One-half the sum of two consecutive odd integers is −12. Find the smallest integer.

Step-by-step explanation:

Let the larger integer = 2x + 1

Let the second integer = 2x - 1

1/2 (2x + 1 + 2x - 1) = - 12               Combine the left

1/2(4x) = - 12                                  Combine 1/2 (4x)

2x = - 12                                         Divide by 2

2x/2 = - 12/2

x = - 6

2x + 1 = 2*-6 + 1 = -11

2x - 1 = 2*-6 - 1  = - 13

Check

-11 + - 13  = - 24

1/2(-24) = -12                               which checks.

(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

the correct answers i think are b. and d.

give me brainliest if it is correct!