tolong ya bantu dari nomor 6 sampai 10 secepatnya makasih
Matematika
dinda1996
Pertanyaan
tolong ya bantu dari nomor 6 sampai 10
secepatnya
makasih
secepatnya
makasih
2 Jawaban
-
1. Jawaban klp
mudah-mudahan dapat dipahami yachh2. Jawaban algebralover
Mapel : Matematika
Materi : Akar Pangkat
Soal No. 6
[tex]=3 \sqrt{32}- \sqrt{18} \\ =3 \sqrt{16\times2}- \sqrt{9\times2} \\ =(3\times4) \sqrt{2}-3 \sqrt{2} \\ =12 \sqrt{2}-3 \sqrt{2} \\ =\boxed{9 \sqrt{2}.....Jawaban\ (D)} [/tex]
Soal No. 7
[tex]=(81^{ \frac{3}{4} })^{ \frac{1}{3} } \\ =((3^4)^{ \frac{3}{4} })^{ \frac{1}{3} } \\ =(3^{4\times \frac{3}{4} })^{ \frac{1}{3} } \\ =3^{(4\times \frac{3}{4} \times \frac{1}{3})}........\text{Pada pangkat, 4 : 4 = 1 dan 3 : 3 = 1 (dicoret)} \\ =3^1 \\ =\boxed{3\ .........Jawaban\ (C)}[/tex]
Soal No. 8
[tex]= \frac{2}{\sqrt{5}-\sqrt{3}}\\ \\= \frac{2}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}\\ \\= \frac{2\times(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}............Penyebut : (a-b)(a+b)=a^2-b^2\\ \\= \frac{2\times(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^2-(\sqrt{3})^2}\\ \\= \frac{2\times(\sqrt{5}+\sqrt{3})}{5-3}\\ \\= \frac{2\times(\sqrt{5}+\sqrt{3})}{2} \\ \\ =\boxed{\sqrt{5}+\sqrt{3}...................Jawaban\ (B)}[/tex]
Soal No. 9
Barisan 9, 3, 1, ............
[tex]a=\bold{9}; r= \frac{U_2}{U_1}= \frac{3}{9}=\bold{\frac{1}{3}}\\ \\ \\U_n=a\times r^{(n-1)}\\ \\U_n=9\times (\frac{1}{3})^{(n-1)}\\ \\U_n=3^2\times3^{-1(n-1)}\\ \\U_n=3^2\times3^{(1-n)}\\ \\U_n=3^{(2+1-n)}\\ \\U_n=3^{(3-n)}\\ \\ U_{(4)}=3^{(3-4)}=3^{(-1)}= \frac{1}{3}\\ \\U_{(5)}=3^{(3-5)}=3^{(-2)}= \frac{1}{3^2}= \frac{1}{9} \\ \\\boxed{\frac{1}{3}\ dan\ \frac{1}{9}.............Jawaban\ (C)}[/tex]
Soal No. 10
[tex]U_{(2)}=10 \\ 10=a +(n-1)b\\10=a +(2-1)b\\10=a+b................(1) \\ \\ U_{(6)}=50 \\ 50=a +(n-1)b\\50=a +(6-1)b\\50=a +5b................(2)\\ \\Eliminasi\ (1)\ dan\ (2)\\ \\10=a+b\\\underline{50=a +5b}\ (-) \\ -40=-4b............kedua\ ruas\ dikali\ (-1)\\ b=40\div4 \\ b=10 \\ \\ 10=a+b \\ 10=a+10 \\ a=10-10 \\ a=0 \\ \\ U_n=a+(n-1)b \\ U_n=0+(n-1)10 \\ U_n=10n-10 \\ \\U_{(10)}=10(10)-10\\U_{(10)}=100-10\\\boxed{U_{(10)}=90.......................Jawaban\ (A)} [/tex]
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