Which are correct representations of the inequality 6x ≥ 3 + 4(2x – 1)? check all that apply. 1 ≥ 2x 6x ≥ 3 + 8x – 4

6x ≥ 3 + 4(2x - 1)

⇔ 6x ≥ 3 + 8x - 4        => remove the parentheses

⇔ 6x - 8x ≥ 3 - 4

⇔ -2x ≥ -1

⇔ 2x ≤ 1 (or 1 ≥ 2x)

⇔ x ≤ 1/2

⇔ x ≤ 0.5

The answer for the question should be:

1 ≥ 2x

6x ≥ 3 + 8x – 4

and the first graph.

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation:

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180