suppose that a square picture frame has sides that vary between 9.9 in and 10.1 in. what range of values is possible for the perimeter P of the picture frame? and how do i express this using a three-part inequality?
i know the perimeter is 39.6 for the 9.9" and 40.4 in for the 10.1". do i need to calcute the 40 in perimeter for sides of 10"?
thanks for the help.
2006-09-16 10:32:29 · 3 answers · asked by shih rips 6 in Science & Mathematics ➔ Mathematics
I think this is the answer: 39.6<=P<=40.4.
Here is my reasoning: if you want all of the values of the parameter such that each side is between 9.9 and 10.1, then you need all possible values of the parameter that are between and include 39.6 and 40.4. The answer simply says that the parameter P can be equal to or greater than the lowest possible value, 39.6, and less than or equal to the highest possible value, 40.4. The inequality is expressed in three parts: 39.6, P, and 40.4. Hope this helps.
2006-09-16 10:51:08 · answer #1 · answered by thomthum2000 2 · 0⤊ 0⤋
Suppose "pf" stands for picture frame and "P" for perimeter
2006-09-16 10:58:18
·
answer #2
·
answered by Nexus 2
·
0⤊
0⤋
for square shapes we know that : P=4*pf
now we have a pf between 9.9 and 10.1 so the perimeter (P) changes between 39.6 and 10.4 as you mentioned.
let say it in math language:
if 9.9
if the perimeter is P
39.6<=P<=40.4
2006-09-16 10:48:38 · answer #3 · answered by raj 7 · 0⤊ 0⤋