Find four consecutive odd integers such that 3 times the largest decreased by 19 is twice the second integer

n, n + 2, n + 4, n + 6 - four consecutive odd integers

3 times the largest decreased by 19 is twice the second integer

3(n + 6) - 19 = 2(n + 2)     use distributive property

(3)(n) + (3)(6) - 19 = (2)(n) + (2)(2)

3n + 18 - 19 = 2n + 4

3n - 1 = 2n + 4      add 1 to both sides

3n = 2n + 5      subtract 2n from both sides

n = 5

n + 2 = 5 + 2 = 7

n + 4 = 5 + 4 = 9

n + 6 = 5 + 6 = 11

answer: 60

step-by-step explanation:

1/2a + (2/3×30)   = 50

1/2a + 20 = 50

1/2a = 30

a = 60

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic