Find the perimeter of the rectangle

P = 2a²b(7a + 3). The perimeter of a rectangle that it has a width of 7a³b and a length of 3a²b is 2a²b(7a + 3).

The key to solve this problem is using the perimeter of the rectangle equation. The perimeter is equal to the sum of all sides.

P = w + l + w + l = 2 (w + l), where w is width and l is length.

With w = 7a³b and l = 3a²b:

P = 2(7a³b + 3a²b) = 2(7a³b) + 2(3a²b) = 14a³b+6a²b

We can factor 14a³b+6a²b as follow:

2a²b(7a + 3)

d

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.