A company dyes two sizes of rugs. A small rug requires 4 hours for dyeing and a medium-size rug requires 6 hours for dyeing. The dyers have to make at least 20 rugs, and they must do it in less than 120 hours. Let x equal small rugs and y equal medium rugs. Which of the following inequalities can be paired with x + y ≥ 20 to create a system that represents this situation? (5 points)

Option (b) is correct.

4x + 6y < 120  inequalities that can be paired with x + y ≥ 20 to create a system that represents the given situation.

Step-by-step explanation:

Given:  A small rug requires 4 hours for dyeing and a medium-size rug requires 6 hours for dyeing. The dyers have to make at least 20 rugs, and they must do it in less than 120 hours.

We have write an inequalities that can be paired with x + y ≥ 20 to create a system that represents the given situation and choose the correct option from given options.

Let x represents the  small rugs and y represents the medium rugs.

Also, x + y ≥ 20  represents the dyers have to make at least 20 rugs.

And it is given that a small rug requires 4 hours for dyeing and a medium-size rug requires 6 hours for dyeing and that they must do it in less than 120 hours.

So, time taken to dye x small rugs = 4x

also, time taken to dye y medium rugs = 6y

And they have maximum of 120 hours.

So this inequality is represented by 4x + 6y < 120

Thus, 4x + 6y < 120  inequalities that can be paired with x + y ≥ 20 to create a system that represents the given situation.

SI=(P*R*T)/100

P=2000

R=1.5

T=6

SI=(2000*1.5*6)/100

=(2000*9)/100

=180

Neil will earn interest of 180

Cost of 7 gallons=$24.50

Cost of 1 gallon=24.50/7=3.5

Cost of 15 gallons=15*3.5=52.5

Cost of 15 gallons will be $52.5

The total nom of  code that can be used is equal to 5+3 = 8

The solution is in the following image

y=2x+15

where y=Value of coin

x=Age in years

Value of coin after 19 years=2*19+15

=$53

Therefore, Value after 19 years=$53

The answer is in the image 

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

Approximately it is aqual to 1300kg