Question

Answer this please!!!!

Answer 1

6y^-1

Explanation:

Step 1: Rewrite 2∛y using fractional exponents:

Taking the cube root of a number is the same as raising it to the 1/3 power.

Thus, 2∛y = 2 * y^1/3 = 2y^1/3

Step 2: Multiply 3y^-4/3 and 2y^1/3 and simplify:

To simplify 3y^-4/3 and 2y^1/3, we multiply bases and add the exponents according to the product rule.

(3 * 2)^(-4/3 + 1/3)

6y^-1

Thus, 6y^-1 is our final answer.

Answer 2

`\sf 6\cdot y^(-1)`

Explanation:

`\sf 3y^{-(4)/(3)}\cdot 2\sqrt[3]{y}`

Converting cube root of y in terms of power:

which is:
`\sf \sqrt[a]{b}=b^(1)/(a)`

`\sf 3y^{-(4)/(3)}\cdot 2\cdot y^(1)/(3)`

Keeping like terms together:

`\sf 3\cdot 2 y^{-(4)/(3)} \cdot y^(1)/(3)`

Simplify like terms:

`\sf 6 y^{-(4)/(3)} \cdot y^(1)/(3)`

Using product rule of indices:

The product rule of indices states that the product of two powers with the same base is equal to the power with the same base and the sum of the exponents

`\sf a^m\cdot a^n= a^(m+n)`

Applying this:

`\sf 6 \cdot y^{-(4)/(3)+(1)/(3)}`

Solve the power of y:

`\sf 6\cdot y^{(-4+1)/(3)}`

`\sf 6\cdot y^{(-3)/(3)}`

`\sf 6\cdot y^(-1)`

So, expression in the form of
`\sf k\cdot y^n` is:

`\sf 6\cdot y^(-1)`

And

`\sf \textsf{In fraction or integer form is:} (6)/(y)`