w − t h e w i d t h l − t h e l e n g t h w = l − 12 t h e p er im e t er o f rec t an g l e : P = 2 ( w + l ) an d P = 156 c m 2 ( w + l ) = 156 → p u t w = l − 12 t o e q u a t i o n : 2 ( l − 12 + l ) = 156 2 ( 2 l − 12 ) = 156 ∣ d i v i d e b o t h s i d es b y 2 2 l − 12 = 78 ∣ a dd 12 t o b o t h s i d es 2 l = 90 ∣ d i v i d e b o t h s i d es b y 2 l = 45 w = 45 − 12 = 33 S o l u t i o n : t h e w i d t h e q u a l 33 c m an d t h e l e n g t h e q u a l 45 c m .
The width of a rectangle is 12 cm less than the length, and the perimeter is 156 cm. To find the width and the length, let's denote the length as L and the width as W . The problem tells us that W = L - 12. The formula for perimeter P of a rectangle is P = 2L + 2W. Plugging in the values, we get 156 = 2L + 2(L - 12).
Simplify the equation: 156 = 2L + 2L - 24. Combine like terms: 156 = 4L - 24. Add 24 to both sides: 180 = 4L. Divide by 4: L = 45 cm. Now find W by substituting L into W = L - 12: W = 45 - 12 = 33 cm.
So, the length of the rectangle is 45 cm and the width is 33 cm.
The length of the rectangle is determined to be 45 cm, while the width is 33 cm.
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Jawaban:2x² - 5x - 12Penjelasan dengan langkah-langkah:= (2x + 3)(x - 4)= (2x.x) + (2x.-4) + (3.x) + (3.-4)= 2x² - 8x + 3x - 12= 2x² - 5x - 12[tex]\boxed{ \red{ \boxed{\pink{\mathcal{M \frak{ ilana} \purple{ \tt01}}}}}} [/tex]