Mrs. Mireles needs to hire landscapers to maintain her yard. Company L charges $250 per month plus $30 per hour. Company S charges $200 per month plus $35 per hour. Which inequality can be used to find x, the minimum number of hours each company would have to work so that the total charges from Company L is less than the total charges from Company S?

Step-by-step explanation:

Step one:

given information

Company L charges

monthly charge= $250

hourly charge= $30

let the total charge be y

and the number of hours be x

y=30x+2501

Company S charges

monthly charge= $200

hourly charge= $35

let the total charge be y

and the number of hours be x

y=35x+2002

Step two:

equating 1 and 2

30x+250=35x+200

collect like terms

30x-35x=200-250

-5x=-50

x=50/5

x=10

10 hours

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2