Mathematics, 11.10.2019 03:30 ambercombs

Find the area of the following shape. you must show all work to receive credit. figure abcdef is shown. a is located at negative 2, 6. b is located at negative 5, 2. c is located at negative 2, negative 3. d is located at 4, negative 3. e is located at 5, 3. f is located at 1, 3.

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In a analytical geometry, there is a technique where you can find the area of any polygon as long as you know the coordinates of all its corner points. First, you need to plot the points in a cartesian plane as what is shown at the top of the attached picture. Then, you make a matrix wherein the x coordinates are placed in the first row, and the y coordinates in the second row (shown as the black box in the picture). Make sure that you arrange the points in order (whether clockwise or counterclockwise).

Once you create the matrix, you find its determinant. This is done by multiplying the numbers diagonally. The equation will be: arrow down - arrow up. The solution is as follows:

Determinant = [(-2×2) + (-5×3) + (-2×3) + (4×3) + (5×3)] - [(6×-5) + (2×-2) + (3×4) + (3×5) + (3×1)] = 6

Then, we use the equation for area of any polygon: A = 1/2*determinant
A = 1/2*6 = 3 square units

Thus, the shape has an area of 3 square units.

Answer from: slycooper99

I just took the test and the following is the correct answer. IDK what the other guy was thinking lol.

In order to find the area of this irregular shape, we have to divide it into triangles and rectangles, find the areas of each shape, and then add them together.

First, put a point (G) at 4,3 then remove the new triangle DEG. GD = 6 and EG = 1. 6 times 1 = 1, then 6 divided by 2 equals 3... the area of DEG is 3 square units.

Next, put a point (H) at -2,3 then remove the square CDGH. Side CD = 6 and side DG = 6. 6 times 6 = 36... the area of CDGH is 36 square units.

Next, remove triangle AFH. Side HA = 3 and side FH = 3. 3 times 3 = 9 and 9 divided by 2 = 4.5... the area of AFH is 4.5 square units.

Next, put a point (I) at -2,2 the remove the new triangle ABI. Side IA = 4 and side BI = 3. 4 times 3 = 12 and 12 divided by 2 = 6... the area of ABI is 6 square units.

Next, remove triangle BCI, Side IB = 3 and side CI = 5. 3 times 5 = 15 and 15 divided by 2 = 7.5... the area of BCI is 7.5

Lastly, we need to add the areas of each individual shape to find the area of the larger irregular shape shown above. The area of DEG is 3 square units, the area of CDGH is 36 square units, the area of AFH is 4.5 square units, the area of ABI is 6 square units, and the area of BCI is 7.5.

3+36+4.5+6+7.5= 57

The area of ABCDEF = 57 square units.

Answer from: Quest

b. the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

Answer from: Quest

step-by-step explanation:

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