4. The length of the side of a rectangle is 1 more foot than twice its width. The perimeter is 36 ft. Find the dimensions. W +

Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter9: Polynomial And Rational Functions
Section9.4: Graphing Polynomial Functions
Problem 43PS
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4. The length of the side of a rectangle is
1 more foot than twice its width. The
perimeter is 36 ft. Find the dimensions.
6
2W+]
6
12
= 2(XW
The dimension is b/ 12 inches
24
3
48
16-347/16
= 34=7/16-6
et using interval notation, then graph the solu
6. 2(4y - 3) ≥ 7(y + 4)
8y-227/7y + - 12/²²
6274 28
-28
84-347/74
18
2W+1
·bw + 2 = 7
=3434
Transcribed Image Text:4. The length of the side of a rectangle is 1 more foot than twice its width. The perimeter is 36 ft. Find the dimensions. 6 2W+] 6 12 = 2(XW The dimension is b/ 12 inches 24 3 48 16-347/16 = 34=7/16-6 et using interval notation, then graph the solu 6. 2(4y - 3) ≥ 7(y + 4) 8y-227/7y + - 12/²² 6274 28 -28 84-347/74 18 2W+1 ·bw + 2 = 7 =3434
4
5.7--x < 9
5
7-4x <
5
+9
-2-4x
5
=ㅛ
5x
-9
6. 2(4y - 3) ≥ 7(y + 4)
84-6274 + 28
J-28 -28
89-34>174
18
18
16-34716
= 34 16-6
=3434
Transcribed Image Text:4 5.7--x < 9 5 7-4x < 5 +9 -2-4x 5 =ㅛ 5x -9 6. 2(4y - 3) ≥ 7(y + 4) 84-6274 + 28 J-28 -28 89-34>174 18 18 16-34716 = 34 16-6 =3434
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