SOLUTION: 1.The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.
2. Solve and graph the solution
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-> SOLUTION: 1.The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.
2. Solve and graph the solution
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Question 85591: 1.The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.
2. Solve and graph the solution set. 3x > 7x + 20
3.Solve and graph the solution set. 3(x – 4) > 7x – 9
You can put this solution on YOUR website! 1.
Let w=width of rectangle
Then the length=2w+2
Perimeter=2w+2l, so:
52=2w+2(2w+2) get rid of parens
52=2w+4w+4 subtract 4 from both sides
52-4=2w+4w+4-4 collect like terms
48=6w divide both sides by 6
w=8 cm--------------------width of rectangle
2w+2=2*8+2=18cm-------------------length of rectangle
CK
52=2w+2l
52=2*8+2*18
52=16+36
52=52
2.
3x>7x+20 subtract 7x from both sides
3x-7x>20 collect like terms
-4x>20 divide both sides by -4 (remember the inequality sign will reverse)
x<-5 You Graph. Hint: got to x=-5 on the x-y axis and draw a vertical line. x will always be to the left of that line.
3.
3(x–4)>7x–9 get rid of parens
3x-12>7x-9 subtract 7x and add 12 to both sides
3x-12+12-7x>7x-7x-9+12 collect like terms
-4x>3 divide both sides by -4(inequality sign reverses, again)
x<-3/4 You graph--same as before
