Complete the coordinate proof of the theorem.

given: abcd is a parallelogram.

prove: the diagonals of abcd bisect each other.

parallelogram in quadrant 1 of the coordinate plane with vertices at a begin ordered pair 0 comma 0 end ordered pair, b begin ordered pair a comma 0 end ordered pair, c begin ordered pair a plus c comma b end ordered pair and d.

enter your answers in the boxes.
the coordinates of parallelogram abcd are a(0, 0) , b(a, 0) , c(a+c, b) , and d(
,
).

the coordinates of the midpoint of ac are ( a+c2 ,
).

the coordinates of the midpoint of bd are (
, b2 ).

the midpoints of the diagonals have the same coordinates.

therefore, the diagonals of abcd bisect each other.

the answer is b i just did that

step-by-step explanation:

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2

the first one, all those points on the line just make a strait line.

20%

step-by-step explanation: