Question

Please help, this is algebra and rational export lesson

Answer 1

Answer: The correct option is the second one, counting from the top.

Explanation:

For a rectangle of length L, and width W, the area can be calculated as:

A = L*W

In this case, we know that:

`L = √(x)`

and

`W = \sqrt[3]{x^2}`

First, we need to remember the relations;

`\sqrt[n]{x^m} = x^(m/n)`

`x^m*x^n = x^(m + n)`

`(x^n)/(x^m) = x^(n - m)`

a) Now we can calculate the area of the rectangle as:

`A = L*W = √(x) *\sqrt[3]{x^2} = (x^(1/2))*(x^(2/3)) = x^(1/2 + 2/3) = x^(3/6 + 4/6) = x^(1/6 + 1)`

And we can write that last part as:

`x^(1/6 + 1) = x*x^(1/6) = x*\sqrt[6]{x}`

b) Now we want to find the ratio between the width and the length:

`(W)/(L) = \frac{\sqrt[3]{x^2} }{√(x) } = (x^(2/3))/(x^(1/2)) = x^(2/3 - 1/2) = x^(4/6 - 3/6) = x^(1/6) = \sqrt[6]{x}`

Now, if x = 1, the ratio will be equal to 1.

if x > 1, the ratio will be larger than 1.

if 0 < x < 1, the ratio will be smaller than 1.

The correct option is the second one, counting from the top.