Square of (1\4A+1\4B)^2

4a = 2⋅(2a)⋅1 4 a = 2 ⋅ ( 2 a) ⋅ 1

Step-by-step explanation:

Factor 4a^2-4a+1. 4a2 − 4a + 1 4 a 2 - 4 a + 1. Rewrite 4a2 4 a 2 as (2a)2 ( 2 a) 2. (2a)2 − 4a+1 ( 2 a) 2 - 4 a + 1. Rewrite 1 1 as 12 1 2. (2a)2 − 4a+12 ( 2 a) 2 - 4 a + 1 2. Check that the middle term is two times the product of the numbers being squared in the first term and third term. 4a = 2⋅(2a)⋅1 4 a = 2 ⋅ ( 2 a) ⋅ 1.

step-by-step explanation: h: (- 3, 6)

m: (1, - 2)

o: (7, 6)

r: (- 9, - 2)

midpoint of hm: (- 1, 2)

midpoint of or: (- 1, 2)

therefore, b is the midpoint of both diagonals bisect each other !

to prove the diagonals are perpendicular, show that the product of the slopes = - 1

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph