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nydimaria [60]
3 years ago
10

Which graph best represents the feasibility region for the system shown below?

Mathematics
2 answers:
krek1111 [17]3 years ago
8 0
<h3>Answer: Correct option is 3rd option. </h3>

Step-by-step explanation: Given inequalities

x≥ 0

y≤ 8

y ≥ x and

y ≥ -\frac{1}{2}x +6.

Let us graph the lines on the calculators first.

x≥ 0 : First inequality line would be a solid line and we would shade right side of y-axis because we have greater than or equal to symbol.

y≤ 8 : Second inequality line would be a solid line and we would shade down the line because we have less than or equal to symbol.

y ≥ x : Third inequality line would be a solid line and we would shade up of the line because we have greater than or equal to symbol.

y ≥ -\frac{1}{2}x +6 : Fourth inequality line would be a solid line and we would shade up of the line because we have greater than or equal to symbol.

From the graph we can see that middle portion is the common feasible region.

<h3>Therefore, correct option is 3rd option. </h3>

Vadim26 [7]3 years ago
5 0

The second graph has the right picture


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Arrange the cones in order from lease volume to greatest volume
bearhunter [10]

Answer:

Volume of the cone in ascending order.

V_{2}=270\pi\ units^{3}

cone with DIAMETER of 18 & height of 10

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cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let V_{2}. V_{3}. and\  V_{4}. be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

d_{1} = 20\ and\ h_{1}=12

d_{2} = 18\ and\ h_{2}=10

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Solution:

The volume of the cone is given below.

V=\pi r^{2} \frac{h}{3}----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for d_{1} = 20\ and\ h_{1}=12

r_{1} = \frac{d_{1}}{2}

r_{1} = \frac{20}{2}=10\ units

V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}

V_{1}=\pi (10)^{2} \frac{12}{3}

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Similarly, for volume of the cone for d_{2} = 18\ and\ h_{2}=10

r_{2} = \frac{d_{2}}{2}

r_{2} = \frac{18}{2}=9\ units

V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}

V_{2}=\pi (9)^{2} \frac{10}{3}

V_{2}=\pi\times 81\times \frac{10}{3}

V_{2}=\pi\times 27\times 10

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Similarly, for volume of the cone for r_{3} = 10\ and\ h_{3}=9

V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}

V_{3}=\pi (10)^{2} \frac{9}{3}

V_{3}=\pi\times 100\times 3

V_{3}=\pi\times 300

V_{3}=300\pi\ units^{3}

Similarly, for volume of the cone for r_{4} = 11\ and\ h_{4}=9

V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}

V_{4}=\pi (11)^{2} \frac{9}{3}

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V_{2}=270\pi\ units^{3}

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