A trapezoid with a right angle and an area greater than 40 square inches

see the attached figure.

Step-by-step explanation:

We need to draw a trapezoid with a right angle and an area greater than 40 square inches

See the attached figure

The area of the trapezoid = 0.5 ( base_1 + base_2 ) * height

So, as shown ABCD is a trapezoid with a right angle at A

base_1 = 14 inches , base_2 = 10 inches and the height = 5 inches

So, the area = 0.5 ( 14 + 10 ) * 5 = 0.5 * 24 * 5 = 60 square inches > 40

Note: we can use any numbers other than 5 , 10 , 14

for example: 5 , 8 , 12 in this case the area = 0.5 * (8+12) * 5 = 50 > 40

The point is choose any numbers which gives area > 40

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

there is nothing below so i cannot

step-by-step explanation: