Trapezoid ABCD is graphed in a coordinate plane. What is the area of the trapezoid?
B
2
✓x
10 square units
O 12 square units
O 20 square units
O 24 square units
--5
32-1
2
3
4
5
x
С
А
2
D

  (b)  12 square units

Step-by-step explanation:

There are a couple of ways to go at this. One is to find the lengths of the base and height, then use the usual formula. Another is to make use of Pick's theorem, which finds the area by counting grid points.

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We notice that the parallel bases both have a slope of 1 (1 unit up for 1 unit to the right). That means the distance between grid point crossings is √2. The longer base can be seen to have a length of 4√2, while the shorter base has a length of 2√2. The perpendicular distance between the bases is also 2√2.

Using the formula for the area of a trapezoid, we find the area to be ...

  A = 1/2(b1 +b2)h

  A = (1/2(4√2 +2√2)(2√2) = 1/2(6√2)(2√2) = 1/2(12)(2) = 12 . . . square units

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Pick's theorem tells you the area is ...

  A = i +b/2 -1

where i is the number of interior grid points, b is the number of grid points on the boundary line.

Counting grid points, we find ...

  i = 9

  b = 8

Then the area is ...

  A = 9 +8/2 -1 = 9 + 4 -1 = 12 . . . square units

all but the second one

step-by-step explanation:

cuz that one goes over 180

mark brainliest

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: