Hello send the details for the area of each shape. sincerely, leanoriaia

This type of problems are solved by partitioning the figure into rectangles, whose areas we can find easily.

Picture 1)

From the point denoted by E, draw two altitudes towards the sides AG and AC.

Since ED is 3, then BC is 3. Thus, AB=4-3=1. 

Since EF is 1, then HG is 1, so AH=3-1=2.

The area of the figure is equal to the area of rectangle ACDH + the area of the square HEFG.

That is, the area of the figure is 4*2+1*1=8+1=9 (square in).

Picture 2)

As in the first part, draw the altitudes EB and EH from point E.

GF is 5, so AB is 5. This means that BC=7-5=2 (in).

FE is 3, so GH is 3, that is HA=6-3=3.

Thus, Area(figure)=Area(ABFG)+Area(BCDE)=5*6+2*3=30+6=36 (square in.)

9 (square in); 36 (square in.)

Polar coordinates is 8,9

Ifound this on google “a skewed (non-symmetric) distribution is a distribution in which there is no such mirror-imaging. a "skewed right" distribution is one in which the tail is on the right side. a "skewed left" distribution is one in which the tail is on the left side. the above histogram is for a distribution that is skewed right.” i hope this you for what your looking for