From df8cea4ec64df05c0cf2694183269c1ecf49cdea Mon Sep 17 00:00:00 2001 From: James Su Date: Tue, 3 Sep 2019 22:27:21 +0000 Subject: [PATCH] Update Final_Exam_Study_Sheet.md --- Grade 9/Math/MPM1DZ/Final_Exam_Study_Sheet.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/Grade 9/Math/MPM1DZ/Final_Exam_Study_Sheet.md b/Grade 9/Math/MPM1DZ/Final_Exam_Study_Sheet.md index 2806b32..623a2d9 100644 --- a/Grade 9/Math/MPM1DZ/Final_Exam_Study_Sheet.md +++ b/Grade 9/Math/MPM1DZ/Final_Exam_Study_Sheet.md @@ -362,7 +362,7 @@ x, & \text{if } x > 0\\ |Shape|Maximum Area|Minimum Perimeter| |:----|:-----------|:----------------| - |4-sided rectangle|A rectangle must be a square to maximaze the area for a given perimeter. The length is equal to the width
$`A = lw`$
$`A_{max} = (w)(w)`$
$`A_{max} = w^2`$|A rectangle must be a square to minimaze the perimeter for a given area. The length is equal to the width.
$`P = 2(l+w)`$
$`P_{min} = 2(w)(w)`$
$`P_{min} = 2(2w)`$
$`P_{min} = 4w`$| + |4-sided rectangle|A rectangle must be a square to maximaze the area for a given perimeter. The length is equal to the width
$`A = lw`$
$`A_{max} = (w)(w)`$
$`A_{max} = w^2`$|A rectangle must be a square to minimaze the perimeter for a given area. The length is equal to the width.
$`P = 2(l+w)`$
$`P_{min} = 2(w + w)`$
$`P_{min} = 2(2w)`$
$`P_{min} = 4w`$| |3-sided rectangle|$`l = 2w`$
$`A = lw`$
$`A_{max} = 2w(w)`$
$`A_{max} = 2w^2`$|$`l = 2w`$
$`P = l+2w`$
$`P_{min} = 2w+2w`$
$`P_{min} = 4w`$|