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plis tlg di jwb hiks, buka dulu dehhnomer 10-16 ajaa

Berikut ini adalah pertanyaan dari sarinawang818 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Plis tlg di jwb hiks, buka dulu dehh
nomer 10-16 ajaa pls plss plssss . .​
plis tlg di jwb hiks, buka dulu dehhnomer 10-16 ajaa pls plss plssss . .​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

NOMOR 10.

(C). \frac{1}{18}

PEMBAHASAN :

{(3 \sqrt{2})}^{ - 2} = \frac{1}{{(3 \sqrt{2})}^{2}} = \frac{1}{(3 \sqrt{2})(3 \sqrt{2})} = \frac{1}{9 \sqrt{4}} = \frac{1}{9 \times 2} = \frac{1}{18}

NOMOR 11.

(C). 54

PEMBAHASAN :

\sqrt[3]{216} \times \sqrt{81} = 6 \times 9 = 54

NOMOR 12.

(D). 32 \sqrt{3} \: (2 - 3 \sqrt{2})

PEMBAHASAN :

4 \sqrt{32} \: ( \sqrt{24} - 3 \sqrt{12})

= 4 \sqrt{16 \times 2} \: ( \sqrt{4 \times 6} - 3 \sqrt{4 \times 3})

= 4 \times 4 \sqrt{2} \: (2 \sqrt{6} - 3 \times 2 \sqrt{3})

= 16 \sqrt{2} \: (2 \sqrt{6} - 6 \sqrt{3})

= 32 \sqrt{12} - 96 \sqrt{6}

= 32 \sqrt{4 \times 3} - 96 \sqrt{6}

= 32 \times 2 \sqrt{3} - 96 \sqrt{6}

= 32 \sqrt{3} \: (2 - 3 \sqrt{2})

NOMOR 13.

(B). 48 {a}^{7} {b}^{4}

PEMBAHASAN :

{3a}^{3} \times {(2ab)}^{4} = {3a}^{3} \times {2}^{4} {a}^{4} {b}^{4} = {3a}^{3} \times 16 {a}^{4} {b}^{4}

= 3 \times 16 {a}^{3 + 4} {b}^{4} = 48 {a}^{7} {b}^{4}

NOMOR 14.

(A). \frac{2}{3} \sqrt{3}

PEMBAHASAN :

\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{2 \sqrt{3} }{ \sqrt{9} } = \frac{2}{3} \sqrt{3}

NOMOR 15.

(B). {3}^{ - 5}

PEMBAHASAN :

\frac{1}{243} = {(243)}^{ - 1} = {({3}^{5})}^{ - 1} = {3}^{5 \times ( - 1)} = {3}^{ - 5}

NOMOR 16.

(C). \frac{2}{3} {p}^{6} {qr}^{2}

PEMBAHASAN :

(\frac{{p}^{5} {q}^{3} {r}^{3}}{ {4qr}^{5}}) \times (\frac{8pr}{ {3qr}^{ - 3}}) = \frac{8 {p}^{5 + 1} {q}^{3} {r}^{3 + 1}}{4 \times 3 {q}^{1 + 1} {r}^{5 + ( - 3)}}

= \frac{8 {p}^{6} {q}^{3} {r}^{4}}{12 {q}^{2} {r}^{2}} = \frac{2}{3} {p}^{6} {q}^{3 - 2} {r}^{4 - 2} = \frac{2}{3} {p}^{6} {qr}^{2}

NOMOR 10.(C). [tex]\frac{1}{18} [/tex]PEMBAHASAN :[tex]{(3 \sqrt{2})}^{ - 2} = \frac{1}{{(3 \sqrt{2})}^{2}} = \frac{1}{(3 \sqrt{2})(3 \sqrt{2})} = \frac{1}{9 \sqrt{4}} = \frac{1}{9 \times 2} = \frac{1}{18} [/tex]NOMOR 11.(C). [tex]54[/tex]PEMBAHASAN :[tex]\sqrt[3]{216} \times \sqrt{81} = 6 \times 9 = 54 [/tex]NOMOR 12.(D). [tex]32 \sqrt{3} \: (2 - 3 \sqrt{2}) [/tex]PEMBAHASAN :[tex]4 \sqrt{32} \: ( \sqrt{24} - 3 \sqrt{12}) [/tex][tex]= 4 \sqrt{16 \times 2} \: ( \sqrt{4 \times 6} - 3 \sqrt{4 \times 3}) [/tex][tex]= 4 \times 4 \sqrt{2} \: (2 \sqrt{6} - 3 \times 2 \sqrt{3}) [/tex][tex]= 16 \sqrt{2} \: (2 \sqrt{6} - 6 \sqrt{3}) [/tex][tex] = 32 \sqrt{12} - 96 \sqrt{6} [/tex][tex] = 32 \sqrt{4 \times 3} - 96 \sqrt{6} [/tex][tex] = 32 \times 2 \sqrt{3} - 96 \sqrt{6} [/tex][tex] = 32 \sqrt{3} \: (2 - 3 \sqrt{2})[/tex]NOMOR 13.(B). [tex]48 {a}^{7} {b}^{4} [/tex]PEMBAHASAN :[tex]{3a}^{3} \times {(2ab)}^{4} = {3a}^{3} \times {2}^{4} {a}^{4} {b}^{4} = {3a}^{3} \times 16 {a}^{4} {b}^{4} [/tex][tex] = 3 \times 16 {a}^{3 + 4} {b}^{4} = 48 {a}^{7} {b}^{4} [/tex]NOMOR 14.(A). [tex] \frac{2}{3} \sqrt{3} [/tex]PEMBAHASAN :[tex]\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{2 \sqrt{3} }{ \sqrt{9} } = \frac{2}{3} \sqrt{3} [/tex]NOMOR 15.(B). [tex] {3}^{ - 5} [/tex]PEMBAHASAN :[tex]\frac{1}{243} = {(243)}^{ - 1} = {({3}^{5})}^{ - 1} = {3}^{5 \times ( - 1)} = {3}^{ - 5} [/tex]NOMOR 16.(C). [tex]\frac{2}{3} {p}^{6} {qr}^{2} [/tex]PEMBAHASAN :[tex](\frac{{p}^{5} {q}^{3} {r}^{3}}{ {4qr}^{5}}) \times (\frac{8pr}{ {3qr}^{ - 3}}) = \frac{8 {p}^{5 + 1} {q}^{3} {r}^{3 + 1}}{4 \times 3 {q}^{1 + 1} {r}^{5 + ( - 3)}} [/tex][tex]= \frac{8 {p}^{6} {q}^{3} {r}^{4}}{12 {q}^{2} {r}^{2}} = \frac{2}{3} {p}^{6} {q}^{3 - 2} {r}^{4 - 2} = \frac{2}{3} {p}^{6} {qr}^{2} [/tex]NOMOR 10.(C). [tex]\frac{1}{18} [/tex]PEMBAHASAN :[tex]{(3 \sqrt{2})}^{ - 2} = \frac{1}{{(3 \sqrt{2})}^{2}} = \frac{1}{(3 \sqrt{2})(3 \sqrt{2})} = \frac{1}{9 \sqrt{4}} = \frac{1}{9 \times 2} = \frac{1}{18} [/tex]NOMOR 11.(C). [tex]54[/tex]PEMBAHASAN :[tex]\sqrt[3]{216} \times \sqrt{81} = 6 \times 9 = 54 [/tex]NOMOR 12.(D). [tex]32 \sqrt{3} \: (2 - 3 \sqrt{2}) [/tex]PEMBAHASAN :[tex]4 \sqrt{32} \: ( \sqrt{24} - 3 \sqrt{12}) [/tex][tex]= 4 \sqrt{16 \times 2} \: ( \sqrt{4 \times 6} - 3 \sqrt{4 \times 3}) [/tex][tex]= 4 \times 4 \sqrt{2} \: (2 \sqrt{6} - 3 \times 2 \sqrt{3}) [/tex][tex]= 16 \sqrt{2} \: (2 \sqrt{6} - 6 \sqrt{3}) [/tex][tex] = 32 \sqrt{12} - 96 \sqrt{6} [/tex][tex] = 32 \sqrt{4 \times 3} - 96 \sqrt{6} [/tex][tex] = 32 \times 2 \sqrt{3} - 96 \sqrt{6} [/tex][tex] = 32 \sqrt{3} \: (2 - 3 \sqrt{2})[/tex]NOMOR 13.(B). [tex]48 {a}^{7} {b}^{4} [/tex]PEMBAHASAN :[tex]{3a}^{3} \times {(2ab)}^{4} = {3a}^{3} \times {2}^{4} {a}^{4} {b}^{4} = {3a}^{3} \times 16 {a}^{4} {b}^{4} [/tex][tex] = 3 \times 16 {a}^{3 + 4} {b}^{4} = 48 {a}^{7} {b}^{4} [/tex]NOMOR 14.(A). [tex] \frac{2}{3} \sqrt{3} [/tex]PEMBAHASAN :[tex]\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{2 \sqrt{3} }{ \sqrt{9} } = \frac{2}{3} \sqrt{3} [/tex]NOMOR 15.(B). [tex] {3}^{ - 5} [/tex]PEMBAHASAN :[tex]\frac{1}{243} = {(243)}^{ - 1} = {({3}^{5})}^{ - 1} = {3}^{5 \times ( - 1)} = {3}^{ - 5} [/tex]NOMOR 16.(C). [tex]\frac{2}{3} {p}^{6} {qr}^{2} [/tex]PEMBAHASAN :[tex](\frac{{p}^{5} {q}^{3} {r}^{3}}{ {4qr}^{5}}) \times (\frac{8pr}{ {3qr}^{ - 3}}) = \frac{8 {p}^{5 + 1} {q}^{3} {r}^{3 + 1}}{4 \times 3 {q}^{1 + 1} {r}^{5 + ( - 3)}} [/tex][tex]= \frac{8 {p}^{6} {q}^{3} {r}^{4}}{12 {q}^{2} {r}^{2}} = \frac{2}{3} {p}^{6} {q}^{3 - 2} {r}^{4 - 2} = \frac{2}{3} {p}^{6} {qr}^{2} [/tex]NOMOR 10.(C). [tex]\frac{1}{18} [/tex]PEMBAHASAN :[tex]{(3 \sqrt{2})}^{ - 2} = \frac{1}{{(3 \sqrt{2})}^{2}} = \frac{1}{(3 \sqrt{2})(3 \sqrt{2})} = \frac{1}{9 \sqrt{4}} = \frac{1}{9 \times 2} = \frac{1}{18} [/tex]NOMOR 11.(C). [tex]54[/tex]PEMBAHASAN :[tex]\sqrt[3]{216} \times \sqrt{81} = 6 \times 9 = 54 [/tex]NOMOR 12.(D). [tex]32 \sqrt{3} \: (2 - 3 \sqrt{2}) [/tex]PEMBAHASAN :[tex]4 \sqrt{32} \: ( \sqrt{24} - 3 \sqrt{12}) [/tex][tex]= 4 \sqrt{16 \times 2} \: ( \sqrt{4 \times 6} - 3 \sqrt{4 \times 3}) [/tex][tex]= 4 \times 4 \sqrt{2} \: (2 \sqrt{6} - 3 \times 2 \sqrt{3}) [/tex][tex]= 16 \sqrt{2} \: (2 \sqrt{6} - 6 \sqrt{3}) [/tex][tex] = 32 \sqrt{12} - 96 \sqrt{6} [/tex][tex] = 32 \sqrt{4 \times 3} - 96 \sqrt{6} [/tex][tex] = 32 \times 2 \sqrt{3} - 96 \sqrt{6} [/tex][tex] = 32 \sqrt{3} \: (2 - 3 \sqrt{2})[/tex]NOMOR 13.(B). [tex]48 {a}^{7} {b}^{4} [/tex]PEMBAHASAN :[tex]{3a}^{3} \times {(2ab)}^{4} = {3a}^{3} \times {2}^{4} {a}^{4} {b}^{4} = {3a}^{3} \times 16 {a}^{4} {b}^{4} [/tex][tex] = 3 \times 16 {a}^{3 + 4} {b}^{4} = 48 {a}^{7} {b}^{4} [/tex]NOMOR 14.(A). [tex] \frac{2}{3} \sqrt{3} [/tex]PEMBAHASAN :[tex]\frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{ \sqrt{3} }{ \sqrt{3} } = \frac{2 \sqrt{3} }{ \sqrt{9} } = \frac{2}{3} \sqrt{3} [/tex]NOMOR 15.(B). [tex] {3}^{ - 5} [/tex]PEMBAHASAN :[tex]\frac{1}{243} = {(243)}^{ - 1} = {({3}^{5})}^{ - 1} = {3}^{5 \times ( - 1)} = {3}^{ - 5} [/tex]NOMOR 16.(C). [tex]\frac{2}{3} {p}^{6} {qr}^{2} [/tex]PEMBAHASAN :[tex](\frac{{p}^{5} {q}^{3} {r}^{3}}{ {4qr}^{5}}) \times (\frac{8pr}{ {3qr}^{ - 3}}) = \frac{8 {p}^{5 + 1} {q}^{3} {r}^{3 + 1}}{4 \times 3 {q}^{1 + 1} {r}^{5 + ( - 3)}} [/tex][tex]= \frac{8 {p}^{6} {q}^{3} {r}^{4}}{12 {q}^{2} {r}^{2}} = \frac{2}{3} {p}^{6} {q}^{3 - 2} {r}^{4 - 2} = \frac{2}{3} {p}^{6} {qr}^{2} [/tex]

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Last Update: Sun, 30 Oct 22
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