Find the percent of change to the nearest tenth of a percent for centerville
Lakeside in 2000: 51,342 2010: 45,863
Centerville in 2000: 37,065 2010: 26,102
Riverview in 2000: 41,726 2010: 63,017

The percent change for centerville is 30% (rounded off to nearest tenth of a percent)

Step-by-step explanation:

Given details-

Centerville during the year 2000-37,065

Centerville during the year 2010-26,102

Per cent change to the nearest tenth of a percent

total change during the time gap of 10 years= Centerville during 2000- Centerville during 2010

⇒ total change during the time gap of 10 years=37,065-26,102

⇒ total change during the time gap of 10 years=10,963

⇒ per cent change= (total change/Centerville during 2000)*100

⇒ per cent change= (10,963/37,065)*100

⇒ 29.57776%

⇒ 30% (The answer has been rounded)

any options?

step-by-step explanation:

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1