Write a coordinate proof. Given: Rectangle ABCD has vertices A(0,4), B(6,4), C(6,0), D(0,0). E is the midpoint of line DC. F is the midpoint of line DA. Prove: The area of rectangle DEGF is one fourth the area of rectangle ABCD. Please help I've been trying for a while

The answer is below

Step-by-step explanation:

Distance between two points

The area of rectangle = length * breadth

The area of rectangle ABCD = |AB| * |BC|

The area of rectangle ABCD = |AB| * |BC| = 6 * 4 = 24

E(x, y) is midpoint of line DC. Its coordinate is:

x = (6 + 0) / 2 = 3; y = (0 + 0) / 2 =0

The coordinate of E = (3, 0)

F(a, b) is midpoint of line DA. Its coordinate is:

a = (0 + 0) / 2 = 0; b = (4 + 0) / 2 = 2

The coordinate of E = (0, 2)

The area of rectangle DEGF = |DE| * |DF|

The area of rectangle ABCD = |DE| * |DF| = 3 * 2 = 6

Therefore, area of rectangle DEGF is one fourth the area of rectangle ABCD

what the question so we can you out

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.