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ki77a [65]
4 years ago
9

The area of a rectangular field is represented by the expression 14x-3x^2+2y. The width of the field is represented by the expre

ssion 5x-7x^2+7y. How much greater is the length of the field than the width?
9x+4x^2-5y
9x-10x^2-5y
19x+4x^2+9y
19x-10x^2+9y
Mathematics
2 answers:
xxTIMURxx [149]4 years ago
6 0
The answer is 9x+4x^2-5y.
Subtract the length by the width.
Travka [436]4 years ago
5 0

Answer:

9x + 4x^2 - 5y

Step-by-step explanation:

It is given in the question that the length of a rectangular field is represented by the expression 14x - 3x^2 + 2y and width of the field by 5x - 7x^2 + 7y

Now we have to tell how much greater is the length of the field than the width.

That means length is greater than width and we have to subtract width from length.

Length - width = (14x - 3x^2 + 2y) - (5x - 7x^2+ 7y)

                        = 14x - 5x - 3x^2 + 7x^2 + 2y - 7y

                        = 9x + 4x^2 - 5y

Therefore expression which represents the difference between length and width of the field will be (9x + 4x^2 - 5y)

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\bf \begin{array}{lllll}
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