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Y_Kistochka [10]
2 years ago
6

Solve the following inequality using both the graphical and algebraic approach: 0.5 x + 3 greater-than-or-equal-to 2 x minus 1.5

Graph A On a coordinate plane, a line goes through (0, 2) and (4, 4). Another line goes through (0, 0) and (2, 6). The lines intersect at (1, 1.5). Graph B On a coordinate plane, a line goes through (0, 3) and (4, 4). Another line goes through (1, 0) and (4, 6). The lines intersect at (3, 4). a. x greater-than-or-equal-to 3 Graph A b. x less-than-or-equal-to 3 Graph A c. x greater-than-or-equal-to 3 Graph B d. x less-than-or-equal-to 3 Graph B
Mathematics
2 answers:
Nana76 [90]2 years ago
8 0

Answer:

D

Step-by-step explanation:just took the test

Liono4ka [1.6K]2 years ago
5 0

Answer:

d. x les-than-or-equal-to 3 Graph B

Graph B expressed as follows; The graph on the coordinate plane, a line goes through (0, 3) and (4, 5). Another line goes through (1, 0.5), and (4, 6.5). The lines intersect at (3, 4.5)

Step-by-step explanation:

The given inequality is expressed as follows;

0.5·x + 3 ≥ 2·x - 1.5

Let y₁ = 0.5·x + 3, and y₂ = 2·x - 1.5, we get;

For x = -1, 0, 1, 2, 3, 4, 5, 6

y₁ = 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6

y₂ = -3.5, -1.5, 0.5, 2.5, 4.5, 6.5, 8.5, 10.5

From the given data, the lines intersect at (3, 4.5)

The graph on the coordinate plane, a line goes through (0, 3) and (4, 5). Another line goes through (1, 0.5), and (4, 6.5). The lines intersect at (3, 4.5)

Please find attached the required inequality created with MS Excel

Therefore, we have;

3 + 1.5 ≥ 2·x - 0.5·x

4.5 ≥ 1.5·x

∴ 3 ≥ x

x ≤ 3

Therefore, with (typographical) correction, the best option is x les-than-or-equal-to 3 Graph B

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Step-by-step explanation:

Let us revise the sine rule

In ΔABC:

  • \frac{AB}{sin(C)}=\frac{BC}{sin(A)}=\frac{AC}{sin(B)}
  • AB is opposite to ∠C
  • BC is opposite to ∠A
  • AC is opposite to ∠B

Let us use this rule to solve the problem

In ΔABC:

∵ m∠A = 45°

∵ m∠C = 30°

- The sum of measures of the interior angles of a triangle is 180°

∵ m∠A + m∠B + m∠C = 180

∴ 45 + m∠B + 30 = 180

- Add the like terms

∴ m∠B + 75 = 180

- Subtract 75 from both sides

∴ m∠B = 105°

∵ \frac{AB}{sin(C)}=\frac{BC}{sin(A)}

∵ AB = 5\sqrt{2}

- Substitute AB and the 3 angles in the rule above

∴ \frac{5\sqrt{2}}{sin(30)}=\frac{BC}{sin(45)}

- By using cross multiplication

∴ (BC) × sin(30) = 5\sqrt{2} × sin(45)

∵ sin(30) = 0.5 and sin(45) = \frac{1}{\sqrt{2}}

∴ 0.5 (BC) = 5

- Divide both sides by 0.5

∴ BC = 10 units

∵ \frac{AB}{sin(C)}=\frac{AC}{sin(B)}

- Substitute AB and the 3 angles in the rule above

∴ \frac{5\sqrt{2}}{sin(30)}=\frac{AC}{sin(105)}

- By using cross multiplication

∴ (AC) × sin(30) = 5\sqrt{2} × sin(105)

∵ sin(105) = \frac{\sqrt{6}+\sqrt{2}}{4}

∴ 0.5 (AC) = \frac{5+5\sqrt{3}}{2}

- Divide both sides by 0.5

∴ AC = 5+5\sqrt{3} units

BC is 10 units and AC is 5+5\sqrt{3} units

Learn more:

You can learn more about the sine rule in brainly.com/question/12985572

#LearnwithBrainly

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