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Mama L [17]
3 years ago
14

A circular rug has a radius of 3 ft. What is the circumference of the rug? Use 3.14 for mc014-1.jpg.

Mathematics
2 answers:
solniwko [45]3 years ago
7 0

Answer: D: 18.84

Step-by-step explanation:

Pavel [41]3 years ago
3 0
Hello!

We must use a certain formula to find the circumference of anything circular.

The formula is:

C = 2 × pi × r

We know that pi = 3.14 approximately and the radius is 3 feet.

Substitute:

C = 2 × 3.14 × 3

C = 6.28 × 3

C = 18.84 ft

ANSWER:

The circumference of the circular rug is 18.84 ft. (Last option)
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Solve the linear programming problem. Minimize and maximize Upper P equals negative 20 x plus 30 y Subject to 2 x plus 3 y great
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Answer:

Maximum = 540 at (6,14)

Minimum = 300 at (0,10) or (12,2).

Step-by-step explanation:

The given linear programming problem is

Minimize and maximize: P = 20x + 30y

Subject to constraint,

2x+3y\ge 30            .... (1)

2x+y\le 26            .... (2)

-2x+3y\le 30            .... (3)

x,y\geq 0

The related equation of given inequalities are

2x+3y=30

2x+y=26

-2x+3y=30

Table of values are:

For inequality (1).

x      y

0     10

15     0

For inequality (2).

x      y

0     26

13     0

For inequality (3).

x      y

0     10

15     0

Pot these ordered pairs on a coordinate plane and connect them draw the corresponding related line.

Check each inequality by (0,0).

2(0)+3(0)\ge 30\Rightarrow 0\ge 30    False

2(0)+(0)\le 26\Rightarrow 0\le 26     True

-2(0)+3(0)\le 30\Rightarrow 0\le 30    True

It means (0,0) is included in the shaded region of inequality (2) and (3), and (0,0) is not included in the shaded region of inequality (1).

From the below graph it is clear that the vertices of feasible region are (0,10), (6,14) and (12,2).

Calculate the values of objective function on vertices of feasible region.

Point           P = 20x + 30y

(0,10)           P = 20(0) + 30(10) = 300

(6,14)           P = 20(6) + 30(14) = 540

(12,2)           P = 20(12) + 30(2) = 300

It means objective function is maximum at (6,14) and minimum at (0,10) or (12,2).

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