A rectangle has length 2y and width y+ 2.
Its perimeter is less than 50 cm. What are
possible values for y ?
Write inequality?
perimeter=2(2y+y+2)
perimeter= 6y+4
so, 6y+4<50
6y<46
y<46/6
length<46/3
width<46/6+2
check that.
To find the possible values for y, we can start by writing an inequality statement based on the given information.
The perimeter of a rectangle is calculated by adding the length of all four sides. In this case, the length of two opposite sides is 2y, and the length of the other two opposite sides is y + 2. Therefore, the perimeter (P) can be expressed as:
P = 2(2y) + 2(y + 2)
Simplifying this expression, we get:
P = 4y + 2y + 4
P = 6y + 4
The question states that the perimeter is less than 50 cm. So, we set up the inequality:
6y + 4 < 50
To isolate the variable y, we subtract 4 from both sides:
6y < 46
Finally, we divide both sides of the inequality by 6:
y < 46/6
y < 23/3
Therefore, the possible values for y are any real numbers less than 23/3. In interval notation, it can be written as (-∞, 23/3).