The rectangle has a length of 4x+3 and a width of 3x. Show your work.

A. Find the perimeter.


B. Find the area.


C. Find the perimeter and area if x = 8.

Respuesta :

Answer:

  A.  P = 14x+6

  B.  A = 12x^2 +9x

  C.  P = 118; A = 840

Step-by-step explanation:

A. The perimeter is twice the sum of length and width:

  P = 2(L +W) = 2((4x+3) +(3x)) = 2(7x +3)

  P = 14x +6 . . . . the perimeter of the rectangle

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B. The area is the product of length and width:

  A = LW = (4x +3)(3x)

  A = 12x^2 +9x . . . . . the area of the rectangle

__

C. When x = 8, these values are ...

  P = 14·8 +6 = 118 . . . . . perimeter in units

  A = 12·8^2 +9·8 = 768 +72 = 840 . . . . . area in square units

Answer:

a) [tex] P = 2(4x+3) +2(3x) =8x +6 +6x = 14x +6[/tex]

b) [tex] A= 12x^2 +9x[/tex]

c) [tex] P = 14*8 +6 = 112+6 = 118[/tex]

[tex] A= 12(8)^2 +9*8 = 840[/tex]

Step-by-step explanation:

We know that the length is 4x+3 and the width is of 3x

Part a

For this case the perimeter is given by:

[tex] P = 2(4x+3) +2(3x) =8x +6 +6x = 14x +6[/tex]

Part b

The area is given by:

[tex] A= (4x+3) (3x)[/tex]

And after multiply we got:

[tex] A= 12x^2 +9x[/tex]

Part c

For this case replacing the value of x =8 we got:

[tex] P = 14*8 +6 = 112+6 = 118[/tex]

[tex] A= 12(8)^2 +9*8 = 840[/tex]