Asked by Al
Lee walks along the edges of a rectangular pool from point A to B to C to D, a distance of 38 metres. Marina walks along the edges of the same pool from B to C to D to A, a distance of 31 metres. What is the perimeter of the pool, in metres?
Answers
Answered by
Steve
If the length (AB) is x and the width (BC) is y, then
2x+y = 38
x+2y = 31
add them up and
3x+3y = 69
the perimeter is 2x+2y = 2/3 * 69 = 46
2x+y = 38
x+2y = 31
add them up and
3x+3y = 69
the perimeter is 2x+2y = 2/3 * 69 = 46
Answered by
kevin
46
Answered by
?
Steve can you please tell the actual answer. Is it 69 or 46?
Answered by
?
38-31=7, so the difference between the width and the length is 7m
So X+(X+7)x2=38 X+2X+2x7=38
38-2x7=3X X=8
The width is 8m, so the length is 8+7=15. 8x2+15x2=46
Answer: 46m
So X+(X+7)x2=38 X+2X+2x7=38
38-2x7=3X X=8
The width is 8m, so the length is 8+7=15. 8x2+15x2=46
Answer: 46m
Answered by
??
Where did the 2/3 come from?
Answered by
Mary Ann
Set up the two linear equations. Solve for one of the variables, x or y. Use substitution to solve for the other variable.
The two equations are:
2x+y=38 and x+2y=31
Solve for y:
y=38-2x
Equations combined:
X+2(38-2x)=31
X+76-4x=31
-3x=-45
x=15
Plug in 15 for x in one of the original equations to solve for y:
2(15)+y=38
30+y=38
y=8
Add two lengths and two widths to get the total perimeter of 46
The two equations are:
2x+y=38 and x+2y=31
Solve for y:
y=38-2x
Equations combined:
X+2(38-2x)=31
X+76-4x=31
-3x=-45
x=15
Plug in 15 for x in one of the original equations to solve for y:
2(15)+y=38
30+y=38
y=8
Add two lengths and two widths to get the total perimeter of 46
Answered by
.
46
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