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Novay_Z [31]
3 years ago
14

HELP! Multiplying radicals. Questions on photo.

Mathematics
1 answer:
DerKrebs [107]3 years ago
5 0
These are 10 questions and 10 answers

1) \sqrt[3]{24}  . \sqrt[3]{45}


Answer: third option 6∛5

Explanation:

24 = 2^3 * 3

45 = 3^2 * 5

=> 24 * 45 = 2^3 * 3^3 * 5

=> (∛24).(∛45) = ∛[ (2^3).(3^3).5 ] = (2)(3)∛5 = 6∛5

2) \sqrt[5]{4x^2} . \sqrt[5]{4x^2}

Answer: second option.

Demostration:

 \sqrt[5]{4x^2} . \sqrt[5]{4x^2} =  \sqrt[5]{4^2x^4} = \sqrt[5]{2^4x^4} = \sqrt[5]{16x^2}

3) \sqrt{10} . \sqrt{10}

Answer: first option 10

Justification:

√10 . √10 = (√10)^2 = √(10^2) = √100 = 10

4) \sqrt[4]{7} . \sqrt[4]{7} . \sqrt[4]{7} . \sqrt[4]{7}

Answer: fourth option: 7

Explanation:

\sqrt[4]{7} . \sqrt[4]{7} . \sqrt[4]{7} . \sqrt[4]{7}= (\sqrt[4]{7^})^4= \sqrt[4]{7^4}=7 ^{4/4}=7^1=7

5) (x \sqrt{7} -3 \sqrt{8}).(x \sqrt{7}-3 \sqrt{8})

Answer: the third option: 7x^2 - 12x√14 + 72

Solution:

Notice that it is the two factors are identical, so this is a perfect square binomial:

(x√7 - 3√8)^2 = (x√7)^2 - 2*(x√7)(3√8) + (3√8)^2 = 7x^2 - 6√(56)x + 72 =

= 7x^2 -(6)(2)x√14 + 72 = 7x^2 - 12x√14 + 72

6) √12 . √18

Answer: the fourth option 6√6

Explanation:

√12 . √18 = √ (2 . 2 . 3 . 2 . 3 . 3) = √ [( 2^3) . (3^3)] = 2 . 3 √6 = 6√6

7) \sqrt{y^3} . \sqrt{y^3}

Answer: first option y^3

Justification:

\sqrt{y^3} . \sqrt{y^3} =( \sqrt{y^3} )^2 =(y^3)^{2/2}=y^3

8) ∛d . ∛d . ∛d

Answer: first option: d

Explanation:

∛d . ∛d . ∛d =     ( \sqrt[3]{d}) ^3 = d{3/3}=d^1=d

9) \sqrt{5x^8y^2} . \sqrt{10x^3} . \sqrt{12y}

Answer: second option

Explanation:

\sqrt{5x^8y^2} . \sqrt{10x^3} . \sqrt{12y} = \sqrt{(5.10.12)x^8y^2x^3y}= \sqrt{600x^{11}y^3} =

=10x^5y \sqrt{6xy}

10) (∛4) . √3

Answer: third option     \sqrt[6]{432}

Explanation:

\sqrt[3]{4} . \sqrt{3} = \sqrt[6]{4^2} . \sqrt[6]{3^3} = \sqrt[6]{16.27} = \sqrt[6]{432}
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