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A rectangle has an area of 14y^3-35y.
a. what are the expressions for the length and the width where one dimension is the GCF?
b. does the rectangle exist with the given dimension if y=2? explain.

Respuesta :

A) The expressions for the length and the width where one dimension is the GCF is:

[tex]Length = 7y\\\\Width = 2y^2 - 5\\\\Or\\\\Length = 2y^2 - 5\\\\Width = 7y[/tex]

B) Yes, the rectangle exist with given dimension if y = 2

Solution:

Given that,

A rectangle has an area of:

[tex]Area = 14y^3-35y[/tex]

Factor out the greatest common factor

Find GCF of 14 and 35

The factors of 14 are: 1, 2, 7, 14

The factors of 35 are: 1, 5, 7, 35

Then the greatest common factor is 7

Find the GCF of variables

[tex]GCF\ of\ y^3\ and\ y = y[/tex]

Therefore,

Factor out 7y

[tex]Area = 7y(2y^2 - 5)[/tex] --------- eqn 1

The area of rectangle is given as:

[tex]Area = length \times width[/tex]

On comparing eqn 1 with above,

[tex]Length = 7y\\\\Width = 2y^2 - 5\\\\Or\\\\Length = 2y^2 - 5\\\\Width = 7y[/tex]

If, y = 2 we have:

[tex]Area = 7y(2y^2 - 5)[/tex]

[tex]Area = 7 \times 2(2(2)^2 - 5)\\\\Area = 7 \times 2(3)\\\\Area = 42[/tex]

Thus the the rectangle exist with the given dimension if y=2

adioabiola

The expressions for the length and the width where one dimension is the GCF are; 7y and (2y²-5).

Area of a Rectangle

A rectangle is a two dimensional figure and provided one of the dimension is the GCF, as described in the task, it follows that;

  • The expressions for the length and width are 7y and (2y²-5) respectively.

Additionally, when y= 2, the area of the rectangle is; 14(2³) - 35(2) = 52.

Read more on area of a rectangle;

https://brainly.com/question/25292087

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