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What is the product of (3 y Superscript negative 4 Baseline) (2 y Superscript negative 4 Baseline)? StartFraction 6 Over y Superscript 8 Baseline EndFraction StartFraction 1 Over 6 y Superscript 6 Baseline EndFraction StartFraction 6 Over y Superscript 16 Baseline EndFraction StartFraction 1 Over 6 y Superscript 16 Baseline EndFraction.

Sagot :

The product of these two expressions 3y⁻⁴ and 2y⁻⁴ is 6/y⁸. Then the correct option is A.

What is multiplication?

It is also known as the product. If the object n is given to m times then we just simply multiply them.

The expressions are as 3y⁻⁴ and 2y⁻⁴.

The product of these two will be

[tex]\rm 3y^{-4} * 2y^{-4}\\\\3*2*y^{-4}*y^{-4}\\\\6*y^{-4-4}\\\\6*y^{-8}\\\\\dfrac{6}{y^{8}}[/tex]

The product of these two expressions 3y⁻⁴ and 2y⁻⁴ is 6/y⁸. Then the correct option is A.

More about the multiplication link is given below.

https://brainly.com/question/19943359

The result of the product expression [tex](3y ^{-4}) (2y^{-4})[/tex] is 6/y^8

How to determine the product?

The product expression is given as:

(3y ^{-4}) (2y^{-4)

Rewrite the expression properly as:

[tex](3y ^{-4}) (2y^{-4})[/tex]

Rewrite the factors of the expression as fractions

[tex]\frac{3}{y ^4} * \frac{2}{y^4}[/tex]

Multiply 3 and 2

[tex]\frac{6}{y ^4*y^4}[/tex]

Apply the law of indices

[tex]\frac{6}{y ^{4+4}}[/tex]

Add 4 and 4

[tex]\frac{6}{y ^{8}}[/tex]

Hence, the product of [tex](3y ^{-4}) (2y^{-4})[/tex] is [tex]\frac{6}{y ^{8}}[/tex]

Read more about products at:

https://brainly.com/question/10873737