What is the area of the composite figure whose vertices have the following coordinates?

(−2, −2) , (4, −2) , (5, 1) , (2, 3) , (−1, 1)

24

Step-by-step explanation:

From the picture you can see that area is sume of area P1 and area P2.

a=6 because its egdes are on point -2 and 4, so it is 6. (See x-ose)

h=3 Edges on -2 and 1 (see y-ose)

k=2 edges on 3 and 1 on y-ose

P1=a*h=6*3=18

P2=a*k/2=6*2/2=6

P=18+6=24

  24 square units

Step-by-step explanation:

The area of a polygon can be computed from its list of coordinates as ...

where there are n coordinate points. Subscripts on y "wrap around" so that ...

The area from these coordinates is ...

  A = (1/2)|-2(-2-1) +4(1-(-2)) +5(3-(-2)) +2(1-1) -1(-2-3)|

  = (1/2)|6 +12 +25 +0 +5|

  = (1/2)(48)

  = 24 . . . . square units

_____

If you graph the points, you see that the figure can be considered to be a parallelogram topped by a triangle. The area of the parallelogram is ...

  A = bh = 6·3 = 18

The area of the triangle is ...

  A = 1/2bh = (1/2)(6)(2) = 6

Then the total area is 18+6 = 24, which agrees with the above calculation.

true

explanation:

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0