A cable company claims that the average household pays $78 a month for a basic cable plan, but it could differ by as much as $20. Write an absolute value inequality to determine the range of basic cable plan costs with this cable company. A.) |x − 78| ≥ 20
B.) |x − 20| ≥ 78
C) |x − 20| ≤ 78
B.) |x − 78| ≤ 20

D) |x − 78| ≤ 20

Step-by-step explanation:

Given,

The monthly charges for a basic cable plan = $ 78,

Also,  it could differ by as much as $20,

So, the maximum charges = $(78 + 20) ,

And, the minimum charges = $(78 - 20),

Let x represents the monthly charges ( in dollars ),

78 - 20 ≤ x ≤ 78 + 20

⇒ 78 - 20 ≤ x and x ≤ 78 + 20

⇒ -20 ≤ x -78 and x-78 ≤ 20

⇒ 20 ≥ -(x-78) and x-78 ≤ 20   ( ∵ a > b ⇒ -a < -b )

⇒ |x-78| ≤ 20

Which is the required absolute value inequality to determine the range of basic cable plan costs,

Option 'D' is correct.

step-by-step explanation:

answer:

total cost for tiles and paints is $924.  

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

given: 1m squared floor tiles costs $3. so,

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

therefore, the combined total cost of tiles and paint is $924.