the perimeter of a rectangle must be greater than 16 cm, let x represent the length of the rectangle. let y represent the width. write and graph the inequality that descripes the different lengths and widths of the rectangle.
how do you graph it?
2006-09-11 17:39:49 · 8 answers · asked by sonovabeeech 1 in Science & Mathematics ➔ Mathematics
perimeter=16cm
semi perimeter=8cm
x+y=8
draw the line x+y=8 by ascribing values to x and finding the corresponding values of y by finding 2 ordered pairs say
(0,8) and (1,7)
now all the points above the line will give you combinations of x and y which will give a perimeter of >16 cm
2006-09-11 17:45:58 · answer #1 · answered by raj 7 · 0⤊ 0⤋
2(x+y) > 16. Lets consider this straight line: 2(x+y) = 16, or
x+y = 8. Its slope is -45 and who passes by the point (0,8) and (8,0).
This is the border of the semiplane. So, you only have to choose which of the 2 is the right one. If you consider x = 0 and y = 0, the result will be 0 for the perimeter, so you have to choose the other one
As a final step, consider that the lengths are greater than 0, so pick the part or the semiplane that is in the first quadrant
The result will then be described in an analytical way by the followin system of equations:
x+y>8
x>0
y>0
Later
Ana
2006-09-12 04:08:40 · answer #2 · answered by Ilusion 4 · 0⤊ 0⤋
x= length of rectangle
y= width of rectangle
perimeterof rectangle = length + width
perimeter= x+y =16
on equating as
when x=0; y=16
when x=1; y=15
when x=2; y=14
when x=3; y=13
when x=4; y=12
when x=5; y=11
when x=6; y=10
when x=7; y=9
when x=8 ; y=8
when x=9; y=7
when x=10; y=6
when x=11; y=5
when x=12; y=4
when x=13; y=3
when x=14; y=2
when x=15; y=1
when x=16; y=0
then plot the graph representing these points
2006-09-11 18:27:57 · answer #3 · answered by azeem 2 · 0⤊ 0⤋
2(x+y)>16 - Is is inequality
This comes from the form of 2(l+w)=perimiter
As for graphing the inequality, it would be a straight-dotted line with intercepts at x=8 and y=8 where the area above the line is shaded, and only the area in the first quadrent above the line is shaded.
Hope this helps!
2006-09-11 17:45:12 · answer #4 · answered by priestlake22 2 · 0⤊ 0⤋
2x +2y>16 or x+y>8
x>0 and y>0 and x>y
When y=1 then x>7
When y=2 then x>6
When y=3 then x>5
When y=4 then x>4
2006-09-11 17:49:55 · answer #5 · answered by Amar Soni 7 · 0⤊ 0⤋
Put a dot on the y axis at 16. Put another dot on the x axis at 16. Draw a straight line between those two points.
All acceptable values for x and y lie 'above' that line and in the 1'st quadrant.
Doug
2006-09-11 17:45:10 · answer #6 · answered by doug_donaghue 7 · 0⤊ 0⤋
First, graph the equation as though the fewer than sign is an equals sign. 2nd, p.c.. a factor, any factor, on the two factor of the line. (frequently (0,0) is the least complicated one to apply.) enter the coordinates of this comparable factor into the equation, with the unique much less-than sign. If the equation is valid for the factor, coloration the factor of the line(or curve) that contains the factor. If no longer, coloration the different factor of the line (or curve).
2016-11-07 03:48:19 · answer #7 · answered by ? 4 · 0⤊ 0⤋
2x+2y>16
2y>-2x+16
y>-x+8
graph that line, except make it dotted and then shade in everything above it
2006-09-11 17:45:09 · answer #8 · answered by dan 4 · 0⤊ 0⤋