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Please help
write on paper or something
please
this is my finals

Please help write on paper or something please this is my finals-example-1
User Zkminusck
by
8.0k points

2 Answers

3 votes

Answer:

6.Ans;


√(18) = √(9) √(2) = 3 √(2)

___o___o__


- √(48) = - √(8) √(6) = - √(4) √(2) √(2) √(3) = - 2 √(4) √(3) = - 2 * 2 √(3) = - 4 √(3)

__o___o___


√(75) = √(15) √(5) = √(5) √(3) √(5) = 5 √(3)

___o___o__


\sqrt{ (30)/(49) } = ( √(30) )/( √(49) ) = ( √(5) √(6) )/( √(7) √(7) ) = ( √(5) √(6) )/(7) = ( √(30) )/(7)

___o___o___


\sqrt{ (10)/(121) } = ( √(10) )/( √(121) ) = ( √(5) √(2) )/( √(11) √(11) ) = ( √(10) )/(11)

___o___o__


\sqrt[3]{40} = \sqrt[3]{8} \sqrt[3]{5} = 2 \sqrt[3]{5}

___o___o___


\sqrt[3]{54} = \sqrt[3]{27} \sqrt[3]{2} = 3 \sqrt[3]{2}

___o___o___


- \sqrt[3]{128} = - \sqrt[3]{64} \sqrt[3]{2} = - 4 \sqrt[3]{2}

___o__o__


\sqrt[3]{192} = \sqrt[3]{64} \sqrt[3]{3} = 4 \sqrt[3]{3}

__o_o__


\sqrt[3]{ \frac{3m}{8 {n}^(3) } } = \frac{ \sqrt[3]{3m} }{ \sqrt[3]{8 {n}^(3) } } = \frac{ \sqrt[3]{3m} }{2n}

___o__o___


\sqrt[3]{16 {a}^(5) } = \sqrt[3]{8 {a}^(3) } \sqrt[3]{2 {a}^(2) } = 2a \sqrt[3]{2 {a}^(2) }

___o__o___


\sqrt[4]{80} = \sqrt[4]{16} \sqrt[4]{5} = 2 \sqrt[4]{5}

___o___o___


\sqrt[4]{ (32)/(256) } = \frac{ \sqrt[4]{32} }{ \sqrt[4]{256} } = \frac{ \sqrt[4]{16} \sqrt[4]{2} }{ \sqrt[4]{16} \sqrt[4]{16} } = \frac{ \sqrt[4]{2} }{ \sqrt[4]{16} } = \frac{ \sqrt[3]{2} }{2}

__o__o__


\sqrt[4]{ \frac{4 {x}^(2) }{81 {y}^(4) } } = \frac{ \sqrt[4]{4 {x}^(2) } }{ \sqrt[4]{81 {y}^(4) } } = ( √(2x) )/(3y)

___o___o___


- \sqrt[5]{ (3)/(243) } = - \frac{ \sqrt[5]{3} }{ \sqrt[5]{243} } = - \frac{ \sqrt[5]{3} }{3}

__o___o__


\sqrt[5]{ \frac{375}{32 {x}^(6) } } = \frac{ \sqrt[5]{375} }{ \sqrt[5]{32 {x}^(6) } } = \frac{ \sqrt[5]{375 {x}^(4) } }{2 {x}^(2) }

__________o__________o_____________

7.Ans ;


\sqrt{36 {a}^(2) {b}^(3) } = 6ab √(b)

__o_o___


\sqrt{27 {a}^(4) {b}^(3) } = 3 {a}^(2) b √(3b)

___o___o__


\sqrt{72 {x}^(5) {y}^(2) } = 6 {x}^(2)y √(2x)

___o__o__


- \sqrt{ {112}^(3) {b}^(4) } = - 448 {b}^(2) √(7)

__o__o___


\sqrt{80 {m}^(4) {n}^(3) } = 4 {m}^(2)n √(5n)

___o___o___


\sqrt{64 {x}^(2) {y}^(3) } = 8xy √(y)

__o__o___


\sqrt[3]{16 {m}^(3) {n}^(3) } = 2mn \sqrt[3]{2}

__o__o__


\sqrt[3]{ - 54 {x}^(4) {b}^(3) } = - 3xb \sqrt[3]{2x}

___o__o__


- \sqrt[3]{128 {a}^(5) {y}^(3) } = - 4ay \sqrt[3]{2 {a}^(2) }

__o__o__


\sqrt[3]{24 {p}^(3) {q}^(5) } = 2pq \sqrt[3]{3 {q}^(2) }

___o__o_


\sqrt[4]{81 {x}^(4) {y}^(2) } = 3x √(y)

__o__o__


- \sqrt[4]{256 {a}^(4) {b}^(5) } = - 4ab \sqrt[4]{b}

I hope I helped you^_^

User HpTerm
by
7.8k points
1 vote

you can use this to attempt others

Please help write on paper or something please this is my finals-example-1
User Alex Kuhl
by
8.1k points

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