given that (y+2) (y+3) and (2y^2+1) are consecutive terms of an arithmetic progression, find the possible value of y

-3/2 and 1

Step-by-step explanation:

Given the  arithmetic sequence  (y+2) (y+3) and (2y²+1), the common difference is gotten by taking the difference in their terms. For example if we have 3 terms T1, T2, T3... the common difference d = T2-T1 = T3-T2

From the  sequence given;

T1 = y+2, T2 = y+3 and T3 = 2y²+1

d = y+3-(y+2) = 2y²+1- (y+3)

open the parenthesis

y+3-y-2 = 2y²+1- y-3

1 = 2y²+1- y-3

1 = 2y²- y-2

2y²- y-2-1 = 0

2y²- y-3 =0

Factorize the resulting expression

2y²- y-3 =0

2y²- 2y+3y-3 =0

2y(y-1)+3(y-1) = 0

(2y+3)(y-1) = 0

2y+3 = 0 and y-1 = 0

2y = -3 and y =1

y = -3/2 and 1

Hence the possible values of y are -3/2 and 1

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,