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garri49 [273]
3 years ago
8

4y+2y+x+2x please help

Mathematics
1 answer:
DochEvi [55]3 years ago
8 0

Answer:

Solution

Solution

3

+

6

Step-by-step explanation:

This is the best of what I got.

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The coordinate plane below represents a city.
alex41 [277]

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Part A) The system of inequalities is

x\geq2  and  y\geq2

Part B) In the procedure

Part C) The schools that Natalie is allowed to attend are A,B and D

Step-by-step explanation:

Part A: Using the graph above, create a system of inequalities that only contain points C and F in the overlapping shaded regions

we have

Points C(2,2), F(3,4)

The system of inequalities could be

x\geq2 -----> inequality A

The solution of the inequality A is the shaded area at the right of the solid line x=2

y\geq2 -----> inequality B

The solution of the inequality B is the shaded area above of the solid line y=2

see the attached figure N 1

Part B: Explain how to verify that the points C and F are solutions to the system of inequalities created in Part A

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities

Verify point C

C(2,2)    

<em>Inequality A</em>

x\geq2 -----> 2\geq2 ----> is true

<em>Inequality B</em>

y\geq2 ------> 2\geq2 ----> is true

therefore

Point C is a solution of the system of inequalities

Verify point D

F(3,4)    

Inequality A

x\geq2 -----> 3\geq2 ----> is true

Inequality B

y\geq2 ------> 4\geq2 ----> is true

therefore

Point D is a solution of the system of inequalities

Part C: Natalie can only attend a school in her designated zone. Natalie's zone is defined by y < −2x + 2. Explain how you can identify the schools that Natalie is allowed to attend.

we have

y < -2x+2

The solution of the inequality is the shaded area below the dotted line y=-2x+2

The y-intercept of the dotted line is the point (0,2)

The x-intercept of the dotted line is the point (1,0)

To graph the inequality, plot the intercepts and shade the area below the dotted line

see the attached figure N 2

therefore

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