you have two equations with two unknowns, length and width let W = width and L = length 2W - 4 = L (1) The perimeter is 2L + 2W 2L + 2W = 34 (2) rearrange (1) and (2) 2W - L = 4 2W +2L = 34 multiply (1) both sides by 2 4W - 2L = 8 2W + 2L =34 add equations 6W = 42 W = 7 solve L from (2) by substituting W = 7 L = 10
8 L
2L is Bigger.
T=1/2l
L = 2w + 1 2l + 2w = 26 2l + l - 1 = 26 l = 9 w = 4
you have two equations with two unknowns, length and width let W = width and L = length 2W - 4 = L (1) The perimeter is 2L + 2W 2L + 2W = 34 (2) rearrange (1) and (2) 2W - L = 4 2W +2L = 34 multiply (1) both sides by 2 4W - 2L = 8 2W + 2L =34 add equations 6W = 42 W = 7 solve L from (2) by substituting W = 7 L = 10
Can 1 and 1/20 be simplified
The width of the room is equal to twice the Length. Suppose Length = L, width = W, and A = areaW = 2L from the information in the questionNow we know area, A is equal to length times widthW*L=A, plug in 2L for W and we get 2L*L=A or 2L^2=ANext, we see that when 6 is subtracted from both length and width A becomes 108 less.So (2L-6)*(L-6)=A-108Multiply (2L-6)*(L-6) out and the result is (2L^2-18L+36) Set that equal to A-108(2L^2-18L+36)=A-108. We found out that A=2L^2 earlier so we can substitute the terms.(2L^2-18L+36)=2L^2-108. Now solve for LSubtract 2L^2 from both sides(-18L+36)=(-108)Subtract 36 from both sides(-18L)=(-144)Divide by (-18)L=8We know W=2L so W=16Now lets test our answer.16*8=128(16-6)*(8-6)=10*2=20128-20=108So the answer of L=8 and W=16 is correct.
Okay. Say, for example if you have a perimeter of 36 inches and your width is 8 inches.Your equation you look like this set up:P = 2L + 2W36 = 2L + 2(8)So,First you would finish multiplying: 36 = 2L + 16Then you would subtract 16 from each side: 36 - 16 = 2L + 16 - 16So it would look like this: 20 = 2LThen you would divide by 2 on each side: 20/2 = 2L/2So your answer would be 10 = L
8 L
2L = 200cL
2L
2L is Bigger.
500ml/1000ml(1l)=1/5
1-2l
T=1/2l
P= 2l+2w