Anna had a solid wooden dowel with a length of 12 inches and a diameter of 1 inch, as shown below. She cut a two-inch portion out of the dowel as indicated. What was the volume, to the nearest tenth of a cubic inch, of the portion of the dowel that Anna cut out?
Anna had a solid wooden dowel with a length of - 1

Answers

Answer 1
Answer:

Answer:

Volume of the portion = 1.6 cubic inches

Step-by-step explanation:

Since, wooden dowel is in the shape of a cylinder,

Formula to be used to find the volume of the part Anna has cut from the dowel will be,

Volume = πr²h

Where r = radius of the cylinder or part of the wooden dowel

h = length of the portion

By substituting the values in the formula,

Volume = π(0.5)²(2)

             = 0.5π

             = 1.57 in³ ≈ 1.6 cubic inch

Therefore, volume of the portion will be 1.6 in³.


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Joey is buying plants for his garden. He wants to have at least twice as many flowering plants as nonflowering plants and a minimum of 36 plants in his garden. Flowering plants sell for $8, and nonflowering plants sell for $5. Joey wants to purchase a combination of plants that minimizes cost. Let x represent the number of flowering plants and y represent the number of nonflowering plants. What are the vertices of the feasible region for this problem?
(0, 0), (0, 36), (24, 12)
(0, 36), (24, 12)
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(24, 12), (36, 0)

Answers

(24, 12) and (36, 0).  The least amount of flowering plants occurs when x=2y, and the largest amount occurs when y=0.  These two points satisfy both conditions and both sum to 36.

Answer: (24, 12), (36, 0)


Step-by-step explanation:

Let x be the number of flowering plants and y be the number of non- flowering plants.

According to the question, we need to minimize the cost of plants.

Minimize:8x+5y

Subject to the constraints,

2y\leq\ x\nx+y\geq36

To find the feasible region find the points of the equation to plot it on graph.

For the first equation 2y=x , at x=0 y=0 and at x=4, y=2

For the second equation x+y=36 , at x=0 y=36 and at x=36, y=0

Thus points for eq (1) are (0,0) and (4,2) and points for equation (2) are (0,36) and (36,0).

Now, plot it on graph, we get the shaded feasible region as shown in the graph.

and we can see the  vertices of the feasible region = (24, 12), (36, 0)

Please help !!!!!!!!!!!!-

Answers

3.(4p−2)(p−4)
=(4p+−2)(p+−4)

=(4p)(p)+(4p)(−4)+(−2)(p)+(−2)(−4)

=4p2−16p−2p+8

=4p2−18p+8

Simplified form - 
4p2−18p+8

4.
(2x2+4x−3)(3x+1)

=(2x2+4x+−3)(3x+1)

=(2x2)(3x)+(2x2)(1)+(4x)(3x)+(4x)(1)+(−3)(3x)+(−3)(1)
=6x3+2x2+12x2+4x−9x−3

=6x3+14x2−5x3

Simplified form - 
6x3+14x2−5x3

Simplify:
8.7+5.8-3.1*1*2.3

Answers

8.7+5.8-3.1*1*2.3 = (8.7 + 5.8) - (3.1 * 2.3)

(14.5) - (7.13)

7.37
I'm assuming that 1*1 means 1 times 1 and not 1^1 so

8.7+5.8-3.1*1*2.3
pemdas so mutiplication first
-3.1*1*2.3=-3.1*2.3=-7.13 since opposite signs multiplied make negative
then addition subtraction
8.7+5.8-7.13
14.50-7.13
7.37

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