|-
| elips || 2 || 0 || 2
|}
; Besar sudut keliling dan sudut pusat
: sudut pusat = 2 x sudut keliling
; Hubungan ketiga sisi segitiga dengan jari-jari lingkaran (lingkaran dalam segitiga siku-siku): <math>r = \frac{a+b-c}{2}</math>
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
a-r + b-r &= c \\
a+b-2r &= c \\
2r &= a+b-c \\
r &= \frac{a+b-c}{2} \\
\end{align}
</math>
</div></div>
; Hubungan dua tali busur dengan jari-jari lingkaran: <math>r = \sqrt{\frac{a^2 + b^2 + c^2 + d^2}{4}}</math>
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\text{ misalkan } p &= \frac{c+d}{2} \text{ dan } q = \frac{a+b}{2} \\
\text{ misalkan persegi panjang dibuat x dan y} & \\
x + p &= d \\
x + \frac{c+d}{2} &= d \\
x &= d - \frac{c+d}{2} \\
x &= \frac{d-c}{2} \\
y + q &= b \\
y + \frac{a+b}{2} &= b \\
y &= b - \frac{a+b}{2} \\
y &= \frac{b-a}{2} \\
y^2 + p^2 &= r^2 \\
(\frac{b-a}{2})^2 + (\frac{c+d}{2})^2 &= r^2 \\
\frac{a^2-2ab+b^2}{4} + \frac{c^2+2cd+d^2}{4} &= r^2 \\
x^2 + q^2 &= r^2 \\
(\frac{d-c}{2})^2 + (\frac{a+b}{2})^2 &= r^2 \\
\frac{c^2-2cd+d^2}{4} + \frac{a^2+2ab+b^2}{4} &= r^2 \\
\text {jumlah dua persamaan tersebut menjadi } & \\
\frac{a^2-2ab+b^2}{4} + \frac{c^2+2cd+d^2}{4} + \frac{c^2-2cd+d^2}{4} + \frac{a^2+2ab+b^2}{4} &= 2r^2 \\
\frac{2a^2+2b^2+2c^2+2d^2}{4} &= 2r^2 \\
\frac{2(a^2+b^2+c^2+d^2)}{4} &= 2r^2 \\
\frac{a^2+b^2+c^2+d^2}{4} &= r^2 \\
a^2+b^2+c^2+d^2 &= 4r^2 \\
r &= \sqrt{\frac{a^2 + b^2 + c^2 + d^2}{4}} \\
\end{align}
</math>
</div></div>
; Lingkaran dalam segitiga: <math>r = \frac{L}{s}</math> (s adalah setengah keliling segitiga)
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\text{ misalkan luas segitiga AOC } L_a &= \frac{ra}{2} \\
\text{ misalkan luas segitiga COB } L_b &= \frac{rb}{2} \\
\text{ misalkan luas segitiga BOC } L_c &= \frac{rc}{2} \\
\text{ maka total luas ketiga segitiga tersebut adalah } L &= L_a + L_b + L_c \\
&= \frac{ra}{2} + \frac{rb}{2} + \frac{rc}{2} \\
&= r \frac{a+b+c}{2} \\
&= r s \\
r &= \frac{L}{s} \\
\end{align}
</math>
</div></div>
; Lingkaran luar segitiga: <math>r = \frac{abc}{4L}</math>
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\text{ lihat posisi kesebangunan } \frac{t}{b} &= \frac{c}{2r} \\
t &= \frac{bc}{2r} \\
L &= \frac{at}{2} \\
&= \frac{abc}{2(2r)} \\
r &= \frac{abc}{4L} \\
\end{align}
</math>
</div></div>
; Garis singgung (satu) lingkaran: <math>d = \sqrt{a^2-b^2}</math>
; Garis singgung persekutuan dalam (dua) lingkaran: <math>d = \sqrt{p^2-(R+r)^2}</math>
; Garis singgung persekutuan luar (dua) lingkaran: <math>d = \sqrt{p^2-(R-r)^2}</math>
keterangan:
: d = panjang singgung persekutuan dalam/luar lingkaran
: p = jarak kedua titik pusat lingkaran
; Teorema Ceva: AF x BD x CE = BF x CD x AE
; Teorema de Pitot: AB + DC = AD + BC (segiempat luar lingkaran)
; Teorema Power de Point: (segiempat dalam lingkaran)
:: AE x EC = BE x ED
; Tali busur pada satu titik di luar lingkaran:
:: AB<sup>2</sup> = AD x AE
:: AB x AC = AD x AE
; Teorema Ptolomeos (segiempat dalam lingkaran)
: AC x BD = AB x DC + AD x BC
; Titik E adalah pertemuan kedua perpanjangan tali busur yang bertemu.
:: maka sudut E = 1/2 (sudut besar-sudut kecil)
; Hubungan besar sudut, luas juring dan panjang busur
: <math>\frac{\text{besar sudut AOB }}{\text{besar sudut COD }} = \frac{\text{luas juring AOB }}{\text{luas juring COD }} = \frac{\text{panjang busur AB }}{\text{panjang busur CD }}</math>
[[Kategori:Soal-Soal Matematika]]
qxww8tj5liapttxmwrb94iusn8clh6x
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Bangun datar disebut juga dimensi dua yang memiliki sebidang serta beberapa rusuknya. bangun ini memiliki simetri lipat, simetri putar dan sumbu simetri
{| class="wikitable"
|+
|-
! bangun datar !! simetri lipat !! simetri putar !! sumbu simetri
|-
| persegi || 4 || 4 || 4
|-
| segi-n || || ||
|-
| persegi panjang || 2 || 2 || 2
|-
| segitiga sama sisi || 3 || 3 || 3
|-
| segitiga sama kaki || 1 || 0 || 1
|-
| segitiga siku-siku || 1 || 0 || 1
|-
| lingkaran || takhingga || takhingga || takhingga
|-
| jajar genjang || 0 || 2 || 0
|-
| belah ketupat || 2 || 2 || 2
|-
| trapesium sama kaki || 1 || 0 || 1
|-
| trapesium siku-siku || 0 || 0 || 0
|-
| layang-layang || 1 || 0 || 1
|-
| elips || 2 || 0 || 2
|}
; Besar sudut keliling dan sudut pusat
: sudut pusat = 2 x sudut keliling
; Hubungan ketiga sisi segitiga dengan jari-jari lingkaran (lingkaran dalam segitiga siku-siku): <math>r = \frac{a+b-c}{2}</math>
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
a-r + b-r &= c \\
a+b-2r &= c \\
2r &= a+b-c \\
r &= \frac{a+b-c}{2} \\
\end{align}
</math>
</div></div>
; Hubungan dua tali busur dengan jari-jari lingkaran: <math>r = \sqrt{\frac{a^2 + b^2 + c^2 + d^2}{4}}</math>
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\text{ misalkan } p &= \frac{c+d}{2} \text{ dan } q = \frac{a+b}{2} \\
\text{ misalkan persegi panjang dibuat x dan y} & \\
x + p &= d \\
x + \frac{c+d}{2} &= d \\
x &= d - \frac{c+d}{2} \\
x &= \frac{d-c}{2} \\
y + q &= b \\
y + \frac{a+b}{2} &= b \\
y &= b - \frac{a+b}{2} \\
y &= \frac{b-a}{2} \\
y^2 + p^2 &= r^2 \\
(\frac{b-a}{2})^2 + (\frac{c+d}{2})^2 &= r^2 \\
\frac{a^2-2ab+b^2}{4} + \frac{c^2+2cd+d^2}{4} &= r^2 \\
x^2 + q^2 &= r^2 \\
(\frac{d-c}{2})^2 + (\frac{a+b}{2})^2 &= r^2 \\
\frac{c^2-2cd+d^2}{4} + \frac{a^2+2ab+b^2}{4} &= r^2 \\
\text {jumlah dua persamaan tersebut menjadi } & \\
\frac{a^2-2ab+b^2}{4} + \frac{c^2+2cd+d^2}{4} + \frac{c^2-2cd+d^2}{4} + \frac{a^2+2ab+b^2}{4} &= 2r^2 \\
\frac{2a^2+2b^2+2c^2+2d^2}{4} &= 2r^2 \\
\frac{2(a^2+b^2+c^2+d^2)}{4} &= 2r^2 \\
\frac{a^2+b^2+c^2+d^2}{4} &= r^2 \\
a^2+b^2+c^2+d^2 &= 4r^2 \\
r &= \sqrt{\frac{a^2 + b^2 + c^2 + d^2}{4}} \\
\end{align}
</math>
</div></div>
; Lingkaran dalam segitiga: <math>r = \frac{L}{s}</math> (s adalah setengah keliling segitiga)
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\text{ misalkan luas segitiga AOC } L_a &= \frac{ra}{2} \\
\text{ misalkan luas segitiga COB } L_b &= \frac{rb}{2} \\
\text{ misalkan luas segitiga BOC } L_c &= \frac{rc}{2} \\
\text{ maka total luas ketiga segitiga tersebut adalah } L &= L_a + L_b + L_c \\
&= \frac{ra}{2} + \frac{rb}{2} + \frac{rc}{2} \\
&= r \frac{a+b+c}{2} \\
&= r s \\
r &= \frac{L}{s} \\
\end{align}
</math>
</div></div>
; Lingkaran luar segitiga: <math>r = \frac{abc}{4L}</math>
<div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Pembuktian</div>
<div class="mw-collapsible-content">
<math display="block">
\begin{align}
\text{ lihat posisi kesebangunan } \frac{t}{b} &= \frac{c}{2r} \\
t &= \frac{bc}{2r} \\
L &= \frac{at}{2} \\
&= \frac{abc}{2(2r)} \\
r &= \frac{abc}{4L} \\
\end{align}
</math>
</div></div>
; Garis singgung (satu) lingkaran: <math>d = \sqrt{a^2-b^2}</math>
; Garis singgung persekutuan dalam (dua) lingkaran: <math>d = \sqrt{p^2-(R+r)^2}</math>
; Garis singgung persekutuan luar (dua) lingkaran: <math>d = \sqrt{p^2-(R-r)^2}</math>
keterangan:
: d = panjang singgung persekutuan dalam/luar lingkaran
: p = jarak kedua titik pusat lingkaran
; Teorema Ceva: AF x BD x CE = BF x CD x AE
; Teorema de Pitot: AB + DC = AD + BC (segiempat luar lingkaran)
; Teorema Power de Point: (segiempat dalam lingkaran)
:: AE x EC = BE x ED
; Tali busur pada satu titik di luar lingkaran:
:: AB<sup>2</sup> = AD x AE
:: AB x AC = AD x AE
; Teorema Ptolomeos: (segiempat dalam lingkaran)
:: AC x BD = AB x DC + AD x BC
; Titik E adalah pertemuan kedua perpanjangan tali busur yang bertemu.
:: maka sudut E = 1/2 (sudut besar-sudut kecil)
; Hubungan besar sudut, luas juring dan panjang busur
: <math>\frac{\text{besar sudut AOB }}{\text{besar sudut COD }} = \frac{\text{luas juring AOB }}{\text{luas juring COD }} = \frac{\text{panjang busur AB }}{\text{panjang busur CD }}</math>
[[Kategori:Soal-Soal Matematika]]
4pkl1w9vwq0ftrcwi8tbrzox5njpcih
Resep:Ayam bumbu rujak
100
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[[Berkas:Ayam bumbu rujak.jpg|jmpl|Ayam bumbu rujak]]
'''Ayam bumbu rujak''' adalah sebuah makanan khas [[Jawa]] yang terbuat dari daging ayam yang masih muda dan dibumbui bumbu merah halus. Sajian ayam ini disebut bumbu rujak karena terdapat banyak [[rempah-rempah]] khas rujak, seperti cabai dan gula jawa yang umum dipakai dalam saus rujak buah.<ref>Kementerian Pertanian. 1967. ''Mustikarasa: Buku Masakan Indonesia.'' Jakarta: Kementerian Pertanian</ref>
== Bahan ==
* 1 ekor ayam
== Bumbu ==
* 6 siung bawang merah
* 5 siung bawang putih
* 10 buah cabai merah
* 5 butir kemiri
* 1 ruas jari kunyit
* 1 potong jahe
* 1 batang serai
* 4 sdm minyak goreng
* 4 gelas santan
* 1/2 sdm air asam jawa
* 2 lembar daun salam
* 2 lembar daun jeruk
* gula merah dan garam secukupnya
== Cara Membuat ==
# Bersihkan dan potong ayam menjadi 10 bagian atau sesuai keinginan.
# Lumuri ayam dengan air asam jawa dan 1 sdt garam. Marinasi dan diamkan sekitar 20 menit.
# Haluskan bawang merah, bawang putih, cabai merah, kunyit, dan kemiri.
# Tumis bumbu halus. Lalu, tambahkan daun salam, daun jeruk, serai, dan jahe.
# Tambahkan garam dan gula merah secukupnya, lalu aduk hingga merata
# Masukkan daging ayam yang sudah dipotong dan tambahkan santan.
# Masak dengan api sedang. Aduk supaya santan tidak pecah sampai potongan daging matang dan empuk.
# Hidangkan ayam bumbu rujak selagi hangat.
== Referensi ==
<references />
[[Kategori:WikiCitaRasa]]
[[Kategori:WikiCitaRasa-Bandung]]
[[Kategori:Masakan Jawa]]
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* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{{#babel:plain=1|id-n|en-5|no-3|da-3|jv-1|ar-1|nl-1|fr-1|he-0|ja-0|hr-0|zh-0}}
{{Purge|Buat akun}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{|
|-
|[[Berkas:Lactarius indigo 48568 edit.jpg|jmpl|[[Lactarius indigo]] ]]
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
{{end-col}}
== Referensi ==
[[User:Nunoguevara]]
__NOGLOBAL__
__NOGLOBAL__
ceg22c1oitkut4wqz4ic6yrvykapr6e
108322
108321
2025-07-07T09:32:57Z
Mistsplitter Reforged
41544
108322
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
[[Bruker:Mistsplitter Reforged]]
{{end-col}}
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{{#babel:plain=1|id-n|en-5|no-3|da-3|jv-1|ar-1|nl-1|fr-1|he-0|ja-0|hr-0|zh-0}}
{{Purge|Buat akun}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{|
|-
|[[Berkas:Lactarius indigo 48568 edit.jpg|jmpl|[[Lactarius indigo]] ]]
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
== Referensi ==
[[User:Nunoguevara]]
__NOGLOBAL__
__NOGLOBAL__
9eak1l5g9aot3eo5omhyqaalm54y86k
108323
108322
2025-07-07T09:37:35Z
Mistsplitter Reforged
41544
108323
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{{#babel:plain=1|id-n|en-5|no-3|da-3|jv-1|ar-1|nl-1|fr-1|he-0|ja-0|hr-0|zh-0}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{|
|-
|[[Berkas:Lactarius indigo 48568 edit.jpg|jmpl|[[Lactarius indigo]] ]]
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
{{Purge|Buat akun}}
== Referensi ==
[[User:Nunoguevara]]
__NOGLOBAL__
__NOGLOBAL__
gn7h3ajal6yexqwft3be8w8ooelc66i
108324
108323
2025-07-07T09:38:46Z
Mistsplitter Reforged
41544
108324
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3|jv-1|ar-1|nl-1|fr-1|he-0|ja-0|hr-0|zh-0}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{{Interproyek|
* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{|
|-
|[[Berkas:Lactarius indigo 48568 edit.jpg|jmpl|[[Lactarius indigo]] ]]
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
{{Purge|Buat akun}}
== Referensi ==
[[User:Nunoguevara]]
__NOGLOBAL__
__NOGLOBAL__
kettnechaq8besr496a9lfshdnfdqw1
108325
108324
2025-07-07T09:40:45Z
Mistsplitter Reforged
41544
/* Lihat pula */
108325
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3|jv-1|ar-1|nl-1|fr-1|he-0|ja-0|hr-0|zh-0}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{|
|-
|[[Berkas:Lactarius indigo 48568 edit.jpg|jmpl|[[Lactarius indigo]] ]]
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
{{Purge|Buat akun}}
== Referensi ==
[[User:Nunoguevara]]
__NOGLOBAL__
__NOGLOBAL__
gzg654v268xg776fexiuabnzd1bqv4f
108326
108325
2025-07-07T09:52:31Z
Mistsplitter Reforged
41544
108326
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3|jv-1|ar-1|nl-1|fr-1|he-0|ja-0|hr-0|zh-0}}
[[Berkas:Mandelbrot sequence new.gif|pus|jmpl|320x320px]]
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{|
|-
|[[Berkas:Lactarius indigo 48568 edit.jpg|jmpl|[[Lactarius indigo]] ]]
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
{{Purge|Buat akun}}
== Referensi ==
[[User:Nunoguevara]]
__NOGLOBAL__
__NOGLOBAL__
rtio9l9fmz0sivdto8fmxzssuazernh
108327
108326
2025-07-07T09:53:28Z
Mistsplitter Reforged
41544
/* Lihat pula */
108327
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3|jv-1|ar-1|nl-1|fr-1|he-0|ja-0|hr-0|zh-0}}
[[Berkas:Mandelbrot sequence new.gif|pus|jmpl|320x320px]]
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{|
|-
|
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
{{Purge|Buat akun}}
== Referensi ==
[[User:Nunoguevara]]
__NOGLOBAL__
__NOGLOBAL__
npm4u2bibekrlbzimtma6gbehebd7k5
108329
108327
2025-07-07T09:59:04Z
Mistsplitter Reforged
41544
108329
wikitext
text/x-wiki
== Profil ==
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3|jv-1|ar-1|nl-1|fr-1|he-0|ja-0|hr-0|zh-0}}
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{|
|-
|
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
{{Purge|Buat akun}}
== Referensi ==
[[User:Nunoguevara]]
__NOGLOBAL__
__NOGLOBAL__
dbvxxpaurexhjzd9xc65w0bowf3ya2v
108332
108329
2025-07-07T10:21:22Z
Mistsplitter Reforged
41544
108332
wikitext
text/x-wiki
== Profil ==
{{Interproyek|* [[:w:user:Mistsplitter Reforged|Wikipedia]]
* [[:wikt:user:Mistsplitter Reforged|Wiktionary]]
* [[mw:user:Mistsplitter Reforged|Mediawiki]]
* [[:b:user:Mistsplitter Reforged|Wikibooks]]
* [[:q:user:Mistsplitter Reforged|Wikiquote]]
* [[:s:user:Mistsplitter Reforged|Wikisource]]}}
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3}}
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{|
|-
|
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
{{Purge|Buat akun}}
== Referensi ==
[[User:Nunoguevara]]
[[no:User:mistsplitter Reforged]]
__NOGLOBAL__
__NOGLOBAL__
igx6imf2l0kkkbrb7wqzz9xw1u6137g
108333
108332
2025-07-07T10:26:02Z
Mistsplitter Reforged
41544
108333
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3}}
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{|
|-
|
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:en:User:Mistsplitter_Reforged|Halaman Inggris]]
|-
|[[:de:Benutzer:Mistsplitter_Reforged|Halaman Jerman]]
|-
|[[:fi:Käyttäjä:Mistsplitter_Reforged|Halaman Finlandia]]
|-
|[[:da:Bruger:Mistsplitter_Reforged|Halaman Denmark]]
|-
|[[:is:Notandi:Mistsplitter_Reforged|Halaman Islandia]]
|-
|[[Pengguna:Mistsplitter Reforged/Bak pasir|Bak pasir pribadi]]
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
== Referensi ==
[[User:Nunoguevara]]
[[no:User:mistsplitter Reforged]]
[[en:User:mistsplitter Reforged]]
[[da:User:mistsplitter Reforged]]
[[pl:User:mistsplitter Reforged]]
[[is:User:mistsplitter Reforged]]
[[es:User:mistsplitter Reforged]]
[[pt:User:mistsplitter Reforged]]
[[fr:User:mistsplitter Reforged]]
__NOGLOBAL__
__NOGLOBAL__
smkkb3u88x9bwk1rwcw7frmefc24rtr
108334
108333
2025-07-07T10:28:45Z
Mistsplitter Reforged
41544
/* Lihat pula */
108334
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3}}
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{|
|-
|
|-
|{{purge|[ Slett denne hurtigbufferen ]}}
|-
|[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
|-
|[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
|-
|}
== Referensi ==
[[User:Nunoguevara]]
[[no:User:mistsplitter Reforged]]
[[en:User:mistsplitter Reforged]]
[[da:User:mistsplitter Reforged]]
[[pl:User:mistsplitter Reforged]]
[[is:User:mistsplitter Reforged]]
[[es:User:mistsplitter Reforged]]
[[pt:User:mistsplitter Reforged]]
[[fr:User:mistsplitter Reforged]]
__NOGLOBAL__
__NOGLOBAL__
qsszi4mgwtdgw9930illkj8qo53hnxz
108335
108334
2025-07-07T10:30:36Z
Mistsplitter Reforged
41544
108335
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3}}
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{{purge|[ Slett denne hurtigbufferen ]}}
[[:he:משתמש:Mistsplitter_Reforged/טיוטה|Bak pasir pribadi]] ([[Israel|Ibrani]])
[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
== Referensi ==
[[User:Nunoguevara]]
[[no:User:mistsplitter Reforged]]
[[en:User:mistsplitter Reforged]]
[[da:User:mistsplitter Reforged]]
[[pl:User:mistsplitter Reforged]]
[[is:User:mistsplitter Reforged]]
[[es:User:mistsplitter Reforged]]
[[pt:User:mistsplitter Reforged]]
[[fr:User:mistsplitter Reforged]]
[[sv:User:mistsplitter Reforged]]
[[fi:User:mistsplitter Reforged]]
[[ar:User:mistsplitter Reforged]]
[[nn:User:mistsplitter Reforged]]
[[zh:User:mistsplitter Reforged]]
[[ja:User:mistsplitter Reforged]]
__NOGLOBAL__
__NOGLOBAL__
si3clh1ej6hyd5z6hjqkbgpl02udq2o
108336
108335
2025-07-07T10:32:08Z
Mistsplitter Reforged
41544
108336
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3}}
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{{purge|[ Slett denne hurtigbufferen ]}}
[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
== Referensi ==
[[User:Nunoguevara]]
[[no:User:mistsplitter Reforged]]
[[en:User:mistsplitter Reforged]]
[[da:User:mistsplitter Reforged]]
[[pl:User:mistsplitter Reforged]]
[[is:User:mistsplitter Reforged]]
[[es:User:mistsplitter Reforged]]
[[pt:User:mistsplitter Reforged]]
[[fr:User:mistsplitter Reforged]]
[[sv:User:mistsplitter Reforged]]
[[fi:User:mistsplitter Reforged]]
[[ar:User:mistsplitter Reforged]]
[[nn:User:mistsplitter Reforged]]
[[zh:User:mistsplitter Reforged]]
[[ja:User:mistsplitter Reforged]]
__NOGLOBAL__
__NOGLOBAL__
nrlxpmdrio1xnbc8a4woczyhrjjmdfy
108337
108336
2025-07-07T10:32:33Z
Mistsplitter Reforged
41544
/* Referensi */
108337
wikitext
text/x-wiki
== Profil ==
{{purge|[ OPPRETT ARTIKKEL NÅ ] }}
{{col}}
{{end-col}}
{{#babel:plain=1|id-n|en-5|no-3|da-3}}
{{Pp-protected|small=yes|expiry=1 april 999999|demolevel=sysop=full|demospace=sysop=full}}
== Lihat pula ==
[[Berkas:New-Bouncywikilogo.gif|jmpl|101x101px]]
[[Berkas:New-Bouncywikilogo.gif|jmpl|100x100px]]
{{purge|[ Slett denne hurtigbufferen ]}}
[[mediawikiwiki:User:Mistsplitter_Reforged|Halaman Inggris]] ([[MediaWiki|Mediawiki]])
== Referensi ==
[[User:Nunoguevara]]
[[no:User:mistsplitter Reforged]]
[[en:User:mistsplitter Reforged]]
[[da:User:mistsplitter Reforged]]
[[pl:User:mistsplitter Reforged]]
[[is:User:mistsplitter Reforged]]
[[es:User:mistsplitter Reforged]]
[[pt:User:mistsplitter Reforged]]
[[fr:User:mistsplitter Reforged]]
[[sv:User:mistsplitter Reforged]]
[[fi:User:mistsplitter Reforged]]
[[ar:User:mistsplitter Reforged]]
[[nn:User:mistsplitter Reforged]]
[[zh:User:mistsplitter Reforged]]
[[ja:User:mistsplitter Reforged]]
[[he:User:mistsplitter Reforged]]
__NOGLOBAL__
__NOGLOBAL__
13q227ybf9d954fsg4lgzfsqikkqc2n