P = 100 l = 30 w = 20
5w+2 = 2w+5 5w-2w = 5-2 3w = 3 w = 1
p=2l+2w
'W' and 'L' are independent variables. 'P' is the dependent variable. '2' and '2' are the constants.
To solve for area: A=lw (Area equals length times width) The length and width are both 12. Substitute 12 for l and w. A=12x12 Multiply. A=144 The room has an area of 144 units. To solve for perimeter: P=2l+2w (Perimeter equals two time the length plus two times the width). Substitute 12 for l and w. P=(2x12)+(2x12) Simplify. P=24+24 P=48 The room has a perimeter of 48 units.
2L+2W=P (S2w) Subtract 2W from both sides 2L=P-2W (D2) Divide both sides by 2 L=(P-2W)/2
22
x = (P - 2W) / 2
a = L x W: area equals length times the width. p = 2L + 2W: perimeter equals 2 times the length plus 2 times the width so L = (p - 2W)/2
There is no indication as to what B represents and consequently no possibility of answering the question.
The perimeter of a rectangle is given by (2L plus 2W). If you double either the width or length dimension, then it is four times the original dimension, such as (4L plus 2W) or (2L plus 4W).
First, I am assuming you mean P = 2L + 2W, solve for W. If so, then see below! You solve equations the same way, whether they have multiple variables or not: Distribute, Combine Like Terms, Move the variable you are isolating to one side, Add or Subtract, then Multiply or Divide. Let's show how this equation is very much like an equation having only one variable: 28 = 20 + 2W -20 -20 8 = 2W ÷2 ÷2 4 = W So, if you get stuck, just "fake" the variables you aren't solving for into being random numbers. Solve your "fake" equation like you would any equation. Then use the same steps in your original "variable" equation. P = 2L + 2W -2L -2L P-2L= 2W ÷2 ÷2 (P-2L)/2 = W
P = 100 l = 30 w = 20
Using basic algebra we can solve this problem. First we need to write out the problem:5-32+2w = -7-27+2w = -72w = 20w = 10
length = 48, width = 24 l = 2w 2l + 2w = 144 2l + 2w = 144 2(2w) + 2w = 144 4w + 2w =144 w = 144/6 w = 24 l = 2w l = 2(24) l = 48
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
For a rectangle, P = 2L + 2W (L is length, W is width)