Asked by Sissy
what would the dimensions of a retangle be whose perimeter is 50(P=50)inches when the length is 2x-5 ane the width is unknown
The width must be 25 - (2x-5) = 30-2x, since the sum of the length and width must be half the perimeter.
You do not have enough information to solve for x, or a specific length and width.
I beleive I have figured this one out:
2x+2(2x-5)=50
2x being twice the width
2(2x-5) being twice the length
and 50 being the perimeter
2x=2(2x-5)=50
2x+4x-10=50
6x-10 =50
6x=60
x=10 If you place 10 whwre the x is you get 50 as a result
If the width is x then you are correct. You had only stated that the length is 2x-5, and had not defined x. Your first line should have been
2x + 2x(2x-5) = 50
The rest is fine. The width is 10 and the length is 15
The width must be 25 - (2x-5) = 30-2x, since the sum of the length and width must be half the perimeter.
You do not have enough information to solve for x, or a specific length and width.
I beleive I have figured this one out:
2x+2(2x-5)=50
2x being twice the width
2(2x-5) being twice the length
and 50 being the perimeter
2x=2(2x-5)=50
2x+4x-10=50
6x-10 =50
6x=60
x=10 If you place 10 whwre the x is you get 50 as a result
If the width is x then you are correct. You had only stated that the length is 2x-5, and had not defined x. Your first line should have been
2x + 2x(2x-5) = 50
The rest is fine. The width is 10 and the length is 15
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