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3 years ago

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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What is the value of x? <br> sin41°=cos x<br> Enter your answer in the box. <br> x = °

creativ13 [48]

Answer:

x = 49°

Step-by-step explanation:

The sine of an angle is equal to the cosine of its complement:

x = 90° -41° = 49°

5 0

3 years ago

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

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

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:

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

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

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

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

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

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

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

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

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

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

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

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

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

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

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

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

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

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 $r_{4} = 11\ and\ h_{4}=9$

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

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

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

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

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2 years ago

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vitfil [10]

Answer:

The answer to 9 is 358

Step-by-step explanation:

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3 years ago

Read 2 more answers

high contrast toggle How many window coverings are necessary to span 50 windows if each window covering is 15 windows long?

raketka [301]

We know that
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</span>So,
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the answer is
4 window coverings

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2 years ago

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Answer:

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Step-by-step explanation:

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