The length of a rectangle is 10 feet more than twice the width. The perimeter is 110 feet. Find the dimensions.
2006-10-22 14:13:34 · 5 answers · asked by wil46yang 2 in Education & Reference ➔ Homework Help
1) the dimensions are 60 ft by 50 ft
2) a= 65 and -65
2006-10-22 14:17:01 · answer #1 · answered by a person 5 · 0⤊ 0⤋
1) Where is the question? 2) |a| -2=63 =>|a|=65 =>a=+65 or - 65 Thired question:- Let the width of the rectangle be x ft.Therefore the length= 2x+10 .According to the problem 2(x+2x+10)= 110 => 2(3x+10)=110 => 6x+20=110 =>6x=110-20=90 =>x=90/6=15 .KTherefore the width of the rectngle is 15 ft, and its length is (15X2+10 or)40 ft. ans.
2006-10-22 14:34:28 · answer #2 · answered by alpha 7 · 0⤊ 0⤋
this
1) the dimensions are 60 ft by 50 ft
2) a= 65 and -65
or
15 by 40
or
Let the width be 'x' , then length = 10feet more than twice the width= 2x+10
2006-10-22 14:25:11 · answer #3 · answered by Anonymous · 0⤊ 0⤋
15 by 40
2006-10-22 14:17:54 · answer #4 · answered by Joe S 2 · 0⤊ 0⤋
Let the width be 'x' , then length = 10feet more than twice the width= 2x+10
2006-10-22 14:18:51 · answer #5 · answered by Anonymous · 0⤊ 0⤋